Chapter 3: Ionic Equilibria
In the early years of the twentieth century, a Danish chemist named Søren Sørensen was working at the Carlsberg Laboratory in Copenhagen. The laboratory had been set up to put brewing on a scientific footing, and Sørensen’s task was an unglamorous one: find out why some batches of beer fermented well and others did not.
He soon traced the problem to acidity. The enzymes and proteins doing the work of fermentation were extraordinarily sensitive to how much hydrogen ion the liquid contained. But the numbers he had to work with were awkward. One solution might contain \(0.00000001\) mol of hydrogen ion per litre, another a thousand times more. Writing, comparing and plotting such numbers was clumsy and error-prone.
So in 1909 Sørensen did something simple and powerful. Instead of quoting the concentration itself, he quoted the negative logarithm of that concentration — and called it the pH of the solution. A messy string of zeros became a single tidy number between 0 and 14, as shown in Fig. 3.0.
Diagram
Fig. 3.0 – A brewery vat and a beaker side by side, with a 0–14 pH strip running below them; mark acidic (0–7), neutral (7) and basic (7–14) regions, and label lemon juice, pure water, blood and soap solution at their approximate positions.
That one idea now sits on every soil-testing kit, every swimming pool, every blood report and every shampoo bottle. Behind it lies the chemistry of this chapter: what happens when ions and unionized molecules share the same solution and settle into equilibrium.
Keep these five questions in mind as you read. The chapter will give you everything you need to reason each of them out.
3.1 Introduction
Some substances, when dissolved in water, do not stay as whole molecules. They break up — partly or completely — into charged particles called ions. An ion is simply an atom or group of atoms carrying a positive or negative electrical charge, produced when a neutral particle loses or gains electrons.
In many solutions, this breaking up is not total. A certain fraction of the substance exists as ions, and the rest remains as intact, uncharged molecules. Both forms are present in the same solution at the same time, and they are constantly interconverting. When the rate at which molecules break into ions becomes exactly equal to the rate at which those ions recombine into molecules, the composition of the solution stops changing. Such a balance is called ionic equilibrium.
Building on What You Already Know
You have already met chemical equilibrium in Standard XI. Nothing about it changes here. The same ideas — that equilibrium is dynamic rather than dead, that a reaction quotient settles at a fixed value called the equilibrium constant at a given temperature, and that the position of equilibrium shifts to oppose any stress applied to it — apply word for word to solutions containing ions. In this chapter we take those principles and use them to work out equilibrium constants and the actual concentrations of ions and unionized species present in a solution.
The Four Equilibria Ahead
The chapter is built around four particular ionic equilibria, and it is worth knowing at the outset what they are, because everything that follows fits into one of these four boxes. The route through them is laid out in FC 3.1.
- The equilibrium between \(\mathrm{H^+}\) ions, \(\mathrm{OH^-}\) ions and unionized water molecules. Water itself ionizes very slightly, and this small effect turns out to control the acidity of every aqueous solution.
- The ionization of weak acids and weak bases, which dissociate only partially and therefore always leave undissociated molecules behind.
- The reactions between the ions of a salt and the ions of water — the process that decides whether a salt solution turns out acidic, basic or neutral.
- The equilibrium between a solid salt and its ions in water, which governs how much of a sparingly soluble salt actually dissolves, and when a precipitate appears.
Flowchart
FC 3.1 – Roadmap of ionic equilibria in this chapter: electrolytes in water → strong electrolytes (complete ionization) or weak electrolytes (partial ionization, equilibrium set up) → four equilibria studied: (1) H⁺/OH⁻ with unionized water → pH scale; (2) weak acid/weak base ionization → Ka, Kb, degree of dissociation; (3) salt ions with water ions → hydrolysis, buffers; (4) solid salt with its dissolved ions → solubility product, precipitation, common ion effect.
Notice that all four are equilibria of the same family. Each one involves ions on one side and something unionized — a molecule, or an undissolved solid — on the other. Once you are comfortable writing an equilibrium constant expression for one of them, you can write it for all of them.
3.2 Types of Electrolyte
Dissolve a spoon of common salt in a beaker of water, dip in two electrodes, and the bulb in the circuit glows. Dissolve the same amount of sugar instead, and the bulb stays dark. The difference lies entirely in what the dissolved particles are.
Substances that give rise to ions when dissolved in water are called electrolytes. Because their solutions contain freely moving charged particles, they conduct electricity. Non-electrolytes are substances that do not ionize at all; they simply disperse through the water as whole, uncharged molecules, and their solutions cannot carry a current. Sugar and urea behave this way.
Electrolytes themselves are not all alike. Some hand over nearly all of their particles as ions, while others part with only a small fraction. On this basis, electrolytes are classified into strong and weak electrolytes. The classification is made in dilute aqueous solutions, and it rests on one property only — the extent of ionisation. The difference in behaviour is easy to picture, as drawn in Fig. 3.2.
Diagram
Fig. 3.2 – Three beakers in a conductivity circuit with a bulb: (a) strong electrolyte (NaCl) — solution shown crowded with Na⁺ and Cl⁻ ions, bulb glowing brightly; (b) weak electrolyte (CH₃COOH) — mostly whole molecules with a few CH₃COO⁻ and H⁺ ions, bulb glowing dimly; (c) non-electrolyte (sugar) — only whole molecules, bulb dark.
3.2.1 Strong Electrolyte
Electrolytes that ionize completely, or almost completely, are called strong electrolytes. Once dissolved, almost nothing of the original substance survives as an intact molecule — the solution is essentially a crowd of ions.
Strong acids, strong bases and salts belong to this class. Because the ionization goes practically to completion, we write it with a single arrow pointing forward:
\[\mathrm{NaCl}\text{(s)} \xrightarrow{\ \mathrm{H_2O}\ } \mathrm{Na^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
3.2.2 Weak Electrolyte
Electrolytes that dissociate to a smaller extent in aqueous solution are called weak electrolytes. Weak acids and weak bases belong to this class.
Weak electrolytes dissociate only partially in dilute aqueous solutions. This partial dissociation is exactly what makes them interesting to us: since some molecules break up while others do not, both ions and nonionized molecules are present together, and an equilibrium can be established between them.
To show that the change runs in both directions and settles at a balance, the ionization reaction of a weak electrolyte is written with a double arrow (\(\rightleftharpoons\)) between the ions and the nonionized molecule:
\[\mathrm{CH_3COOH}\text{(aq)} \rightleftharpoons \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H^+}\text{(aq)}\]
The single arrow and the double arrow are therefore not decoration. A single arrow says the reaction is over; a double arrow says an equilibrium has been reached and undissociated molecules are still present.
3.2.3 Degree of Dissociation (\(\alpha\))
Saying that a weak electrolyte dissociates “only slightly” is not good enough for chemistry. We need a number that tells us exactly what fraction of it has broken into ions. That number is the degree of dissociation.
It is denoted by the symbol \(\alpha\) and given by
\[\alpha = \frac{\text{number of moles dissociated}}{\text{total number of moles}} \qquad \text{......(3.1)}\]
Because it is a fraction of moles divided by moles, \(\alpha\) has no unit at all. Its value can only lie between 0 and 1: a value of 0 would mean nothing has dissociated, and a value of 1 would mean every molecule has. For a weak electrolyte, \(\alpha\) is a small number, well below 1.
Fractions such as 0.0005 are awkward to compare at a glance, so the same information is often quoted as a percentage:
\[\text{Percent dissociation} = \alpha \times 100 \qquad \text{.......(3.2)}\]
From Degree of Dissociation to Ion Concentration
The real value of \(\alpha\) is that it lets us calculate how much of each ion is actually floating about in the solution. If ‘\(c\)’ is the molar concentration of an electrolyte, the equilibrium concentration of cation or anion is \((\alpha \times c)\) mol dm\(^{-3}\).
The reasoning behind this is worth spelling out. Out of every mole of electrolyte taken, only the fraction \(\alpha\) has dissociated. If the electrolyte gives one cation and one anion per molecule, then each mole that dissociates yields \(\alpha\) mole of cation and \(\alpha\) mole of anion. Multiply by the concentration \(c\), and you have the concentration of each ion in the solution.
3.3 Acids and Bases
Acids and bases are familiar chemical compounds — far more familiar than their names suggest. Acetic acid is found in vinegar, citric acid in lemons, magnesium hydroxide in antacids, ammonia in household cleaning products. The tartaric acid is present in tamarind paste. These are some acids and bases we come across in everyday life.
Recognising them is easy. Defining them turned out to be surprisingly hard. Over about four decades, chemists produced three successive definitions of what an acid and a base actually are — each one wider than the last, and each one still in use today, because each is convenient for a different kind of problem.
3.3.1 Arrhenius Theory of Acids and Bases
According to this theory acids and bases are defined as follows.
An acid is a substance which contains hydrogen and gives rise to \(\mathrm{H^+}\) ions in aqueous solution. The key idea is that the acidic character does not belong to the molecule sitting in a bottle; it appears when water pulls the molecule apart. For example:
\[\mathrm{HCl}\text{(aq)} \xrightarrow{\ \text{water}\ } \mathrm{H^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
\[\mathrm{CH_3COOH}\text{(aq)} \rightleftharpoons \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H^+}\text{(aq)} \qquad \text{(in water)}\]
Notice the two different arrows. Hydrochloric acid is a strong electrolyte and its ionization is written as complete, while acetic acid is a weak electrolyte and reaches an equilibrium — exactly the distinction drawn in the previous section.
Arrhenius described \(\mathrm{H^+}\) ions in water as bare ions. A bare \(\mathrm{H^+}\) ion is nothing but a proton — a hydrogen nucleus stripped of its only electron — and such a naked, intensely concentrated charge cannot survive on its own in water. It immediately attaches itself to a water molecule. So these ions hydrate in aqueous solutions and are thus represented as hydronium ions \(\mathrm{H_3O^+}\). We herewith conveniently represent them as \(\mathrm{H^+}\).
A base, on the other hand, is a substance that contains OH group and produces hydroxide ions (\(\mathrm{OH^-}\)) ions in aqueous solution. For example,
\[\mathrm{NaOH}\text{(aq)} \longrightarrow \mathrm{Na^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
\[\mathrm{NH_4OH}\text{(aq)} \rightleftharpoons \mathrm{NH_4^{+}}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Where the Arrhenius Picture Runs Out
Arrhenius theory accounts for properties of different acids and bases and is applicable only to aqueous solutions. That restriction is the first limitation: a reaction between an acid and a base carried out in benzene or in the gas phase falls outside the theory altogether, because there is no water to supply the ions.
The second limitation is sharper. The theory does not account for the basicity of \(\mathrm{NH_3}\) and \(\mathrm{Na_2CO_3}\) which do not have OH group. Both of these turn red litmus blue and neutralise acids, so they are unmistakably basic — yet neither molecule contains a hydroxyl group to release. A definition that cannot classify ammonia as a base is a definition that needs widening.
3.3.2 Brønsted–Lowry Theory
J. N. Bronsted and T. M. Lowry (1923) proposed a more general theory known as the Bronsted-Lowry proton transfer theory. Working independently, they shifted the focus away from water. Instead of asking what a substance releases into water, they asked what it does with a proton. According to this theory acids and bases are defined as follows.
An acid is a substance that donates a proton (\(\mathrm{H^+}\)) to another substance.
A base is a substance that accepts a proton (\(\mathrm{H^+}\)) from another substance.
This immediately settles the problem that defeated Arrhenius. Ammonia has no OH group, but it has a lone pair with which it can catch a proton — so it is a base. For example:
\[\mathrm{HCl} + \mathrm{NH_3} \rightleftharpoons \mathrm{NH_4^{+}} + \mathrm{Cl^-}\] \[\text{Acid}_1 \qquad \text{Base}_2 \qquad\ \ \text{Acid}_2 \qquad \text{Base}_1\]
In the above reaction HCl and \(\mathrm{NH_4^{+}}\) are proton donors and act as acids. The \(\mathrm{NH_3}\) and \(\mathrm{Cl^-}\) are proton acceptors and act as bases. Further it follows that the products of the Bronsted-Lowry acid-base reactions are acids and bases.
Conjugate Acid–Base Pairs
Read that last statement again, because it contains something important. An acid does not simply vanish after donating its proton — what is left behind is a species that can take a proton back. That species is a base.
The base produced by releasing the proton from an acid is the conjugate base of that acid. Likewise the acid produced when a base accepts a proton is called the conjugate acid of that base. A pair of an acid and a base differing by a proton is said to be a conjugate acid-base pair.
The ionization of hydrochloric acid in water shows both pairs at once, as labelled in Fig. 3.3:
\[\mathrm{HCl}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_3O^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\] \[\text{Acid}_1 \qquad\ \text{Base}_2 \qquad\ \ \ \text{Acid}_2 \qquad\ \ \text{Base}_1\]
Diagram
Fig. 3.3 – The equation HCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + Cl⁻(aq) labelled Acid₁, Base₂, Acid₂, Base₁, with a linking bracket joining H₂O and H₃O⁺ marked “conjugate acid-base pair₂” and an outer bracket joining HCl and Cl⁻ marked “conjugate acid-base pair₁”.
Here HCl loses a proton to become \(\mathrm{Cl^-}\), so \(\mathrm{Cl^-}\) is the conjugate base of HCl. Water gains that proton to become \(\mathrm{H_3O^+}\), so \(\mathrm{H_3O^+}\) is the conjugate acid of water. A useful working trick follows from this: to write the conjugate base of any species, remove one \(\mathrm{H^+}\) from its formula and make the charge one unit more negative; to write the conjugate acid, add one \(\mathrm{H^+}\) and make the charge one unit more positive.
3.3.3 Lewis Theory
A more generalized acid-base concept was put forward by G.N. Lewis in 1923. Lewis noticed that the proton itself is only a middleman. When a base “accepts a proton,” what physically happens is that the base offers a lone pair of electrons and the proton attaches to it. If that electron pair is the essential part of the story, then any species hungry for an electron pair should behave as an acid — proton or no proton. According to this theory acids and bases are defined as follows.
An acid is any species that accepts a share in an electron pair.
A base is any species that donates a share in an electron pair.
The bond formed in such a reaction is a coordinate bond, since both the shared electrons come from one partner — the base. Two examples are drawn in Fig. 3.4. In the first, a proton accepts the lone pair on nitrogen in ammonia to give the ammonium ion. In the second, boron trifluoride, whose boron atom has only six electrons around it and therefore an empty orbital to fill, accepts the same lone pair to give an adduct in which nitrogen has donated the pair to boron.
Diagram
Fig. 3.4 – Two Lewis acid–base reactions with curly arrows: (a) H⁺ + :NH₃ → [H←NH₃]⁺, showing the lone pair on N attacking H⁺ and the resulting coordinate bond drawn as an arrow from N to H; (b) F₃B + :NH₃ → F₃B←NH₃, showing the lone pair on N donated into the empty orbital of B, with the coordinate bond drawn as an arrow from N to B.
For example, \(\mathrm{BF_3}\), \(\mathrm{AlCl_3}\) and metal cations such as \(\mathrm{Cu^{2+}}\) are Lewis acids because each has a vacancy for an electron pair, while \(\mathrm{NH_3}\), \(\mathrm{H_2O}\) and \(\mathrm{OH^-}\) are Lewis bases because each carries a lone pair to offer.
Each theory is broader than the one before it, and that is the pattern worth carrying away: Arrhenius is a special case of Brønsted–Lowry, which is itself a special case of Lewis. Board and entrance examinations regularly test this in assertion–reason form — for instance, asserting that a species is an acid and giving a reason that belongs to a different theory — so always check which definition a question is working with.
Amphoteric Nature of Water
Look back at the two reactions of water written in this section. In one it lost a proton; in the other it gained one. Water has the ability to act as an acid as well as a base. Such behaviour is known as amphoteric nature of water. For example:
\[\mathrm{H_2O}\text{(l)} + \mathrm{NH_3}\text{(aq)} \rightleftharpoons \mathrm{OH^-}\text{(aq)} + \mathrm{NH_4^{+}}\text{(aq)}\] \[\text{Acid}\]
\[\mathrm{H_2O}\text{(l)} + \mathrm{HCl}\text{(aq)} \rightleftharpoons \mathrm{H_3O^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\] \[\text{Base}\]
\(\mathrm{H_2O}\) acts as an acid towards \(\mathrm{NH_3}\) and as a base towards HCl. Therefore \(\mathrm{H_2O}\) is amphoteric.
Which role water plays is decided entirely by its partner. Faced with a stronger proton donor than itself, water accepts the proton and behaves as a base; faced with a stronger proton acceptor, it gives a proton away and behaves as an acid. Keep this dual character in mind — the next few sections depend on it completely, because water is the one substance present in every solution we shall study.
3.4 Ionisation of Acids and Bases
Acids and bases are classified as strong acids and strong bases, weak acids and weak bases on the basis of their extent of dissociation. This is the same yardstick used earlier for electrolytes in general, applied now specifically to the substances that release \(\mathrm{H^+}\) or \(\mathrm{OH^-}\) ions.
Strong acids and bases are almost completely dissociated in water. For example:
\[\mathrm{HCl}\text{(aq)} \longrightarrow \mathrm{H^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
\[\mathrm{NaOH}\text{(aq)} \longrightarrow \mathrm{Na^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Typical strong acids are HCl, \(\mathrm{HNO_3}\), \(\mathrm{H_2SO_4}\), HBr and HI while typical strong bases may include NaOH and KOH. These are worth committing to memory, because in any problem the moment you recognise a strong acid or strong base you may take the ion concentration to be equal to the concentration of the acid or base taken.
Weak acids and weak bases are partially dissociated in water. The solution of a weak acid or a weak base contains undissociated molecules along with a small number of ions at equilibrium. For example:
\[\mathrm{CH_3COOH}\text{(aq)} \rightleftharpoons \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H^+}\text{(aq)}\]
\[\mathrm{NH_4OH}\text{(aq)} \rightleftharpoons \mathrm{NH_4^{+}}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Note that HCOOH, HF, \(\mathrm{H_2S}\) are examples of weak acids while \(\mathrm{Fe(OH)_3}\), \(\mathrm{Cu(OH)_2}\) are examples of weak bases.
3.4.1 Dissociation Constant of Weak Acids and Weak Bases
Since a weak acid in water reaches a genuine equilibrium, it must have an equilibrium constant — and that constant is the number that tells us, once and for all, how weak the acid is.
The dissociation of a weak acid HA in water is expressed as
\[\mathrm{HA}\text{(aq)} \rightleftharpoons \mathrm{H^+}\text{(aq)} + \mathrm{A^-}\text{(aq)}\]
The equilibrium constant called acid-dissociation constant for this equilibrium is:
\[K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \qquad \text{...........(3.3)}\]
Similarly the dissociation of weak base BOH in water is represented as:
\[\mathrm{BOH}\text{(aq)} \rightleftharpoons \mathrm{B^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
The equilibrium constant called base-dissociation constant for this equilibrium is,
\[K_b = \frac{[\mathrm{B^+}][\mathrm{OH^-}]}{[\mathrm{BOH}]} \qquad \text{...........(3.4)}\]
Thus, the dissociation constant of a weak acid or a weak base is defined as the equilibrium constant for dissociation equilibrium of weak acid or weak base, respectively.
Reading the Value of \(K_a\)
Look at where the terms sit in Eq. (3.3). The ions are in the numerator and the undissociated molecule is in the denominator. So a large value of \(K_a\) means a great deal of ion and very little intact molecule — a comparatively strong acid. A very small \(K_a\) means the opposite: the acid has barely let go of its protons. Comparing \(K_a\) values is therefore the proper way to rank weak acids in order of strength, and questions asking you to arrange acids by strength are really asking you to arrange them by \(K_a\).
Two practical points about these expressions are worth noting. First, the concentration of water does not appear in Eq. (3.3), even though water is the solvent in which the reaction happens; water is present in such enormous excess that its concentration is effectively constant, and a constant is absorbed into \(K_a\) itself. Second, \(K_a\) and \(K_b\) are equilibrium constants, so their values change only with temperature. Adding more acid, or adding water to dilute the solution, does not alter them.
3.4.2 Ostwald’s Dilution Law
Arrhenius concept of acids and bases was expressed quantitatively by F. W. Ostwald in the form of the dilution law in 1888.
The question Ostwald answered is a very practical one. We now have two separate ways of describing a weak electrolyte: the degree of dissociation \(\alpha\) from Section 3.2, and the dissociation constant \(K_a\) from just above. They must be connected, because both describe the same equilibrium. The dilution law is that connection.
a. Weak Acids
Consider an equilibrium of weak acid HA that exists in solution partly as the undissociated species HA and partly \(\mathrm{H^+}\) and \(\mathrm{A^-}\) ions. Then
\[\mathrm{HA}\text{(aq)} \rightleftharpoons \mathrm{H^+}\text{(aq)} + \mathrm{A^-}\text{(aq)}\]
The acid dissociation constant is given by Eq. (3.3)
\[K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\]
Suppose 1 mol of acid HA is initially present in volume \(V\) dm\(^3\) of the solution. At equilibrium the fraction dissociated would be \(\alpha\), where \(\alpha\) is degree of dissociation of the acid. The fraction of an acid that remains undissociated would be \((1 - \alpha)\).
Setting the amounts out in a table makes the next step easy. Each mole of HA that dissociates produces one mole of \(\mathrm{H^+}\) and one mole of \(\mathrm{A^-}\), so both ions appear in the amount \(\alpha\). Dividing every amount by the volume \(V\) converts moles into concentrations.
| \(\mathrm{HA}\text{(aq)} \rightleftharpoons \mathrm{H^+}\text{(aq)} + \mathrm{A^-}\text{(aq)}\) | HA | \(\mathrm{H^+}\) | \(\mathrm{A^-}\) |
|---|---|---|---|
| Amount present at equilibrium/mol | \((1-\alpha)\) | \(\alpha\) | \(\alpha\) |
| Concentration at equilibrium/mol dm\(^{-3}\) | \(\dfrac{1-\alpha}{V}\) | \(\dfrac{\alpha}{V}\) | \(\dfrac{\alpha}{V}\) |
Thus, at equilibrium \([\mathrm{HA}] = \dfrac{1-\alpha}{V}\), mol dm\(^{-3}\),
\[[\mathrm{H^+}] = [\mathrm{A^-}] = \frac{\alpha}{V}\ \text{mol dm}^{-3}.\]
Substituting these in Eq. (3.3)
\[K_a = \frac{(\alpha/V)\,(\alpha/V)}{(1-\alpha)/V} = \frac{\alpha^2}{(1-\alpha)V} \qquad \text{.......(3.5)}\]
If \(c\) is the initial concentration of an acid in mol dm\(^{-3}\) and \(V\) is the volume in dm\(^3\) mol\(^{-1}\) then \(c = 1/V\). Replacing \(1/V\) in Eq. (3.5) by \(c\) we get
\[K_a = \frac{\alpha^2 c}{1-\alpha} \qquad \text{......(3.6)}\]
For the weak acid HA, \(\alpha\) is very small, or \((1 - \alpha) \cong 1\). This approximation is the step students most often skip over, so it is worth being clear about it: if only one molecule in a thousand has dissociated, then \(\alpha = 0.001\) and \((1-\alpha) = 0.999\), which is 1 for all practical purposes. With this Eq. (3.5) and (3.6) reduce.
\[K_a = \alpha^2/V \quad \text{and} \quad K_a = \alpha^2 c \qquad \text{...........(3.7)}\]
Rearranging to make \(\alpha\) the subject — divide by \(c\) and take the square root —
\[\alpha = \sqrt{\frac{K_a}{c}} \quad \text{or} \quad \alpha = \sqrt{K_a . V} \qquad \text{...........(3.8)}\]
The Eq. (3.8) implies that the degree of dissociation of a weak acid is inversely proportional to the square root of its concentration or directly proportional to the square root of volume of the solution containing 1 mol of the weak acid.
The whole chain of reasoning is set out step by step in FC 3.2.
Flowchart
FC 3.2 – Deriving Ostwald’s dilution law for a weak acid: take 1 mol HA in V dm³ → fraction α dissociates → equilibrium amounts HA = (1−α), H⁺ = α, A⁻ = α → divide each by V to get concentrations → substitute into Ka = [H⁺][A⁻]/[HA] → Ka = α²/(1−α)V → put c = 1/V → Ka = α²c/(1−α) → apply weak-electrolyte approximation (1−α) ≅ 1 → Ka = α²c → α = √(Ka/c).
Because \(\alpha\) depends on concentration but \(K_a\) does not, we can watch the two behave differently as a solution is diluted, as plotted in Fig. 3.5.
Diagram
Fig. 3.5 – Two curves for a weak acid plotted against concentration c on the x-axis: (a) degree of dissociation α rising steeply as c falls towards zero; (b) Ka drawn as a flat horizontal line at a constant value; label the axes and mark the point where α approaches 1 at very high dilution.
b. Weak Base
Consider 1 mol of weak base BOH dissolved in \(V\) dm\(^3\) of solution. The base dissociates partially as
\[\mathrm{BOH}\text{(aq)} \rightleftharpoons \mathrm{B^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
The base dissociation constant is
\[K_b = \frac{[\mathrm{B^+}][\mathrm{OH^-}]}{[\mathrm{BOH}]}\]
Let the fraction dissociated at equilibrium is \(\alpha\) and that remains undissociated is \((1 - \alpha)\).
| \(\mathrm{BOH}\text{(aq)} \rightleftharpoons \mathrm{B^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\) | BOH | \(\mathrm{B^+}\) | \(\mathrm{OH^-}\) |
|---|---|---|---|
| Amount present at equilibrium | \((1-\alpha)\) | \(\alpha\) | \(\alpha\) |
| Concentration at equilibrium | \(\dfrac{1-\alpha}{V}\) | \(\dfrac{\alpha}{V}\) | \(\dfrac{\alpha}{V}\) |
At equilibrium,
\[[\mathrm{BOH}] = \frac{1-\alpha}{V}\ \text{mol dm}^{-3}, \qquad [\mathrm{B^+}] = [\mathrm{OH^-}] = \frac{\alpha}{V}\ \text{mol dm}^{-3}.\]
Substitution of these concentrations in Eq. (3.4), gives
\[K_b = \frac{(\alpha/V)\,(\alpha/V)}{(1-\alpha)/V} = \frac{\alpha^2}{(1-\alpha)V} \qquad \text{..........(3.9)}\]
Similar arguments in the case of weak acid, led to
\[K_b = \frac{\alpha^2 c}{(1-\alpha)} \qquad \text{...........(3.10)}\]
\[\alpha = \sqrt{K_b . V}\ ,\qquad \alpha = \sqrt{\frac{K_b}{c}} \qquad \text{...........(3.11)}\]
The degree of dissociation of a weak base is inversely proportional to square root of its concentration and is directly proportional to square root of volume of the solution containing 1 mol of weak base.
The two derivations are deliberately identical in shape. Once you have followed the argument for the acid, the base version needs no new thinking — only the replacement of \(\mathrm{H^+}\) by \(\mathrm{OH^-}\) and of \(K_a\) by \(K_b\).
3.5 Autoionization of Water
We have already seen that water is amphoteric — it can donate a proton and it can accept one. A curious consequence follows. If one water molecule can hand a proton to another substance, and another water molecule can accept a proton, then water can simply do both to itself.
That is exactly what happens. Pure water ionizes to a very small extent. The ionization equilibrium of water is represented as,
\[\mathrm{H_2O}\text{(l)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_3O^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
One molecule acts as the acid and donates a proton; the other acts as the base and accepts it. The proton transfer is shown in Fig. 3.6.
Diagram
Fig. 3.6 – Autoionization of water: two H₂O molecules drawn side by side, a curly arrow from a lone pair on the oxygen of the left molecule to a hydrogen of the right molecule and a second curly arrow showing the O–H bond breaking; products drawn as H₃O⁺ (with ⊕ charge) and OH⁻ (with ⊖ charge), and the whole shown as an equilibrium with a double arrow.
The Equilibrium Constant for Water
The equilibrium constant (\(K\)) for the ionization of water is given by
\[K = \frac{[\mathrm{H_3O^+}][\mathrm{OH^-}]}{[\mathrm{H_2O}]^2} \qquad \text{...........(3.12)}\]
Multiplying both sides by \([\mathrm{H_2O}]^2\) to clear the denominator,
\[\text{or}\quad K[\mathrm{H_2O}]^2 = [\mathrm{H_3O^+}][\mathrm{OH^-}] \qquad \text{...........(3.13)}\]
A majority of \(\mathrm{H_2O}\) molecules are undissociated, consequently concentration of water \([\mathrm{H_2O}]\) can be treated as constant. Then
\([\mathrm{H_2O}]^2 = K'\). Substituting this in Eq. (3.13) we get,
\[K \times K' = [\mathrm{H_3O^+}][\mathrm{OH^-}] \qquad \text{...........(3.14)}\]
\[K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]
where \(K_w = KK'\) is called ionic product of water. The product of molar concentrations of hydronium (or hydrogen) ions and hydroxyl ions at equilibrium in pure water at the given temperature is called ionic product of water.
Notice what this manoeuvre has achieved. Two constants, \(K\) and \(K'\), have been folded into a single new constant \(K_w\), and the concentration of water has disappeared from the expression altogether. What is left is a remarkably simple statement: in water, the product of the two ion concentrations is fixed.
The Value of \(K_w\) at 298 K
In pure water \(\mathrm{H_3O^+}\) ion concentration always equals the concentration of \(\mathrm{OH^-}\) ion. This must be so, because the two ions are produced together, one for one, by the same reaction — there is no other source of either. Thus at 298 K this concentration is found to be \(1.0 \times 10^{-7}\) mol/L.
\[K_w = (1.0 \times 10^{-7})\,(1.0 \times 10^{-7})\]
\[K_w = 1.0 \times 10^{-14} \qquad \text{...........(3.15)}\]
This constancy is the single most useful fact in the whole chapter. \(K_w\) holds not only for pure water but for every dilute aqueous solution, whatever else is dissolved in it. Add acid, and \([\mathrm{H_3O^+}]\) rises — but \([\mathrm{OH^-}]\) must fall in exact proportion so that the product stays at \(1.0 \times 10^{-14}\). Hydroxide ions are never entirely absent from an acidic solution, and hydronium ions are never entirely absent from a basic one; one of them is simply squeezed down to a very small value. Knowing either concentration therefore gives you the other immediately.
3.6 pH Scale
Instead of writing concentration of \(\mathrm{H_3O^+}\) ions in mol dm\(^{-3}\), sometimes it is convenient to express it on the logarithmic scale. This is known as pH scale.
The reason for wanting such a scale becomes obvious the moment you try to compare real solutions. Gastric juice has a hydronium ion concentration of about \(10^{-2}\) mol dm\(^{-3}\), pure water \(10^{-7}\), and a strong alkali as little as \(10^{-13}\). These numbers differ by a factor of ten thousand million million, and no ordinary graph or table can hold them comfortably.
Sorensen in 1909 defined the pH of a solution as the negative logarithm to the base 10, of the concentration of \(\mathrm{H^+}\) ions in solution in mol dm\(^{-3}\). Expressed mathematically as
\[pH = -\log_{10}[\mathrm{H^+}]\]
Why a Logarithm, and Why the Minus Sign
If a mathematical tool from an earlier class needs refreshing, this is the place. The logarithm to base 10 of a number simply asks: to what power must 10 be raised to give this number? Since \(10^{-7}\) is 10 raised to the power \(-7\), we have \(\log_{10}(10^{-7}) = -7\). The logarithm converts an awkward power of ten into a small, easily handled number.
But those logarithms all come out negative, because the concentrations are all less than 1. Sørensen therefore put a minus sign in front, which flips them to positive values. Pure water, with \([\mathrm{H^+}] = 1.0 \times 10^{-7}\), gets a pH of \(+7\) instead of \(-7\).
Two consequences follow, and both are constantly tested. First, a smaller pH means a more acidic solution, because a larger \([\mathrm{H^+}]\) gives a larger negative logarithm, which the minus sign turns into a smaller positive number. Second, each whole unit of pH stands for a tenfold change in hydronium ion concentration — a solution of pH 3 is not slightly more acidic than one of pH 5, it is a hundred times more acidic.
Similarly pOH of a solution can be defined as the negative logarithm to the base 10, of the molar concentration of \(\mathrm{OH^-}\) ions in solution.
Thus,
\[pOH = -\log_{10}[\mathrm{OH^-}] \qquad \text{...........(3.16)}\]
3.6.1 Relationship between pH and pOH
The two scales cannot be independent of each other, because the two ion concentrations are locked together by the ionic product of water.
The ionic product of water is
\[K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]
Now, \(K_w = 1 \times 10^{-14}\) at 298 K and thus
\[[\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}\]
Taking logarithm of both the sides, we write
\[\log_{10}[\mathrm{H_3O^+}] + \log_{10}[\mathrm{OH^-}] = -14\]
The logarithm of a product is the sum of the logarithms, which is why the left-hand side splits into two terms. Multiplying throughout by \(-1\) turns both terms into the quantities we have just defined:
\[-\log_{10}[\mathrm{H_3O^+}] + \{-\log_{10}[\mathrm{OH^-}]\} = 14\]
From Eq. (3.16) and (3.17)
\[pH + pOH = 14 \qquad \text{...........(3.18)}\]
This relation is a great labour-saver. Any problem that gives you one of the four quantities \([\mathrm{H_3O^+}]\), \([\mathrm{OH^-}]\), pH or pOH gives you all four, since \(K_w\) links the concentrations and Eq. (3.18) links the two p-values.
3.6.2 Acidity, Basicity and Neutrality of Aqueous Solutions
1. Neutral solution: For pure water or any aqueous neutral solution at 298 K
\[[\mathrm{H_3O^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}\ \text{M}\]
Hence, \(pH = -\log_{10}[\mathrm{H^+}] = -\log_{10}[1 \times 10^{-7}] = 7\)
2. Acidic solution: In acidic solution, there is excess of \(\mathrm{H_3O^+}\) ions, or \([\mathrm{H_3O^+}] > [\mathrm{OH^-}]\). Hence, \([\mathrm{H_3O^+}] > 1 \times 10^{-7}\) and \(pH < 7\)
3. Basic solution: In basic solution, the excess of \(\mathrm{OH^-}\) ions are present that is \([\mathrm{H_3O^+}] < [\mathrm{OH^-}]\) or \([\mathrm{H_3O^+}] < 1.0 \times 10^{-7}\) with \(pH > 7\).
Note carefully that neutrality does not mean the absence of ions. A neutral solution has plenty of both — it simply has them in equal numbers. The complete picture is summarised on the pH scale in Fig. 3.1.
Diagram
Fig. 3.1 – The pH scale: a horizontal bar numbered 0 to 14 with the region 0–7 marked “Acidic”, the single point 7 marked “Neutral” and 7–14 marked “Basic”; directly below each pH value print the corresponding [H⁺] as 10⁰, 10⁻¹, 10⁻², … down to 10⁻¹⁴ mol dm⁻³.
3.7 Hydrolysis of Salts
A salt is what remains when an acid and a base neutralise each other, so it is tempting to assume that every salt solution must be neutral. Dissolve common salt in water and that assumption holds. Dissolve washing soda, and the solution turns distinctly alkaline; dissolve ammonium chloride, and it turns acidic. Something is clearly happening between the ions of the salt and the water they are dissolved in.
3.7.1 Types of Salts
Since a salt is the product of an acid and a base, and since each of those may be strong or weak, there are exactly four possible combinations. Classifying a salt correctly is the first step in every problem of this section, so it is worth learning to name a salt’s “parents” at sight. These are of four types.
I. Salts derived from strong acid and strong base. For example: NaCl, \(\mathrm{Na_2SO_4}\), \(\mathrm{NaNO_3}\), KCl, \(\mathrm{KNO_3}\).
II. Salts derived from strong acids and weak bases. For example: \(\mathrm{NH_4Cl}\), \(\mathrm{CuSO_4}\), \(\mathrm{NH_4NO_3}\), \(\mathrm{CuCl_2}\).
III. Salts derived from weak acids and strong bases. For example: \(\mathrm{CH_3COONa}\), KCN, \(\mathrm{Na_2CO_3}\).
IV. Salts derived from weak acids and weak bases. For example: \(\mathrm{CH_3COONH_4}\), \(\mathrm{NH_4CN}\).
To identify the parents, split the formula into its cation and anion, then ask which base would supply that cation and which acid would supply that anion. For \(\mathrm{CH_3COONa}\), the \(\mathrm{Na^+}\) comes from NaOH (a strong base) and the \(\mathrm{CH_3COO^-}\) from \(\mathrm{CH_3COOH}\) (a weak acid) — so it is a Type III salt.
3.7.2 Concept of Hydrolysis
When a salt is dissociated in water, it dissociates completely into its constituent ions. The solvent water dissociates slightly as,
\[\mathrm{H_2O}\text{(l)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_3O^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Pure water is neutral and \([\mathrm{H_3O^+}] = [\mathrm{OH^-}]\). If the ions of the salt do not interact with water, the hydronium and hydroxyl ion concentrations remain equal and the solution is neutral.
When one or more of the salt ions react with water, the equality of concentrations of \(\mathrm{H_3O^+}\) and \(\mathrm{OH^-}\) ions is disturbed. The solution does not remain neutral and becomes acidic or basic depending on the type of the salt. Such a reaction between the ions of salt and the ions of water is called hydrolysis of salt.
Hydrolysis of salt is defined as the reaction in which cations or anions or both ions of a salt react with ions of water to produce acidity or alkalinity (or sometimes even neutrality).
What Decides Whether an Ion Will Hydrolyse
Behind all six of the cases that follow lies a single test, and once you have it, none of them needs memorising. Ask what compound an ion of the salt would form if it captured its partner ion from water. If that compound would be a strong electrolyte, it would immediately fall apart again into ions, so nothing has really happened — that ion does not hydrolyse. If the compound would be a weak electrolyte or an unionized molecule, it stays intact, and its partner ion from water is left over in excess. That leftover ion is what makes the solution acidic or basic.
3.7.3 Salts of Strong Acids and Strong Bases
NaCl is a salt of strong acid HCl and a strong base NaOH. When it is dissolved in water, it dissociates completely into its ions.
\[\mathrm{NaCl}\text{(aq)} \longrightarrow \mathrm{Na^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
The ions \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) have no tendency to react with water. This is because the possible products, NaOH and HCl of such reactions are strong electrolytes and dissociate completely in aqueous solutions.
In other words,
\[\mathrm{Na^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)} + \mathrm{H_2O} \longrightarrow \mathrm{HCl}\text{(aq)} + \mathrm{NaOH}\text{(aq)}\] \[\text{(strong acid)}\qquad \text{(strong base)}\] \[\text{[Possible products]}\]
\[\mathrm{HCl}\text{(aq)} + \mathrm{NaOH}\text{(aq)} + \mathrm{H_2O} \longrightarrow \mathrm{H_3O^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)} + \mathrm{Na^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Thus the reactants and the products are the same. This implies that neither the cation nor anion of the salt reacts with water or there is no hydrolysis. Equality \(\mathrm{H_3O^+} = \mathrm{OH^-}\) produced by ionization of water is not disturbed and solution is neutral. It may be concluded that salt of strong acid and strong base does not undergo hydrolysis.
Follow that pair of equations carefully, because they contain a small and rather elegant argument. We assumed that hydrolysis takes place and wrote down what it would give. Those products, being strong electrolytes, at once dissociate — and hand us back exactly the ions we started with. A reaction whose products are identical to its reactants is no reaction at all.
3.7.4 Salts of Strong Acids and Weak Bases
\(\mathrm{CuSO_4}\) is salt of strong acid \(\mathrm{H_2SO_4}\) and weak base \(\mathrm{Cu(OH)_2}\). When \(\mathrm{CuSO_4}\) is dissolved in water, it dissociates completely as,
\[\mathrm{CuSO_4}\text{(aq)} \longrightarrow \mathrm{Cu^{2+}}\text{(aq)} + \mathrm{SO_4^{2-}}\text{(aq)}\]
\(\mathrm{SO_4^{2-}}\) ions of salt have no tendency to react with water because the possible product \(\mathrm{H_2SO_4}\) is strong electrolyte. The reaction of \(\mathrm{Cu^{2+}}\) ions with \(\mathrm{OH^-}\) ions form unionized \(\mathrm{Cu(OH)_2}\). The hydrolytic equilibrium for \(\mathrm{CuSO_4}\) is then written as,
\[\mathrm{Cu^{2+}}\text{(aq)} + 4\,\mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{Cu(OH)_2}\text{(aq)} + 2\,\mathrm{H_3O^+}\text{(aq)}\]
Here only one of the two ions has reacted, and it has consumed hydroxide. That leaves the hydronium ions of water unmatched. Due to the presence of excess of \(\mathrm{H_3O^+}\) ions, the resulting solution of \(\mathrm{CuSO_4}\) becomes acidic and turns blue litmus red.
Why a Copper Sulphate Solution Turns Cloudy
Formation of sparingly soluble \(\mathrm{Cu(OH)_2}\) by hydrolysis makes the aqueous solution of \(\mathrm{CuSO_4}\) turbid. If \(\mathrm{H_2SO_4}\), that is \(\mathrm{H_3O^+}\) ions are added, the hydrolytic equilibrium shifts to the left. A turbidity of \(\mathrm{Cu(OH)_2}\) dissolves to give a clear solution. To get clear solution of \(\mathrm{CuSO_4}\), the addition of \(\mathrm{H_2SO_4}\) would be required. The effect is shown in Fig. 3.7.
Diagram
Fig. 3.7 – Two beakers of copper sulphate solution: (a) a cloudy blue solution labelled “turbid — suspended Cu(OH)₂ formed by hydrolysis”; (b) the same solution after a few drops of dilute H₂SO₄ are added from a dropper, now clear blue, labelled “equilibrium pushed to the left, Cu(OH)₂ dissolves”; show the hydrolytic equilibrium equation between the two beakers with an arrow marked “add H₂SO₄”.
This is a piece of laboratory practice rather than mere theory. Whenever a clear solution of copper sulphate is needed for an experiment, a little dilute sulphuric acid is added at the time of preparation. Le Chatelier’s principle explains why it works: adding \(\mathrm{H_3O^+}\) ions — a product of the hydrolysis — pushes the equilibrium back towards the reactant side, and the suspended hydroxide redissolves.
3.7.5 Salts of Weak Acids and Strong Bases
\(\mathrm{CH_3COONa}\) is a salt of weak acid \(\mathrm{CH_3COOH}\) and strong base NaOH; when dissolved in water, it dissociates completely.
\[\mathrm{CH_3COONa}\text{(aq)} \longrightarrow \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{Na^+}\text{(aq)}\]
Water dissociates slightly as,
\[\mathrm{H_2O}\text{(l)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_3O^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Solution of \(\mathrm{CH_3COONa}\) contains \(\mathrm{Na^+}\), \(\mathrm{H_3O^+}\), \(\mathrm{CH_3COO^-}\), \(\mathrm{OH^-}\). The \(\mathrm{Na^+}\) ions of salt have no tendency to react with \(\mathrm{OH^-}\) ions of water since the possible product of the reaction is NaOH, a strong electrolyte.
On the other hand the reaction of \(\mathrm{CH_3COO^-}\) ions of salt with the \(\mathrm{H_3O^+}\) ions from water produces unionized \(\mathrm{CH_3COOH}\).
\[\mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{CH_3COOH}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
Thus, the hydrolytic equilibrium for \(\mathrm{CH_3COONa}\) is,
\[\mathrm{CH_3COONa}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{CH_3COOH}\text{(aq)} + \mathrm{Na^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
As a result of excess \(\mathrm{OH^-}\) ions produced the solution becomes basic. The solution of \(\mathrm{CH_3COONa}\) is therefore basic.
Notice how exactly this mirrors the copper sulphate case, with the roles reversed. There the cation hydrolysed and locked up hydroxide, leaving excess acid; here the anion hydrolyses and locks up hydronium, leaving excess alkali.
As a general rule the solutions of salts of strong acids and strong bases are neutral, the solutions of salts of strong acids and weak bases are acidic and the solutions of salts of strong bases and weak acids are basic.
A short way to remember it is that the stronger parent wins. If the parent acid was strong and the parent base weak, the acid’s character survives in the solution; if the base was the strong parent, the solution turns basic; if both parents were equally strong, neither wins and the solution is neutral. The whole decision is set out in FC 3.3.
Flowchart
FC 3.3 – Deciding the nature of a salt solution: dissolve salt in water → identify the parent acid and parent base → branch (I) strong acid + strong base → neither ion hydrolyses → neutral, pH = 7 (e.g. NaCl); branch (II) strong acid + weak base → cation hydrolyses → excess H₃O⁺ → acidic, pH < 7 (e.g. NH₄Cl, CuSO₄); branch (III) weak acid + strong base → anion hydrolyses → excess OH⁻ → basic, pH > 7 (e.g. CH₃COONa, Na₂CO₃); branch (IV) weak acid + weak base → both ions hydrolyse → compare Ka and Kb → Ka > Kb acidic / Ka < Kb basic / Ka = Kb neutral.
3.7.6 Salts of Weak Acids and Weak Bases
This is the only case in which both ions of the salt hydrolyse, and so the only case in which the outcome cannot be read off from the parents alone.
When salt BA of weak acid HA and weak base BOH is dissolved in water, it dissociates completely as
\[\mathrm{BA}\text{(aq)} \longrightarrow \mathrm{B^+}\text{(aq)} + \mathrm{A^-}\text{(aq)}\]
The hydrolysis reaction involves the interaction of both the ions of the salt with water,
\[\mathrm{B^+}\text{(aq)} + \mathrm{A^-}\text{(aq)} + \mathrm{H_2O} \rightleftharpoons \mathrm{BOH} + \mathrm{HA}\] \[\text{(weak base)}\quad \text{(weak acid)}\]
The solution may turn out acidic, basic or neutral depending on the relative strength of weak base and weak acid formed in the hydrolysis.
if, \(K_a > K_b\), the solution will be acidic.
if, \(K_a < K_b\), the solution will be basic.
if, \(K_a = K_b\), the solution will be neutral.
The logic is a tug-of-war. The cation’s hydrolysis manufactures \(\mathrm{H_3O^+}\) and the anion’s hydrolysis manufactures \(\mathrm{OH^-}\), both at once. Whichever ion hydrolyses more vigorously wins, and its product is left in excess. Comparing \(K_a\) with \(K_b\) is simply the way of asking which of the two parents holds its ions more tightly.
i. Salt of Weak Acid and Weak Base for which \(K_a > K_b\)
\(\mathrm{NH_4F}\) is a salt of weak acid HF (\(K_a = 7.2 \times 10^{-4}\)) and weak base \(\mathrm{NH_4OH}\) (\(K_b = 1.8 \times 10^{-5}\)). Here, \(K_a\) is greater than \(K_b\). The salt hydrolyses as
\[\mathrm{NH_4^{+}}\text{(aq)} + \mathrm{F^-}\text{(aq)} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4OH}\text{(aq)} + \mathrm{HF}\text{(aq)}\] \[\text{(weak base)}\qquad \text{(weak acid)}\]
The acid HF is slightly stronger than base \(\mathrm{NH_4OH}\). The two ions react with water as
\[\mathrm{NH_4^{+}}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{NH_4OH}\text{(aq)} + \mathrm{H_3O^+}\text{(aq)} \qquad \text{...........(3.19)}\]
\[\mathrm{F^-}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{HF}\text{(aq)} + \mathrm{OH^-}\text{(aq)} \qquad \text{...........(3.20)}\]
The \(\mathrm{NH_4^{+}}\) ions hydrolyse to a slightly greater extent than the \(\mathrm{F^-}\) ions. That means the reaction produces more \(\mathrm{H_3O^+}\) ions than the \(\mathrm{OH^-}\) ions produced in reaction (3.20). In other words, \(\mathrm{NH_4^{+}}\) ions are slightly stronger as acid than \(\mathrm{F^-}\) ions as base. The solution of \(\mathrm{NH_4F}\) is thus only slightly acidic and turns blue litmus red.
Because the two constants differ by only about a factor of forty here, the imbalance is small — hence the repeated word slightly. The solution is acidic, but far less so than a solution of a Type II salt of comparable concentration.
ii. Salt of Weak Acid and Weak Base for which \(K_a < K_b\)
\(\mathrm{NH_4CN}\) is the salt of weak acid HCN (\(K_a = 4.0 \times 10^{-10}\)) and weak base \(\mathrm{NH_4OH}\) (\(K_b = 1.8 \times 10^{-5}\)) showing that \(K_a < K_b\). When \(\mathrm{NH_4CN}\) is dissolved in water, it hydrolyses as
\[\mathrm{NH_4^{+}}\text{(aq)} + \mathrm{CN^-}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{NH_4OH}\text{(aq)} + \mathrm{HCN}\text{(aq)}\] \[\text{(weak base)}\qquad \text{(weak acid)}\]
The base \(\mathrm{NH_4OH}\) is stronger than the acid HCN. The ions of the salt react with water as,
\[\mathrm{NH_4^{+}}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{NH_4OH}\text{(aq)} + \mathrm{H_3O^+}\text{(aq)} \qquad \text{...........(3.21)}\]
\[\mathrm{CN^-}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{HCN}\text{(aq)} + \mathrm{OH^-}\text{(aq)} \qquad \text{...........(3.22)}\]
The \(\mathrm{CN^-}\) ions hydrolyse to a greater extent than \(\mathrm{NH_4^{+}}\) ions. The reaction (3.22) produces more \(\mathrm{OH^-}\) ions than the \(\mathrm{H_3O^+}\) ions produced in reaction (3.21). The solution of \(\mathrm{NH_4CN}\) is basic and turns red litmus blue.
Here the gap between the two constants is enormous — \(K_b\) exceeds \(K_a\) by a factor of about forty thousand — because HCN is an exceptionally weak acid. Its anion is correspondingly eager to take a proton back from water, and the solution comes out clearly basic.
iii. Salt of Weak Acid and Weak Base for which \(K_a = K_b\)
\(\mathrm{CH_3COONH_4}\) is a salt of weak acid, \(\mathrm{CH_3COOH}\) (\(K_a = 1.8 \times 10^{-5}\)) and weak base, \(\mathrm{NH_4OH}\) (\(K_b = 1.8 \times 10^{-5}\)).
When the salt \(\mathrm{CH_3COONH_4}\) is dissolved in water, it undergoes hydrolysis:
\[\mathrm{CH_3COO^-}\text{(aq)} + \mathrm{NH_4^{+}}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{CH_3COOH}\text{(aq)} + \mathrm{NH_4OH}\text{(aq)}\] \[\text{(weak acid)}\qquad\qquad \text{(weak base)}\]
The ions of the salt react with water as
\[\mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{CH_3COOH}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\]
\[\mathrm{NH_4^{+}}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{NH_4OH}\text{(aq)} + \mathrm{H_3O^+}\text{(aq)}\]
As \(K_a = K_b\), the relative strength of acid and base produced in hydrolysis is the same. Therefore, the solution is neutral. Hydrolysis of \(\mathrm{NH_4^{+}}\) produces as many \(\mathrm{H_3O^+}\) ions as that of \(\mathrm{CH_3COO^-}\) produces \(\mathrm{OH^-}\) ions.
This last case is a favourite of examiners, and it is easy to get wrong by answering too quickly. The solution of ammonium acetate is neutral — but not for the reason common salt is neutral. In common salt, no hydrolysis occurs at all. In ammonium acetate, both ions hydrolyse extensively; the solution is neutral only because the two effects are exactly equal and cancel each other out.
3.8 Buffer Solutions
Add a single drop of concentrated hydrochloric acid to a beaker of pure water and its pH plunges. Add the same drop to a litre of blood, and the pH barely moves. Some solutions have a built-in resistance to change in pH, and they are among the most useful solutions in chemistry, biology and industry.
Buffer solution is defined as a solution which resists drastic changes in pH when a small amount of strong acid or strong base or water is added to it.
Note the two words that make this definition honest. The buffer resists drastic changes — its pH does shift a little — and it does so only for a small amount of added acid or base. Pour in enough acid and any buffer will eventually be overwhelmed.
3.8.1 Types of Buffer Solutions
There are two types of buffer solutions. Acidic buffer used to maintain an acidic pH, while basic buffer maintains alkaline pH.
a. Acidic buffer solution: A solution containing a weak acid and its salts with strong base is called an acidic buffer solution.
For example: A solution containing weak acid such as \(\mathrm{CH_3COOH}\) and its salt such as \(\mathrm{CH_3COONa}\) is an acidic buffer solution.
\(pH\) of acidic buffer is given by the equation
\[pH = pK_a + \log_{10}\frac{[\text{salt}]}{[\text{acid}]} \qquad \text{...........(3.23)}\]
where
\[pK_a = -\log_{10}K_a \qquad \text{...........(3.24)}\]
and \(K_a\) is the dissociation constant of the acid.
b. Basic buffer solution: A solution containing a weak base and its salt with strong acid is the basic buffer solution.
For example: A solution containing a weak base such as \(\mathrm{NH_4OH}\) and its salt such as \(\mathrm{NH_4Cl}\) is a basic buffer solution.
The \(pOH\) of basic buffer is given by,
\[pOH = pK_b + \log_{10}\frac{[\text{salt}]}{[\text{base}]} \qquad \text{...........(3.25)}\]
where
\[pK_b = -\log_{10}K_b \qquad \text{...........(3.26)}\]
and \(K_b\) is the dissociation constant for the base. Equations (3.23), (3.24), (3.25) and (3.26) are known to Henderson Hasselbalch equation.
Reading the Henderson–Hasselbalch Equation
Two features of Eq. (3.23) are worth pausing on, because between them they explain almost everything a buffer does.
First, the pH is fixed mainly by \(pK_a\) — a constant belonging to the chosen weak acid. This is why a buffer is designed: to hold a solution near pH 4.7 you pick acetic acid, whose \(pK_a\) is close to that value, and to hold it near pH 9 you pick a different system.
Second, the correction term contains a ratio of concentrations, not the concentrations themselves. When salt and acid are present in equal concentration the ratio is 1, its logarithm is zero, and the buffer sits exactly at \(pH = pK_a\). Fine adjustment is then made by altering the proportion of salt to acid.
3.8.2 Buffer Action
Let us consider sodium acetate - acetic acid buffer. Here sodium acetate is a strong electrolyte which dissociates completely in water producing large concentration of \(\mathrm{CH_3COO^-}\) as follows:
\[\mathrm{CH_3COONa}\text{(aq)} \longrightarrow \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{Na^+}\text{(aq)}\]
On the other hand since the acetic acid is a weak acid, the concentration of undissociated \(\mathrm{CH_3COOH}\) molecules is usually high.
This is the whole secret of a buffer: the solution holds a large reserve of both partners at once — plenty of acid to deal with any incoming base, and plenty of its conjugate base to deal with any incoming acid, as shown in Fig. 3.8.
Diagram
Fig. 3.8 – Buffer action in a sodium acetate–acetic acid buffer: a beaker containing a large reservoir of CH₃COOH molecules and CH₃COO⁻ ions; an arrow on the left labelled “add H⁺” pointing to CH₃COO⁻ with the product CH₃COOH shown, and an arrow on the right labelled “add OH⁻” pointing to CH₃COOH with the products CH₃COO⁻ and H₂O shown; a pH meter beside the beaker reading almost the same value before and after.
If a strong acid is added to this solution the added \(\mathrm{H^+}\) ions will be consumed by the conjugate base \(\mathrm{CH_3COO^-}\) present in large concentration. Similarly, if small amount of base is added, the added \(\mathrm{OH^-}\) ions will be neutralized by the large concentration of acetic acid as shown in the following reactions:
\[\underset{\text{(large concentration)}}{\mathrm{CH_3COO^-}\text{(aq)}} + \underset{\text{(added acid)}}{\mathrm{H^+}\text{(aq)}} \longrightarrow \mathrm{CH_3COOH}\text{(aq)}\]
\[\underset{\text{(large concentration)}}{\mathrm{CH_3COOH}\text{(aq)}} + \underset{\text{(added base)}}{\mathrm{OH^-}\text{(aq)}} \longrightarrow \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
Look closely at what these two equations accomplish. In each case a strongly ionized species — free \(\mathrm{H^+}\) or free \(\mathrm{OH^-}\) — is converted into a weak, largely undissociated one. The added ions have not been destroyed; they have been locked into a form that hardly ionizes. The acid or base added thus can not change the \([\mathrm{H^+}]\) or \([\mathrm{OH^-}]\) concentrations and, \(pH\) of the buffer remains unchanged. The complete sequence is set out in FC 3.4.
Flowchart
FC 3.4 – Buffer action of a CH₃COOH/CH₃COONa buffer: buffer contains large reserves of CH₃COOH and CH₃COO⁻ → branch (a) a few drops of strong acid added → H⁺ ions released → captured by CH₃COO⁻ reserve → converted to weakly ionized CH₃COOH → free [H⁺] almost unchanged → pH almost unchanged; branch (b) a few drops of strong base added → OH⁻ ions released → neutralised by CH₃COOH reserve → converted to CH₃COO⁻ + H₂O → free [OH⁻] almost unchanged → pH almost unchanged; branch (c) water added → both concentrations fall equally → salt/acid ratio unchanged → pH unchanged.
Dilution does not have any effect on \(pH\) of buffer. This is because the concentration ratio term in Eq. (3.23) and Eq. (3.25) remains the same. The dilution does not change this ratio.
3.8.3 Properties of Buffer Solution
The \(pH\) of a buffer solution does not change appreciably
- by addition of small amount of either strong acid or strong base,
- on dilution, or
- when it is kept for long time.
The third property is the one most easily overlooked, and it is what makes buffers practical. A standard solution whose pH drifted overnight would be useless for calibrating instruments or storing biological samples; a buffer holds its value on the shelf.
3.8.4 Applications of Buffer Solution
Buffer solution finds extensive applications in a variety of fields. Some of its applications are given.
i. In biochemical system: \(pH\) of blood in our body is maintained at 7.36 - 7.42 due to (\(\mathrm{HCO_3^{-}} + \mathrm{H_2CO_3}\)) buffer. A mere change of 0.2 \(pH\) units can cause death. The saline solution used for intravenous injection must contain buffer system to maintain the proper \(pH\) of the blood.
ii. Agriculture: The soils get buffered due to presence of salts such as carbonate, bicarbonate, phosphates and organic acids. The choice of fertilizers depends upon \(pH\) of soil.
iii. Industry: Buffers play an important role in paper, dye, ink, paint and drug industries.
iv. Medicine: Penicillin preparations are stabilized by addition of sodium citrate as buffer. When citric acid is added to milk of magnesia (\(\mathrm{Mg(OH)_2}\)), magnesium citrate is formed, which is a buffer.
v. Analytical chemistry: In qualitative analysis, a \(pH\) of 8 to 10 is required for precipitation of cations IIIA group. It is maintained with the use of (\(\mathrm{NH_4OH} + \mathrm{NH_4Cl}\)) buffer.
That last application is one you will meet in the laboratory. Group IIIA cations are precipitated as their hydroxides, and this succeeds only within a narrow band of alkalinity — too little \(\mathrm{OH^-}\) and they do not precipitate, too much and cations of the next group come down as well and spoil the separation. The ammonium hydroxide–ammonium chloride buffer holds the solution in exactly that band.
3.9 Solubility Product
So far every equilibrium in this chapter has been between species dissolved in water. We now turn to an equilibrium of a different kind — one between a solid sitting at the bottom of a beaker and the ions it has released into the liquid above it.
3.9.1 Solubility Equilibria
Hereafter we confine our attention to sparingly soluble compounds that is, compounds those dissolve only slightly in water.
Suppose some powdered sparingly soluble salt such as AgCl is put into water and stirred vigorously. A very small amount of AgCl dissolves in water to form its saturated solution. Most of the salt remains undissolved. Thus, solid AgCl is in contact with its saturated solution.
AgCl is a strong electrolyte. Hence the quantity of AgCl that dissolves in water dissociates completely into its constituent ions, \(\mathrm{Ag^+}\) and \(\mathrm{Cl^-}\).
This point is easy to misread, so it is worth stating plainly. “Sparingly soluble” and “weak electrolyte” are not the same thing. Very little silver chloride dissolves — but the little that does dissolve breaks up completely into ions. The equilibrium here is not between molecules and ions; it is between the solid and its ions.
A dynamic equilibrium exists between undissolved solid AgCl and the dissolved ions, \(\mathrm{Ag^+}\) and \(\mathrm{Cl^-}\), in the saturated solution. This equilibrium, called solubility equilibrium, is represented as:
\[\mathrm{AgCl}\text{(s)} \rightleftharpoons \mathrm{Ag^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
The word dynamic is not decoration. At the surface of the solid, ions are continually leaving the crystal and entering the solution, while other ions are continually returning from the solution and rejoining the crystal, as shown in Fig. 3.9. Once the two rates are equal, the amount of solid stops changing — though the traffic in both directions never stops.
Diagram
Fig. 3.9 – Solubility equilibrium of AgCl: a beaker with a layer of undissolved white AgCl crystals at the bottom and a small number of Ag⁺ and Cl⁻ ions in the solution above; draw one arrow from the crystal surface into the solution labelled “dissolution” and one arrow from the solution back to the crystal labelled “precipitation”, with a note that the two rates are equal at equilibrium.
From Equilibrium Constant to Solubility Product
The expression for its equilibrium constant is:
\[K = \frac{[\mathrm{Ag^+}][\mathrm{Cl^-}]}{[\mathrm{AgCl}]} \qquad \text{...........(3.27)}\]
The concentration of undissolved solid AgCl is constant we may write
\[[\mathrm{AgCl}] = \text{constant} = K'\]
Substituting in Eq. (3.27) we write
\[K = \frac{[\mathrm{Ag^+}][\mathrm{Cl^-}]}{K'}\]
\[K \times K' = [\mathrm{Ag^+}][\mathrm{Cl^-}]\]
The product of \(K \times K'\) is another constant and is called solubility product, that is the product of concentrations of ions in a saturated solution. It is denoted by \(K_{sp}\).
\[K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}]\]
For the general salt solubility equilibrium
\[\mathrm{B_xA_y}\text{(s)} \rightleftharpoons x\,\mathrm{B^{y+}}\text{(aq)} + y\,\mathrm{A^{x-}}\text{(aq)}\]
The solubility product is
\[K_{sp} = [\mathrm{B^{y+}}]^x [\mathrm{A^{x-}}]^y \qquad \text{...........(3.28)}\]
Thus, in the saturated solution of sparingly soluble salt the product of equilibrium concentrations of the constituent ions raised to the power equal to their respective coefficients in the balanced equilibrium expression at a given temperature is called solubility product.
Two conditions in that definition are exam-critical and are the ones most often dropped. The concentrations must be the equilibrium concentrations, that is, those in a saturated solution; and the value holds only at a stated temperature, since \(K_{sp}\) is an equilibrium constant.
Writing \(K_{sp}\) Expressions
Consider following examples. In each case, write the balanced equilibrium first, then simply carry the coefficients up as powers.
\[\mathrm{BaSO_4}\text{(s)} \rightleftharpoons \mathrm{Ba^{2+}}\text{(aq)} + \mathrm{SO_4^{2-}}\text{(aq)}\] \[K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]\]
\[\mathrm{CaF_2}\text{(s)} \rightleftharpoons \mathrm{Ca^{2+}}\text{(aq)} + 2\,\mathrm{F^-}\text{(aq)}\] \[K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2\]
\[\mathrm{Bi_2S_3}\text{(s)} \rightleftharpoons 2\,\mathrm{Bi^{3+}}\text{(aq)} + 3\,\mathrm{S^{2-}}\text{(aq)}\] \[K_{sp} = [\mathrm{Bi^{3+}}]^2[\mathrm{S^{2-}}]^3\]
\[\mathrm{Ca_3(PO_4)_2}\text{(s)} \rightleftharpoons 3\,\mathrm{Ca^{2+}}\text{(aq)} + 2\,\mathrm{PO_4^{3-}}\text{(aq)}\] \[K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2\]
3.9.2 Relationship between Solubility and Solubility Product
The solubility of a compound is the amount in grams that dissolves per unit volume (which may be 100 mL or 1L of its saturated solution).
Molar solubility: The number of moles of a compound that dissolve to give one litre of saturated solution is called its molar solubility.
\[\text{molar solubility (mol/L)} = \frac{\text{solubility in g/L}}{\text{molar mass in g/mol}}\]
Solubility is measured in the laboratory in grams per litre; \(K_{sp}\) is expressed in terms of ion concentrations in moles per litre. Molar solubility is the bridge between the two, and dividing by the molar mass is how one crosses it.
Consider once again the solubility equilibrium for \(\mathrm{B_xA_y}\),
\[\mathrm{B_xA_y}\text{(s)} \rightleftharpoons x\,\mathrm{B^{y+}}\text{(aq)} + y\,\mathrm{A^{x-}}\text{(aq)}\]
The solubility product is given by Eq. (3.28):
\[K_{sp} = [\mathrm{B^{y+}}]^x [\mathrm{A^{x-}}]^y\]
If \(S\) is the molar solubility of the compound, the equilibrium concentrations of the ions in the saturated solution will be
\[[\mathrm{B^{y+}}] = xS\ \text{mol/L}\]
\[[\mathrm{A^{x-}}] = yS\ \text{mol/L}\]
These two lines follow directly from the balanced equation. Each formula unit that dissolves releases \(x\) ions of \(\mathrm{B^{y+}}\) and \(y\) ions of \(\mathrm{A^{x-}}\), so if \(S\) moles dissolve per litre, the ion concentrations are \(xS\) and \(yS\) respectively.
From Eq. (3.28)
\[K_{sp} = [xS]^x[yS]^y = x^x y^y S^{x+y} \qquad \text{...........(3.29)}\]
For example:
- For AgBr,
\[\mathrm{AgBr}\text{(s)} \rightleftharpoons \mathrm{Ag^+}\text{(aq)} + \mathrm{Br^-}\text{(aq)}\]
Here, \(x = 1\), \(y = 1\)
\[\therefore\ K_{sp} = S \times S = S^2\]
- For \(\mathrm{PbI_2}\),
\[\mathrm{PbI_2}\text{(s)} \rightleftharpoons \mathrm{Pb^{2+}}\text{(aq)} + 2\,\mathrm{I^-}\text{(aq)}\] \[x = 1,\ y = 2\]
Therefore, \(K_{sp} = (1)^1(2)^2 S^{1+2} = 4S^3\)
- \(\mathrm{Al(OH)_3}\),
\[\mathrm{Al(OH)_3}\text{(s)} \rightleftharpoons \mathrm{Al^{3+}}\text{(aq)} + 3\,\mathrm{OH^-}\text{(aq)}\] \[x = 1,\ y = 3\]
\[K_{sp} = (1)^1(3)^3 S^{1+3} = 27S^4\]
Notice how quickly the numerical factor grows — 1, then 4, then 27 — as the number of ions increases. This is why two salts with similar \(K_{sp}\) values can have quite different solubilities if they give different numbers of ions, and why \(K_{sp}\) values may be compared directly only between salts of the same type.
3.9.3 Condition of Precipitation
We can now answer a very practical question: when two solutions are mixed, will a precipitate appear or not?
Ionic product (\(IP\)) of an electrolyte is defined in the same way as solubility product (\(K_{sp}\)). The only difference is that the ionic product expression contains concentration of ions under any condition whereas expression of \(K_{sp}\) contains only equilibrium concentrations. If,
\(IP = K_{sp}\) ; the solution is saturated and solubility equilibrium exists.
\(IP > K_{sp}\) ; the solution is supersaturated and hence precipitation of the compound will occur.
If \(IP < K_{sp}\), the solution is unsaturated and precipitation will not occur.
The comparison makes intuitive sense. \(K_{sp}\) is the largest ion product the solution can carry and still remain clear. If mixing has pushed the ion product past that ceiling, the solution is holding more ions than it can, and the excess comes down as a solid until the product falls back to \(K_{sp}\). If the product is below the ceiling, the solution can hold still more, so nothing precipitates. The test is summarised in FC 3.5.
Flowchart
FC 3.5 – Will a precipitate form? Mix the two solutions → correct each concentration for the dilution caused by mixing (moles unchanged, total volume larger) → write the balanced solubility equilibrium of the possible salt → calculate IP with the corrected concentrations → compare with Ksp → branch (a) IP = Ksp → saturated, equilibrium exists, no further precipitation; branch (b) IP > Ksp → supersaturated → precipitate forms; branch (c) IP < Ksp → unsaturated → no precipitate.
The step that decides most precipitation problems is not the comparison itself but the one before it. When two solutions are mixed, each is diluted by the other: the number of moles of each ion is unchanged, but the total volume has increased, so every concentration must be recalculated before \(IP\) is worked out. Both of the worked problems below turn on exactly this point.
3.10 Common Ion Effect
Every equilibrium met so far in this chapter can be pushed about, and the simplest way to push one is to add more of something that is already taking part in it. When the added substance supplies an ion that the equilibrium itself produces, the effect is both predictable and extremely useful.
Suppressing the Ionization of a Weak Acid
Consider a solution of weak acid \(\mathrm{CH_3COOH}\) and its soluble ionic salt \(\mathrm{CH_3COONa}\).
\(\mathrm{CH_3COOH}\) is weak acid, dissociates only slightly in solution
\[\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-}\text{(aq)} + \mathrm{H^+}\text{(aq)}\]
\(\mathrm{CH_3COONa}\) being a strong electrolyte dissociates almost completely in solution.
\[\mathrm{CH_3COONa}\text{(aq)} \longrightarrow \mathrm{CH_3COO^-} + \mathrm{Na^+}\]
Both the acid and the salt produce \(\mathrm{CH_3COO^-}\) ions in solution. \(\mathrm{CH_3COONa}\) dissociates completely. Therefore it provides high concentration of \(\mathrm{CH_3COO^-}\) ions. The acetate ion is thus the common ion — the one species that both compounds put into the same solution, as shown in Fig. 3.10.
Diagram
Fig. 3.10 – Common ion effect in a CH₃COOH/CH₃COONa solution: on the left the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ drawn with a few ions; on the right the same solution after adding CH₃COONa, drawn crowded with CH₃COO⁻ ions from the salt, a bold arrow marked “equilibrium shifts left” and a note that the number of free H⁺ ions has fallen.
According to Le-Chatelier principle, the addition of \(\mathrm{CH_3COO^-}\) from \(\mathrm{CH_3COONa}\) to the solution of \(\mathrm{CH_3COOH}\) shifts equilibrium of dissociation of \(\mathrm{CH_3COOH}\) to left. Thus reverse reaction is favoured in which \(\mathrm{CH_3COO^-}\) combines with \(\mathrm{H^+}\) to form unionised \(\mathrm{CH_3COOH}\). Hence dissociation of \(\mathrm{CH_3COOH}\) is supressed due to presence of \(\mathrm{CH_3COONa}\) containing a common \(\mathrm{CH_3COO^-}\) ion.
The common ion effect states that the ionisation of a weak electrolyte is supressed in presence of a strong electrolyte containing an ion common to the weak electrolyte.
Common ion effect is a special case of Le-Chatelier’s principle in which the stress applied to an equilibrium system is an increase in the concentration of one of the product (ions). The effect of this stress is reduced by shifting the equilibrium to the reactant side.
That sentence is the whole of the theory in one line, and it is worth learning in exactly that form. Nothing new has been introduced in this section — only Le Chatelier’s principle, applied to an equilibrium in which the “product” whose concentration is raised happens to be an ion. The chain of reasoning is laid out in FC 3.6.
Flowchart
FC 3.6 – Common ion effect: weak electrolyte at equilibrium (e.g. CH₃COOH ⇌ CH₃COO⁻ + H⁺) → add a strong electrolyte with a common ion (CH₃COONa) → salt dissociates completely → concentration of the common ion (CH₃COO⁻) rises sharply → stress on the equilibrium → by Le Chatelier’s principle the equilibrium shifts to the reactant side → common ion combines with H⁺ to give unionized CH₃COOH → degree of dissociation of the weak acid falls, [H⁺] falls → but Ka is unchanged, since it is an equilibrium constant.
You have, in fact, already met the common ion effect at work. An acidic buffer is nothing other than a weak acid whose ionization has been suppressed by the common ion supplied by its salt — which is exactly why such a solution holds a large reserve of undissociated acid and of its conjugate base at the same time.
3.10.1 Common Ion Effect and Solubility
\[\mathrm{OH^-}\text{(aq)} + \mathrm{HCO_3^{-}}\text{(aq)} \longrightarrow \mathrm{CO_3^{2-}}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
Solubility product of \(\mathrm{CaCO_3}\) is very low (\(K_{sp} = 4.5 \times 10^{-9}\)). Addition of lime makes \(IP >> K_{sp}\) which results in the precipitation of \(\mathrm{CaCO_3}\) and thereby removal of hardness.
The presence of a common ion also affects the solubility of a sparingly soluble salt. Consider, the solubility equilibrium of AgCl,
\[\mathrm{AgCl}\text{(s)} \rightleftharpoons \mathrm{Ag^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\]
The solubility product of AgCl is
\[K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}]\]
Suppose \(\mathrm{AgNO_3}\) is added to the saturated solution of AgCl. The salt \(\mathrm{AgNO_3}\) being a strong electrolyte dissociates completely in the solution.
\[\mathrm{AgNO_3}\text{(aq)} \longrightarrow \mathrm{Ag^+}\text{(aq)} + \mathrm{NO_3^{-}}\text{(aq)}\]
The dissociation of AgCl and \(\mathrm{AgNO_3}\) produce a common \(\mathrm{Ag^+}\) ion. The concentration of \(\mathrm{Ag^+}\) ion in the solution increases owing to complete dissociation of \(\mathrm{AgNO_3}\). According to Le-chatelier’s principle the addition of \(\mathrm{Ag^+}\) ions from \(\mathrm{AgNO_3}\) to the solution of AgCl shifts the solubility equilibrium of AgCl from right to left. The reverse reaction in which AgCl precipitates, is favoured until the solubility equilibrium is re-established. This is shown in Fig. 3.11.
Diagram
Fig. 3.11 – Effect of a common ion on solubility: (a) a beaker of saturated AgCl solution with a little solid AgCl at the bottom and a few Ag⁺ and Cl⁻ ions in solution; (b) the same beaker after adding AgNO₃ solution, now crowded with extra Ag⁺ ions and with a visibly thicker layer of solid AgCl at the bottom; a bold arrow between them labelled “add AgNO₃ — equilibrium shifts left, more AgCl precipitates”.
The value of \(K_{sp}\) however, remains the same since it is an equilibrium constant. The solubility of a sparingly soluble compound, thus decreases with the presence of a common ion in solution.
This last pair of sentences contains a distinction that examiners return to again and again, so read them together. The solubility has fallen; the solubility product has not. \(K_{sp}\) is fixed at a given temperature, and it is precisely because it is fixed that the solubility must fall: if \([\mathrm{Ag^+}]\) is forced up, then \([\mathrm{Cl^-}]\) must come down so that their product still equals \(K_{sp}\), and that lowered chloride ion concentration is the reduced solubility.
3.11 Summary
Ionic equilibrium is the equilibrium established in a solution between the ions and the unionized molecules of a dissolved substance. The principles of chemical equilibrium studied earlier apply to it unchanged.
Four ionic equilibria are studied in this chapter: \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions with unionized water; the ionization of weak acids and weak bases; the reaction of salt ions with the ions of water; and a solid salt with its ions in water.
Electrolytes give ions in water and conduct electricity; non-electrolytes do not ionize and exist as molecules.
Strong electrolytes (strong acids, strong bases, salts) ionize completely or almost completely and are written with a single arrow. Weak electrolytes (weak acids, weak bases) dissociate only partially, set up an equilibrium, and are written with a double arrow.
Degree of dissociation, \(\alpha = \dfrac{\text{number of moles dissociated}}{\text{total number of moles}}\), and percent dissociation \(= \alpha \times 100\). It has no unit and lies between 0 and 1.
For an electrolyte of molar concentration \(c\), the equilibrium concentration of the cation or anion is \((\alpha \times c)\) mol dm\(^{-3}\).
Arrhenius: an acid contains hydrogen and gives \(\mathrm{H^+}\) in aqueous solution; a base contains OH and gives \(\mathrm{OH^-}\). The theory applies only to aqueous solutions and cannot explain the basicity of \(\mathrm{NH_3}\) and \(\mathrm{Na_2CO_3}\).
The bare \(\mathrm{H^+}\) ion hydrates in water to give the hydronium ion \(\mathrm{H_3O^+}\); \(\mathrm{H^+}\) is written for convenience.
Brønsted–Lowry: an acid is a proton donor, a base is a proton acceptor. Species differing by one proton form a conjugate acid–base pair, and every such reaction has two acids and two bases.
Lewis: an acid accepts a share in an electron pair, a base donates a share in an electron pair. This is the widest definition and covers species with no hydrogen at all, such as \(\mathrm{BF_3}\).
Water is amphoteric: it acts as an acid towards \(\mathrm{NH_3}\) and as a base towards HCl.
Strong acids (HCl, \(\mathrm{HNO_3}\), \(\mathrm{H_2SO_4}\), HBr, HI) and strong bases (NaOH, KOH) are almost completely dissociated. Weak acids (HCOOH, HF, \(\mathrm{H_2S}\), \(\mathrm{CH_3COOH}\)) and weak bases (\(\mathrm{Fe(OH)_3}\), \(\mathrm{Cu(OH)_2}\), \(\mathrm{NH_4OH}\)) are partially dissociated.
Dissociation constants: \(K_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\) and \(K_b = \dfrac{[\mathrm{B^+}][\mathrm{OH^-}]}{[\mathrm{BOH}]}\). A larger \(K_a\) means a stronger acid; both depend only on temperature.
Ostwald’s dilution law relates the two descriptions: \(K_a = \dfrac{\alpha^2}{(1-\alpha)V} = \dfrac{\alpha^2 c}{1-\alpha}\), and for a weak electrolyte \((1-\alpha) \cong 1\), giving \(K_a = \alpha^2 c\) and \(\alpha = \sqrt{K_a/c} = \sqrt{K_a . V}\).
The same treatment for a weak base gives \(K_b = \dfrac{\alpha^2 c}{(1-\alpha)}\) and \(\alpha = \sqrt{K_b/c} = \sqrt{K_b . V}\).
In both cases the degree of dissociation is inversely proportional to the square root of the concentration, and directly proportional to the square root of the volume containing 1 mol of electrolyte.
Water ionizes slightly: \(\mathrm{H_2O}\text{(l)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_3O^+}\text{(aq)} + \mathrm{OH^-}\text{(aq)}\).
Since \([\mathrm{H_2O}]\) is effectively constant, it is absorbed into the constant, giving the ionic product of water \(K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]\).
In pure water the two ion concentrations are equal; at 298 K each is \(1.0 \times 10^{-7}\) mol/L, so \(K_w = 1.0 \times 10^{-14}\). This relation holds for every dilute aqueous solution, so knowing one ion concentration gives the other.
\(pH = -\log_{10}[\mathrm{H^+}]\) and \(pOH = -\log_{10}[\mathrm{OH^-}]\). Both are dimensionless; one pH unit stands for a tenfold change in ion concentration, and pH falls as acidity rises.
Taking logarithms of \(K_w\) gives \(pH + pOH = 14\) at 298 K.
At 298 K: neutral solution — \([\mathrm{H_3O^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}\) M, \(pH = 7\); acidic solution — \([\mathrm{H_3O^+}] > 1 \times 10^{-7}\) M, \(pH < 7\); basic solution — \([\mathrm{H_3O^+}] < 1.0 \times 10^{-7}\) M, \(pH > 7\).
Hydrolysis of salt is the reaction in which the cation, the anion, or both ions of a salt react with the ions of water to produce acidity or alkalinity (or sometimes neutrality).
An ion hydrolyses only if the compound it would form is a weak electrolyte; if that compound would be a strong electrolyte, it dissociates again and no net change occurs.
Salt of strong acid + strong base (NaCl): no hydrolysis, solution neutral.
Salt of strong acid + weak base (\(\mathrm{CuSO_4}\), \(\mathrm{NH_4Cl}\)): the cation hydrolyses, excess \(\mathrm{H_3O^+}\), solution acidic. Hydrolysis of \(\mathrm{CuSO_4}\) makes its solution turbid with \(\mathrm{Cu(OH)_2}\); adding \(\mathrm{H_2SO_4}\) shifts the equilibrium back and clears it.
Salt of weak acid + strong base (\(\mathrm{CH_3COONa}\), \(\mathrm{Na_2CO_3}\)): the anion hydrolyses, excess \(\mathrm{OH^-}\), solution basic.
Salt of weak acid + weak base: both ions hydrolyse, and the outcome depends on the constants — \(K_a > K_b\) acidic (\(\mathrm{NH_4F}\)), \(K_a < K_b\) basic (\(\mathrm{NH_4CN}\)), \(K_a = K_b\) neutral (\(\mathrm{CH_3COONH_4}\)).
A buffer resists drastic changes in pH when a small amount of strong acid, strong base or water is added to it.
An acidic buffer is a weak acid with its salt of a strong base (\(\mathrm{CH_3COOH} + \mathrm{CH_3COONa}\)); a basic buffer is a weak base with its salt of a strong acid (\(\mathrm{NH_4OH} + \mathrm{NH_4Cl}\)).
Henderson–Hasselbalch equation: \(pH = pK_a + \log_{10}\dfrac{[\text{salt}]}{[\text{acid}]}\) with \(pK_a = -\log_{10}K_a\); and \(pOH = pK_b + \log_{10}\dfrac{[\text{salt}]}{[\text{base}]}\) with \(pK_b = -\log_{10}K_b\).
Buffer action: added \(\mathrm{H^+}\) is consumed by the large concentration of conjugate base, and added \(\mathrm{OH^-}\) is neutralised by the large concentration of weak acid, so free \([\mathrm{H^+}]\) and \([\mathrm{OH^-}]\) hardly change. Dilution leaves the salt-to-acid ratio unchanged and so leaves the pH unchanged.
Properties: the pH does not change appreciably on adding a small amount of strong acid or strong base, on dilution, or on long standing.
Applications: blood is held at pH 7.36–7.42 by the \(\mathrm{HCO_3^{-}} + \mathrm{H_2CO_3}\) buffer, and a change of 0.2 pH units can cause death; soils are buffered by carbonates, bicarbonates, phosphates and organic acids, so the choice of fertilizer depends on soil pH; buffers are used in the paper, dye, ink, paint and drug industries; penicillin preparations are stabilized with sodium citrate, and citric acid with milk of magnesia gives magnesium citrate, a buffer; qualitative analysis uses an \(\mathrm{NH_4OH} + \mathrm{NH_4Cl}\) buffer to hold pH 8–10 for precipitating group IIIA cations.
A sparingly soluble salt in contact with its saturated solution sets up a dynamic solubility equilibrium, e.g. \(\mathrm{AgCl}\text{(s)} \rightleftharpoons \mathrm{Ag^+}\text{(aq)} + \mathrm{Cl^-}\text{(aq)}\). The little that dissolves dissociates completely.
Since the concentration of the solid is constant, it is absorbed into the constant, giving the solubility product. For \(\mathrm{B_xA_y}\text{(s)} \rightleftharpoons x\,\mathrm{B^{y+}} + y\,\mathrm{A^{x-}}\), \(K_{sp} = [\mathrm{B^{y+}}]^x[\mathrm{A^{x-}}]^y\).
\(K_{sp}\) is the product of equilibrium ion concentrations in a saturated solution, each raised to its coefficient, at a given temperature.
Molar solubility is the number of moles dissolving to give one litre of saturated solution: \(\text{molar solubility} = \dfrac{\text{solubility in g/L}}{\text{molar mass in g/mol}}\).
In terms of molar solubility \(S\): \(K_{sp} = x^x y^y S^{x+y}\) — so \(K_{sp} = S^2\) for AgBr, \(4S^3\) for \(\mathrm{PbI_2}\) and \(27S^4\) for \(\mathrm{Al(OH)_3}\).
The ionic product \(IP\) has the same form but uses concentrations under any condition. If \(IP = K_{sp}\) the solution is saturated; if \(IP > K_{sp}\) it is supersaturated and precipitation occurs; if \(IP < K_{sp}\) it is unsaturated and no precipitate forms.
The ionisation of a weak electrolyte is suppressed in the presence of a strong electrolyte containing an ion common to it — adding \(\mathrm{CH_3COONa}\) to \(\mathrm{CH_3COOH}\) pushes the ionization equilibrium back towards undissociated acid.
It is a special case of Le Chatelier’s principle: the stress is an increase in the concentration of one of the product ions, and the equilibrium shifts to the reactant side to relieve it.
A common ion also lowers the solubility of a sparingly soluble salt. Adding \(\mathrm{AgNO_3}\) to saturated AgCl raises \([\mathrm{Ag^+}]\), shifts the solubility equilibrium to the left and precipitates more AgCl, while \(K_{sp}\) itself stays the same because it is an equilibrium constant.
Flowchart
FC 3.7 – The chapter at a glance: substance dissolved in water → non-electrolyte (no ions) or electrolyte → strong electrolyte (complete ionization, no equilibrium) or weak electrolyte (partial ionization, equilibrium set up) → branch A: weak acid/weak base → Ka, Kb and Ostwald’s dilution law α = √(K/c) → [H⁺] or [OH⁻] → with Kw = 1.0 × 10⁻¹⁴ → pH, pOH and pH + pOH = 14; branch B: salt dissolved → identify parent acid and base → hydrolysis → neutral / acidic / basic solution; branch C: weak acid or weak base + its salt → buffer → Henderson–Hasselbalch equation; branch D: sparingly soluble salt → solubility equilibrium → Ksp and molar solubility → compare IP with Ksp → precipitation or no precipitation; a common ion added to branch A or branch D → Le Chatelier’s principle → ionization suppressed / solubility lowered, K unchanged.
3.12 Objective Questions
- The \(pH\) of \(10^{-8}\) M of HCl is
- 8
- 7
- less than 7
- greater than 7
- Which of the following solution will have pH value equal to 1.0?
- 50 mL of 0.1M HCl + 50 mL of 0.1M NaOH
- 60 mL of 0.1M HCl + 40 mL of 0.1M NaOH
- 20 mL of 0.1M HCl + 80 mL of 0.1M NaOH
- 75 mL of 0.2M HCl + 25 mL of 0.2M NaOH
- Which of the following is a buffer solution?
- \(\mathrm{CH_3COONa}\) + NaCl in water
- \(\mathrm{CH_3COOH}\) + HCl in water
- \(\mathrm{CH_3COOH}\) + \(\mathrm{CH_3COONa}\) in water
- HCl + \(\mathrm{NH_4Cl}\) in water
- The solubility product of a sparingly soluble salt AX is \(5.2 \times 10^{-13}\). Its solubility in mol dm\(^{-3}\) is
- \(7.2 \times 10^{-7}\)
- \(1.35 \times 10^{-4}\)
- \(7.2 \times 10^{-8}\)
- \(13.5 \times 10^{-8}\)
- Blood in human body is highly buffered at \(pH\) of
- 7.4
- 7.0
- 6.9
- 8.1
- The conjugate base of \([\mathrm{Zn(H_2O)_4}]^{2+}\) is
- \([\mathrm{Zn(H_2O)_4}]^{2-}\mathrm{NH_3}\)
- \([\mathrm{Zn(H_2O)_3}]^{2-}\)
- \([\mathrm{Zn(H_2O)_3OH}]^{+}\)
- \([\mathrm{Zn(H_2O)H}]^{3+}\)
- For \(pH > 7\) the hydronium ion concentration would be
- \(10^{-7}\) M
- \(< 10^{-7}\) M
- \(> 10^{-7}\) M
- \(\geq 10^{-7}\) M
3.13 Check Your Concepts
A student writes the ionization of hydrochloric acid with a double arrow and that of acetic acid with a single arrow. Identify the error in each case and explain what the two kinds of arrow are meant to convey.
Silver chloride is described as a strong electrolyte although only a very small quantity of it dissolves in water. Explain why there is no contradiction between these two statements.
The concentration of water does not appear in the expression for \(K_a\) of a weak acid, nor does the concentration of the solid appear in the expression for \(K_{sp}\). Explain the reasoning that permits both omissions.
Ammonia contains no hydroxyl group, yet its aqueous solution turns red litmus blue. Explain this behaviour using the Brønsted–Lowry theory, and state why the Arrhenius theory cannot account for it.
Boron trifluoride contains no hydrogen at all, and therefore cannot donate a proton. On which theory is it still classed as an acid, and what structural feature of the molecule is responsible?
Write the conjugate acid and the conjugate base of the hydrogen carbonate ion, \(\mathrm{HCO_3^{-}}\), and explain what this pair of answers tells you about the ion.
Assertion: the conjugate base of a strong acid is a very weak base. Reason: a strong acid gives up its proton almost completely in aqueous solution. State whether each statement is correct, and whether the reason correctly explains the assertion.
When a solution of a weak acid is diluted, its degree of dissociation increases but its dissociation constant does not change. Explain how both statements can be true at the same time.
Two weak monobasic acids of the same molar concentration have dissociation constants \(1.8 \times 10^{-5}\) and \(4.0 \times 10^{-10}\). Without calculating, state which solution has the lower pH and justify your answer.
Arrange the following in order of increasing acid strength, giving your reason: HCN (\(K_a = 4.0 \times 10^{-10}\)), HF (\(K_a = 7.2 \times 10^{-4}\)) and \(\mathrm{CH_3COOH}\) (\(K_a = 1.8 \times 10^{-5}\)).
The statement “pure water has a pH of 7” is complete only when one further condition is added. State the missing condition and explain why the value of the pH of pure water depends on it.
Explain why a solution can never contain hydronium ions without also containing hydroxide ions, however strongly acidic it is.
Match each salt in column A with the nature of its aqueous solution in column B, and in each case name the ion, if any, that undergoes hydrolysis.
| Column A (salt) | Column B (nature of solution) |
|---|---|
| a. \(\mathrm{KNO_3}\) | i. acidic |
| b. \(\mathrm{CuCl_2}\) | ii. basic |
| c. \(\mathrm{CH_3COOK}\) | iii. neutral |
| d. \(\mathrm{NH_4CN}\) |
Both a solution of NaCl and a solution of \(\mathrm{CH_3COONH_4}\) are neutral, but for entirely different reasons. Distinguish clearly between the two cases.
A freshly made solution of copper sulphate is turbid. Write the hydrolytic equilibrium responsible, and explain in terms of Le Chatelier’s principle why adding dilute sulphuric acid clears it.
A mixture of hydrochloric acid and sodium chloride is not a buffer, although it contains a weak-acid salt of a strong base by one common description. Explain why it fails to buffer, and state what would have to be changed to make a working buffer.
A sodium acetate–acetic acid buffer is diluted to twice its volume. Predict what happens to its pH, and support your prediction using the Henderson–Hasselbalch equation.
A few drops of sodium hydroxide solution are added to an acetic acid–sodium acetate buffer. Write the equation for the reaction that takes place, and explain why the pH of the solution scarcely alters.
Under what condition does the pH of an acidic buffer become exactly equal to the \(pK_a\) of the weak acid? Explain how this is used when choosing a buffer system for a required pH.
Distinguish between ionic product and solubility product, and state the three conditions relating them that decide whether a precipitate will appear.
Assertion: the solubility of AgCl decreases when a little solid NaCl is added to its saturated solution. Reason: the value of \(K_{sp}\) of AgCl decreases in the presence of chloride ions. State whether each statement is correct, and whether the reason correctly explains the assertion.
In gravimetric analysis a precipitate is washed with a dilute solution containing one of its own ions rather than with pure water. Explain the chemical reasoning behind this practice.
Explain how the common ion effect is applied when group III A cations are precipitated as hydroxides using \(\mathrm{NH_4OH}\) in the presence of \(\mathrm{NH_4Cl}\). Name the common ion and state which equilibrium is suppressed.
Hard water containing calcium ions is softened by adding lime, which itself supplies calcium ions. Explain this apparent contradiction with the help of the relevant equations and the value of \(K_{sp}\) of calcium carbonate.
Why cations are Lewis acids?
Why is KCl solution neutral to litmus?
How are basic buffer solutions prepared?
Write one property of a buffer solution.
The dissociation of \(\mathrm{H_2S}\) is suppressed in the presence of HCl. Name the phenomenon.
Classify the following buffers into different types:
- \(\mathrm{CH_3COOH} + \mathrm{CH_3COONa}\)
- \(\mathrm{NH_4OH} + \mathrm{NH_4Cl}\)
- Sodium benzoate + benzoic acid
- \(\mathrm{Cu(OH)_2} + \mathrm{CuCl_2}\)
What are acids and bases according to Arrhenius theory?
What is meant by conjugate acid-base pair?
Label the conjugate acid-base pair in the following reactions
- \(\mathrm{HCl} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{Cl^-}\)
- \(\mathrm{CO_3^{2-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{OH^-} + \mathrm{HCO_3^{-}}\)
Write a reaction in which water acts as a base.
Ammonia serves as a Lewis base whereas \(\mathrm{AlCl_3}\) is Lewis acid. Explain.
Derive the relation \(pH + pOH = 14\).
Aqueous solution of sodium carbonate is alkaline whereas aqueous solution of ammonium chloride is acidic. Explain.
Define degree of dissociation. Derive Ostwald’s dilution law for the \(\mathrm{CH_3COOH}\).
Define \(pH\) and \(pOH\). Derive relationship between \(pH\) and \(pOH\).
What is meant by hydrolysis? A solution of \(\mathrm{CH_3COONH_4}\) is neutral. Why?
Dissociation of HCN is suppressed by the addition of HCl. Explain.
Derive the relationship between degree of dissociation and dissociation constant in weak electrolytes.
Sulfides of cation of group II are precipitated in acidic solution (\(\mathrm{H_2S}\) + HCl) whereas sulfides of cations of group IIIB are precipitated in ammoniacal solution of \(\mathrm{H_2S}\). Comment on the relative values of solubility product of sulfides of these.
Solubility of a sparingly soluble salt get affected in presence of a soluble salt having one common ion. Explain.
Take two test tubes and label them as A and B. Add zinc filings in both the test tubes. In the test tube labelled A add 5 mL of 1M HCl and in test B 5 mL of acetic acid. Keep the test tubes on the stand. Note down your observations.
- Do you see any effervescence coming from the two test tubes?
- Which gas is evolved?
- How do you identify the gas?
- What is the relative rate at which the gas is evolved in the two test tubes?
- Based on your observations comment on the strength of acids used.
3.14 Practice With Numericals
A weak monobasic acid is 0.6 % dissociated in its 0.05 M aqueous solution. Calculate its degree of dissociation, its dissociation constant, and the concentration of \(\mathrm{H^+}\) ions at equilibrium.
The dissociation constant of a weak monobasic acid is \(6.4 \times 10^{-7}\). Calculate its degree of dissociation and its percent dissociation in a 0.04 M solution.
A weak monoacidic base has \(K_b = 1.8 \times 10^{-5}\). Calculate its degree of dissociation in (i) 0.04 M solution and (ii) 0.16 M solution, and state how the two values illustrate Ostwald’s dilution law.
A weak monobasic acid is 8 % dissociated in its 0.02 M solution. Calculate its percent dissociation in 0.08 M solution at the same temperature.
Calculate the pH and the pOH of a 0.002 M solution of HCl at 298 K, assuming complete dissociation.
Calculate the pH and the pOH of a 0.004 M solution of KOH at 298 K, assuming complete dissociation.
At 298 K a solution has \([\mathrm{OH^-}] = 5.0 \times 10^{-5}\) mol dm\(^{-3}\). Calculate its hydronium ion concentration, its pH and its pOH, and state whether the solution is acidic or basic.
The pH of a soft drink is 3.4. Calculate its hydronium ion concentration and its hydroxide ion concentration at 298 K.
Calculate the pH of a 0.05 M solution of acetic acid at 298 K, given \(K_a = 1.8 \times 10^{-5}\).
A weak monoacidic base is 0.09 % dissociated in its 0.05 M solution. Calculate the concentration of \(\mathrm{OH^-}\) ions, the pOH and the pH of the solution.
Calculate the pH of a buffer solution containing 0.40 mol of \(\mathrm{CH_3COOH}\) and 0.20 mol of \(\mathrm{CH_3COONa}\) per litre, given \(K_a = 1.8 \times 10^{-5}\) for acetic acid.
Calculate the pH of a buffer solution containing 0.25 mol of \(\mathrm{NH_4OH}\) and 0.50 mol of \(\mathrm{NH_4Cl}\) per litre at 298 K, given \(K_b = 1.8 \times 10^{-5}\) for \(\mathrm{NH_4OH}\).
A buffer is to be prepared from formic acid (\(K_a = 1.8 \times 10^{-4}\)) and sodium formate. Calculate the ratio of the concentration of the salt to that of the acid required to give the buffer a pH of 4.00.
The molar solubility of silver chromate, \(\mathrm{Ag_2CrO_4}\), in water at 298 K is \(6.5 \times 10^{-5}\) mol dm\(^{-3}\). Calculate its solubility product.
The solubility product of calcium fluoride, \(\mathrm{CaF_2}\), at 298 K is \(3.2 \times 10^{-11}\). Calculate its molar solubility.
The solubility product of \(\mathrm{BaSO_4}\) at 298 K is \(1.1 \times 10^{-10}\). Calculate its solubility in mol dm\(^{-3}\) and in g dm\(^{-3}\). (Molar mass of \(\mathrm{BaSO_4}\) = 233 g mol\(^{-1}\))
Equal volumes of \(4.0 \times 10^{-3}\) M \(\mathrm{AgNO_3}\) and \(4.0 \times 10^{-3}\) M NaCl are mixed at 298 K. Given \(K_{sp}\) of AgCl \(= 1.8 \times 10^{-10}\), calculate the ionic product of the mixture and state whether a precipitate of AgCl will form.
25.0 cm\(^3\) of 0.010 M \(\mathrm{Ca(NO_3)_2}\) is mixed with 25.0 cm\(^3\) of 0.010 M NaF at 298 K. Given \(K_{sp}\) of \(\mathrm{CaF_2} = 3.2 \times 10^{-11}\), calculate the ionic product of \(\mathrm{CaF_2}\) in the mixture and state, with reasoning, whether it will precipitate.
The solubility product of AgCl at 298 K is \(1.8 \times 10^{-10}\). Calculate its molar solubility in 0.020 M \(\mathrm{AgNO_3}\) solution, and state how many times smaller it is than the solubility of AgCl in pure water.
Write the expression relating the solubility product to the molar solubility \(S\) for each of the following salts: (i) \(\mathrm{BaSO_4}\) (ii) \(\mathrm{PbI_2}\) (iii) \(\mathrm{Bi_2S_3}\) (iv) \(\mathrm{Ca_3(PO_4)_2}\).
Dissociation constant of acetic acid is \(1.8 \times 10^{-5}\). Calculate percent dissociation of acetic acid in 0.01 M solution.
The \(pH\) of a solution is 6.06. Calculate its \(\mathrm{H^+}\) ion concentration.
Calculate the \(pH\) of 0.01 M sulphuric acid.
Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid.
pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.
In NaOH solution \([\mathrm{OH^-}]\) is \(2.87 \times 10^{-4}\). Calculate the \(pH\) of solution.
The \(pH\) of rain water collected in a certain region of Maharashtra on particular day was 5.1. Calculate the \(\mathrm{H^+}\) ion concentration of the rain water and its percent dissociation.