Chapter 2: Solutions
In the 1880s a Dutch chemist named Jacobus Henricus van ’t Hoff was turning over a deceptively simple question: what exactly happens to a liquid when something dissolves in it?
Experimenters before him had measured the pressure that builds up when a sugar solution is separated from pure water by a thin membrane — a membrane that lets water through but holds the sugar back. Water pushes across into the solution, and the pressure climbs until the flow finally stops. Van ’t Hoff studied these measurements and noticed something remarkable. The pressure depended on how many sugar molecules were dissolved and on the temperature — but not at all on what the dissolved substance happened to be. Written out, the relation looked almost exactly like the equation for an ideal gas. Dissolved particles scattered through a solvent were behaving rather like gas particles scattered through empty space.
Then came the awkward part. For salts such as sodium chloride, the measured pressure came out roughly twice what the equation predicted. Instead of discarding the equation, van ’t Hoff introduced a correction factor, i. The reason soon became clear: one formula unit of sodium chloride does not stay as one particle in water — it separates into two ions.
In 1901 van ’t Hoff was awarded the first Nobel Prize in Chemistry. The set-up that started this line of work is sketched in Fig. 2.0.
Diagram
Fig. 2.0 – Two connected compartments separated by a semipermeable membrane: pure water on the left, sugar solution on the right, arrows showing net water movement into the solution and the solution level rising in a vertical tube; label the membrane, the rising column, and mark that sugar particles cannot cross.
Keep these questions in mind as you work through the chapter — each one is answered by the ideas developed in the sections ahead.
2.1 Introduction
You are already familiar with mixtures. A mixture is a combination of two or more substances that have been physically brought together without combining chemically. The air around you is a mixture of gases, a piece of rock is a mixture of two or more minerals, and so forth.
Because the substances in a mixture are not chemically joined, each component keeps its own chemical identity, and the components can be present in any proportion. This is what separates a mixture from a compound, in which elements are chemically combined in a fixed ratio.
Two Kinds of Mixtures
Recall that mixtures are of two kinds. In a homogeneous mixture the mixing of the components is uniform, so every portion of the mixture has the same composition and looks the same throughout.
In a heterogeneous mixture the mixing is nonuniform. Different portions have different compositions, and the separate parts can often be seen or settled out. The difference is easiest to appreciate at the particle level, as drawn in Fig. 2.11.
Diagram
Fig. 2.11 – Two magnified boxes side by side: (a) homogeneous mixture showing solute particles evenly spread among solvent particles, (b) heterogeneous mixture showing clumped particles and a visible boundary; label both, and mark “same composition everywhere” under (a).
We studied in Standard XI that homogeneous mixtures are classified according to the size of their constituent particles, as colloids or as true solutions. In this chapter we deal with the properties of homogeneous mixtures, and especially of true solutions.
The size of the dispersed particles is what draws the line between the two. In a true solution the dissolved particles are of molecular or ionic size, smaller than about 1 nm, so they never settle on standing and cannot be separated by ordinary filtration. In a colloid the dispersed particles are much larger — roughly 1 nm to 1000 nm — large enough to scatter a beam of light, though still too small to settle quickly. This classification scheme is set out in FC 2.1.
Flowchart
FC 2.1 – Classification of matter by mixing: matter → mixture → branch 1: homogeneous mixture → sub-branch: true solution (particles < 1 nm) and colloid (particles ~1–1000 nm); branch 2: heterogeneous mixture (e.g. suspension, particles > 1000 nm, settle on standing). Mark “this chapter” against true solution.
Where Solutions Show Up
Solutions are commonly found in all life processes. The body fluids that carry nutrients, ions and dissolved gases around a living organism are all solutions. Solutions are also important for industrial processes, and for many other areas of work.
There is a chemical reason why so much of chemistry happens in solution. When a substance dissolves, its particles are separated from one another and spread through the solvent, so they can meet the particles of another dissolved substance freely and react. In a heap of dry solid, only the particles at the surfaces can touch each other.
Solute, Solvent and Binary Solutions
The solution is a homogeneous mixture of two or more pure substances.
Within that mixture, the component that dissolves the other and is generally present in the larger amount is called the solvent; the component that gets dissolved, generally present in the smaller amount, is called the solute. In a sugar-and-water solution, water is the solvent and sugar is the solute.
A true solution consists of a solvent and one or more solutes. In this chapter we explore the properties of binary true solutions containing only one solute — that is, solutions made of exactly two components.
Keeping to binary solutions is not a limitation but a simplification. The relationships you are about to meet involve the amount of solute relative to the solvent, and with only two components those relationships stay clean enough to see the chemistry behind them.
2.2 Types of Solutions
We generally think of a solution as either a solid dissolved in a liquid, such as sugar in water, or as a mixture of two liquids. But there are other types of solutions as well, and some of them are less obvious than these two.
The reason is that the solute and the solvent may each be in any of the three states of matter — solid, liquid or gas. The solutions thus may involve any combination of these three states of their components. Since either component can be in any one of three states, there are three possibilities for the solute multiplied by three for the solvent. This gives rise to nine types of solutions depending on the states of solute and solvent. These are summarised in Table 2.1.
Table 2.1 : Types of solutions
| No. | State of solute | State of solvent | Some examples |
|---|---|---|---|
| 1 | Solid | Liquid | Sea water, benzoic acid in benzene, sugar in water |
| 2 | Solid | Solid | Metal alloys such as brass, bronze. |
| 3 | Solid | Gas | Iodine in air |
| 4 | Liquid | Liquid | Gasoline, ethanol in water. |
| 5 | Liquid | Solid | Amalgams of mercury with metals as mercury in silver |
| 6 | Liquid | Gas | Chloroform in nitrogen |
| 7 | Gas | Liquid | Carbonated water (\(\mathrm{CO_2}\) in water), oxygen in water. |
| 8 | Gas | Solid | \(\mathrm{H_2}\) in palladium |
| 9 | Gas | Gas | Air (\(\mathrm{O_2}\), \(\mathrm{N_2}\), Ar and other gases) |
Reading the table
Notice that the physical state of the solution as a whole is decided by the solvent, not by the solute. Brass is a solid because its solvent, copper, is a solid; soda water is a liquid because its solvent, water, is a liquid, even though the dissolved carbon dioxide is a gas.
Some entries repay a second look. Iodine in air is a solid dissolved in a gas, which is possible because iodine sublimes — it passes directly into the vapour state and mixes uniformly with the air. Hydrogen in palladium is a gas dissolved in a solid; the metal takes hydrogen into the spaces of its lattice. An amalgam is a solution of a metal in mercury, and mercury in silver is one such example.
The focus of this chapter
Our focus, in this chapter, will be on the solution of a solid in a liquid, with some attention to a solution of a gas in a liquid. The solvent in most of the cases will be water. These are the solutions you meet most often in the laboratory and in living systems, and they are also the ones whose properties can be measured and predicted most reliably.
Before going further, it is worth having those units clearly in mind, because every quantitative relationship in this chapter depends on saying exactly how much solute is present in a stated amount of solvent or solution.
2.3 Capacity of Solution to Dissolve Solute
Chemists need to know the capacity of solutions to dissolve a solute. A solvent will not accept an unlimited quantity of solute, and knowing where the limit lies decides how a solution is prepared, how strong it can be made, and whether solid will appear later.
Suppose that a solute is added to a solvent. It dissolves readily at first. The dissolution then slows down as more solute is added. If we continue the addition of solute, the dissolution stops. The solution at this point is said to be saturated.
What is happening when dissolution appears to stop
It would be wrong to picture the crystal at the bottom of a saturated solution as simply sitting there, doing nothing. A dynamic equilibrium can be reached where the number of solute molecules leaving the crystal to pass into solution is equal to the number returning from the solution to the crystal. The word dynamic is the important one: both processes continue, but at the same rate, so no net change is seen from outside. What is drawn in Fig. 2.12 is a balance of two opposing processes, not a halt.
Diagram
Fig. 2.12 – Magnified view of a crystal resting at the bottom of a beaker of saturated solution, with arrows leaving the crystal surface labelled “dissolution” and an equal number of arrows returning to it labelled “crystallization”; note alongside: equal rates, no net change.
The two opposing processes are written together as
\[\text{Solute} + \text{Solvent} \underset{\text{crystallization}}{\overset{\text{dissolution}}{\rightleftharpoons}} \text{Solution}\]
The forward change is dissolution, in which particles leave the solid and enter the solvent; the reverse change is crystallization, in which dissolved particles rejoin the solid. The double arrow records that both go on at once.
Saturated, unsaturated and supersaturated solutions
A saturated solution contains maximum amount of solute dissolved in a given amount of solvent at a given temperature. The temperature must be stated, because the same solvent will usually hold a different quantity of the same solute when it is warmer or cooler.
A solution holding less solute than this limit is described as unsaturated. Such a solution will go on dissolving more solute if some is added, which is exactly what you observe in the early stage of the addition described above.
A solution containing greater than the equilibrium amount of solute is said to be supersaturated solution. Such solutions are unstable. Holding more solute than the equilibrium amount, they are, in a sense, carrying a load they should not be able to carry, and they hold it only until something disturbs them. The precipitation occurs by the addition of a tiny crystal of solute and the supersaturated solution changes to saturated solution — the added crystal gives the dissolved particles a ready-made surface to settle on, and the excess solute comes out of solution at once.
2.4 Solubility
Once you accept that a solvent can hold only so much solute, the natural next question is: how much? That quantity has a name of its own. The solubility of a solute is its amount per unit volume of saturated solution at a specific temperature.
Because a saturated solution holds as much solute as it possibly can, the solubility of a solute is its maximum concentration, and it is expressed in the concentration units \(\text{mol L}^{-1}\). Quoting a solubility without stating the temperature is meaningless, since the same solvent generally holds a different quantity of the same solute when warmer or cooler.
2.4.1 Factors Affecting Solubility
The extent to which a substance dissolves in a solvent varies greatly with different substances. Sodium chloride pours generously into water while chalk barely dissolves at all; both are ionic solids, so the difference cannot be a matter of chemical class alone.
Three factors decide the outcome: it depends on the nature of solute and solvent, temperature and pressure. We take them in turn.
i. Nature of Solute and Solvent
Generally the compounds with similar chemical character are more readily soluble in each other than those with entirely different chemical characters. The saying that ‘like dissolves like’ guides to predict the solubility of a solute in a given solvent.
Thus, substances having similar intermolecular forces are likely to be soluble in each other. Generally polar solutes dissolve in polar solvents. This is because in these, solute-solute, solute-solvent and solvent-solvent interactions are all of similar magnitude.
Why should similar magnitude matter so much? For a solute to dissolve, the particles of the solute must be pulled apart from one another, and the solvent particles must also be pushed aside to make room. Both cost energy. That cost is paid back only if the new solute–solvent attractions that form are of comparable strength. When the three kinds of interaction are of similar magnitude, the exchange is a fair one and dissolution proceeds.
For example, \(\mathrm{NaCl}\) dissolves in water. The strong ion-dipole interactions of \(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\) ions with water molecules, hydrogen bonding between water molecules and ion-ion attractions between \(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\) ions are comparable. The polar water molecule turns its negative oxygen end towards the sodium ion and its positive hydrogen ends towards the chloride ion, and this surrounding of ions by solvent molecules is what draws the ions out of the crystal, as shown in Fig. 2.13.
Diagram
Fig. 2.13 – A corner of an NaCl crystal lattice at the left with water molecules pulling ions away: show a Na⁺ ion surrounded by water molecules with their O atoms pointing inward, and a Cl⁻ ion surrounded by water molecules with their H atoms pointing inward; label ion–dipole interaction and mark the partial charges δ+ and δ− on the water molecule.
At the other extreme, nonpolar organic compounds like cholesterol dissolves in nonpolar solvent such as benzene. Here the forces on both sides are weak dispersion forces, again well matched to one another. Water cannot dissolve cholesterol, because to make room for it, water would have to break its own strong hydrogen bonds and would receive only feeble attractions in return.
Sugar dissolves in water! At first sight this looks like an exception, since sugar is a covalent molecular solid rather than an ionic one. The dissolution of sugar in water is due to intermolecular hydrogen bonding between sugar and water. A sugar molecule carries several \(\mathrm{-OH}\) groups, and each of these can hydrogen bond to water exactly as water molecules hydrogen bond to one another.
ii. Effect of Temperature on Solubility
How solubility of substance changes with temperature depends on enthalpy of solution. Dissolution, like any other change, either absorbs heat from the surroundings or releases heat to them, and it is this that decides which way warming will push the solubility.
Many solids, for example \(\mathrm{KCl}\), dissolve in water by endothermic process, that is, with the absorption of heat. When temperature is increased by adding heat to the system, the solubility of substance increases according to the Le-Chatelier principle. Addition of heat causes a stress on the saturated solution. This stress favours endothermic process.
It helps to picture heat as though it were one of the reactants. For an endothermic dissolution we may write it as
\[\text{Solute} + \text{Solvent} + \text{heat} \rightleftharpoons \text{Solution}\]
Supplying more heat is then like supplying more of a reactant, and the equilibrium shifts forward — more solute dissolves.
On the other hand, when the substance dissolves in water by an exothermic process its solubility decreases with an increase of temperature. Here heat appears on the product side, so adding heat pushes the equilibrium backwards and solute comes out of solution. The substances such as \(\mathrm{CaCl_2}\) and \(\mathrm{Li_2SO_4 \cdot H_2O}\) dissolve in water releasing heat.
It is important to understand that there is no direct correlation between solubility and exothermicity or endothermicity. For example, dissolution of \(\mathrm{CaCl_2}\) in water is exothermic and that of \(\mathrm{NH_4NO_3}\) is endothermic. The solubility of these increases with the temperature. In other words, the argument above tells you which way the equilibrium is pushed, but the observed solubility of a real salt is the outcome of several competing energy terms, and it must finally be settled by experiment rather than by prediction.
Figure 2.1 shows the result of experimental determination of solubilities of some ionic solids in water at various temperatures.
Diagram
Fig. 2.1 – Graph of solubility (g solute per 100 g H₂O, axis 0 to 250) against temperature (°C, axis 0 to 100), showing curves labelled KNO₃, NaNO₃, NaBr, KBr, KCl and NaCl rising with temperature at different steepnesses, and a curve for Na₂SO₄ that falls with rising temperature.
Following are some experimental observations from Fig 2.1
Solubilities of \(\mathrm{NaBr}\), \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) change slightly with temperature.
Solubilities of \(\mathrm{KNO_3}\), \(\mathrm{NaNO_3}\) and \(\mathrm{KBr}\) increase appreciably with increasing temperature.
Solubility of \(\mathrm{Na_2SO_4}\) decreases with increase of temperature.
The solubility of gases in water usually decreases with increase of temperature. When gases are dissolved in water, the gas molecules in liquid phase are condensed. The condensation is an exothermic process. The solubility of gases in water must decrease as temperature is raised. This is why a glass of cold water left standing in a warm room slowly grows tiny bubbles on its inner wall — dissolved air is being driven out as the water warms.
iii. Effect of Pressure on Solubility
Pressure has no effect on the solubilities of solids and liquids as they are incompressible. Squeezing a liquid or a solid barely changes the spacing of its particles, so there is nothing for the pressure to act upon.
However pressure greatly affects solubility of gases in liquids. A gas above a solution is highly compressible, and raising the pressure crowds more gas molecules against the liquid surface, so more of them enter the liquid. The solubility of gases increases with increasing pressure. The quantitative relationship between gas solubility in a liquid and pressure is given by Henry’s law.
Thus,
\[S \propto P \qquad \text{or} \qquad S = K_{\mathrm{H}}P \qquad \ldots\ldots(2.1)\]
where, \(S\) is the solubility of the gas in \(\text{mol L}^{-1}\), \(P\) is the pressure of the gas in bar over the solution. \(K_{\mathrm{H}}\), the proportionality constant is called Henry’s law constant.
The unit of \(K_{\mathrm{H}}\) follows directly from rearranging Eq. (2.1).
\[\text{Units of } K_{\mathrm{H}}: \quad K_{\mathrm{H}} = \frac{S}{P} = \frac{\text{mol L}^{-1}}{\text{bar}} = \text{mol L}^{-1}\text{ bar}^{-1}\]
When \(P = 1\) bar, \(K_{\mathrm{H}} = S\). Thus, \(K_{\mathrm{H}}\) is the solubility of the gas in a liquid when its pressure over the solution is 1 bar. The constant therefore has a plain physical meaning, and it is different for every gas–solvent pair and changes with temperature.
Exceptions to Henry’s Law
Gases like \(\mathrm{NH_3}\) and \(\mathrm{CO_2}\) do not obey Henry’s law. The reason is that these gases react with water.
\[\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^{+}} + \mathrm{OH^{-}}\]
\[\mathrm{CO_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2CO_3}\]
Because of these reactions, \(\mathrm{NH_3}\) and \(\mathrm{CO_2}\) gases have higher solubilities than expected by Henry’s law. Henry’s law assumes that the dissolved gas remains as gas molecules in the liquid. When a gas is chemically converted into something else on dissolving, those converted molecules are removed from the dissolved-gas count, so still more gas enters the liquid to take their place, and the measured solubility runs above the value the law predicts.
\[\mathrm{Hb} + 4\,\mathrm{O_2} \rightleftharpoons \mathrm{Hb(O_2)_4}\]
This is the same effect at work in the body: because the oxygen is chemically bound rather than merely dissolved, blood carries far more oxygen than plain water ever could at the same pressure.
2.5 Vapour Pressure of Solutions of Liquids in Liquids
Consider a binary solution of two volatile liquids \(A_1\) and \(A_2\). Both are volatile, which means both evaporate readily at ordinary temperatures, so both will be found in the space above the liquid.
When the solution is placed in a closed container, both the liquids vaporize. Eventually an equilibrium is established between the vapour and liquid phases — molecules keep leaving the liquid surface and keep returning to it, and once the two rates match, the pressure above the liquid settles at a steady value.
Both the components are present in the vapour phase. The partial pressure of the components are related to their mole fractions in the solution. This relationship is given by Raoult’s law.
2.5.1 Raoult’s Law
Raoult’s law states that the partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.
Suppose that for a binary solution of two volatile liquids \(A_1\) and \(A_2\), \(P_1\) and \(P_2\) are their partial vapour pressures and \(x_1\) and \(x_2\) are their mole fractions in solution. Then according to Raoult’s law, we write
\[P_1 = x_1 P_1^{0} \quad \text{and} \quad P_2 = x_2 P_2^{0} \qquad \text{......(2.2)}\]
where \(P_1^{0}\) and \(P_2^{0}\) are vapour pressures of pure liquids \(A_1\) and \(A_2\), respectively. The superscript zero is the signal that the liquid is pure — undiluted by anything else.
Total pressure above the solution
According to Dalton’s law of partial pressures, the total pressure \(P\) above the solution is the sum of the partial pressures of the two components:
\[P = P_1 + P_2 = P_1^{0}x_1 + P_2^{0}x_2 \qquad \text{......(2.3)}\]
Since \(x_1 = 1 - x_2\), the Eq. (2.2) can also be written as
\[P = P_1^{0}(1 - x_2) + P_2^{0}x_2\]
\[= P_1^{0} - P_1^{0}x_2 + P_2^{0}x_2\]
\[= (P_2^{0} - P_1^{0})\,x_2 + P_1^{0} \qquad \text{......(2.4)}\]
It is worth pausing on why \(x_1\) was replaced by \(1 - x_2\). In a binary solution the two mole fractions must add up to one, so fixing one of them fixes the other. Writing everything in terms of a single variable \(x_2\) turns Eq. (2.3) into the equation of a straight line, \(y = mx + c\), with slope \((P_2^{0} - P_1^{0})\) and intercept \(P_1^{0}\).
Because \(P_1^{0}\) and \(P_2^{0}\) are constants, a plot of \(P\) versus \(x_2\) is a straight line as shown in Fig. 2.2. The figure also shows the plots of \(P_1\) versus \(x_1\) and \(P_2\) versus \(x_2\) according to the equations 2.2. These are straight lines passing through the origin, since a component that is absent contributes no vapour pressure at all.
Diagram
Fig. 2.2 – Variation of vapour pressure with mole fraction of solute: vapour pressure on the y-axis, mole fraction on the x-axis (x₂ increasing left to right, x₁ decreasing); mark P₁⁰ on the left axis and P₂⁰ on the right axis; draw the straight line labelled P_total = P₁ + P₂ joining them, plus the two dashed straight lines through the origin, P₁ falling and P₂ rising; label the three lines I, II and III as in the original.
Reading the three lines of Fig. 2.2 at their end points makes them easy to remember:
For the \(P\) versus \(x_2\) straight line, \(P = P_1^{0}\) at \(x_2 = 0\) and \(P = P_2^{0}\) at \(x_2 = 1\).
For the \(P_1\) versus \(x_1\) straight line, \(P_1 = 0\) at \(x_1 = 0\) and \(P_1 = P_1^{0}\) at \(x_1 = 1\).
For the \(P_2\) versus \(x_2\) straight line, \(P_2 = 0\) at \(x_2 = 0\) and \(P_2 = P_2^{0}\) at \(x_2 = 1\).
In each case the reasoning is the same: when a component’s mole fraction is zero it is not present and contributes nothing, and when its mole fraction is one the liquid is pure and it contributes its full vapour pressure.
Composition of vapour phase
Notice that the vapour above a solution need not have the same composition as the liquid below it. The more volatile component escapes more readily, so it is over-represented in the vapour.
The composition of vapour in equilibrium with the solution can be determined by Dalton’s law of partial pressures. If we take \(y_1\) and \(y_2\) as the mole fractions of the two components in the vapour, then
\[P_1 = y_1 P \quad \text{and} \quad P_2 = y_2 P\]
where \(P_1\) and \(P_2\) are the partial pressures of the two components in the vapour and \(P\) is the total vapour pressure. Each component’s share of the total pressure is simply its share of the vapour, which is what \(y_1\) and \(y_2\) record. Take care to keep the two symbols apart: \(x\) always refers to the liquid, \(y\) always to the vapour.
2.5.2 Ideal and Nonideal Solutions
Raoult’s law, as stated above, describes a limiting behaviour. Real mixtures follow it well in some cases and poorly in others, and solutions are classified accordingly.
1. Ideal solutions
Ideal solutions obey Raoult’s law over the entire range of concentrations — not merely when one component is dilute, but at every composition from pure \(A_1\) to pure \(A_2\).
No heat is evolved or absorbed when two components forming an ideal solution are mixed. Thus, the enthalpy of mixing is zero.
\[\Delta_{\text{mix}}H = 0\]
- There is no volume change when two components forming an ideal solution are mixed. Thus the volume of an ideal solution is equal to the sum of the volumes of the two components taken for mixing.
\[\Delta_{\text{mix}}V = 0\]
In an ideal solution solvent–solute, solute–solute and solvent–solvent molecular interactions are comparable.
The vapour pressure of an ideal solution always lies between the vapour pressures of the pure components, as shown in Fig. 2.2.
These five statements are not five separate facts but one fact seen from five sides. If a molecule of \(A_1\) feels the same attraction whether its neighbour is \(A_1\) or \(A_2\), then mixing changes nothing that the molecules can detect: no energy is taken in or given out, no contraction or expansion occurs, and each component escapes into the vapour exactly in proportion to its mole fraction — which is Raoult’s law.
It is important to understand that perfectly ideal solutions are uncommon, and solutions such as benzene + toluene behave nearly ideal. Benzene and toluene are close relatives in structure and size, so the forces between unlike molecules are much like those between like ones.
2. Nonideal solutions
These solutions do not obey Raoult’s law over the entire range of concentrations.
The vapour pressures of these solutions can be higher or lower than those of pure components.
Deviation from Raoult’s law: these solutions show two types of deviation from Raoult’s law. Which type appears depends on one question — are the solute–solvent attractions weaker or stronger than the attractions the molecules already had among themselves?
A. Positive deviations from Raoult’s law
The solutions in which solute–solvent intermolecular attractions are weaker than those between solute–solute molecules and solvent–solvent molecules exhibit positive deviations.
If the new attractions are weaker, molecules are held less firmly in the liquid than they were before mixing, so they escape more easily. The vapour pressures of such solutions are higher than those of pure components as shown in Fig. 2.3.
Diagram
Fig. 2.3 – Positive deviations from Raoult’s law: vapour pressure versus mole fraction, with the dashed straight lines showing ideal behaviour for P₁ and P₂ and for the total pressure, and the solid curves for the actual solution lying above the dashed lines; label the upper solid curve “vapour pressure of solution” and mark the axes x₁ = 0 to x₁ = 1 and x₂ = 1 to x₂ = 0.
The solutions of ethanol and acetone, and of carbon disulphide and acetone, show positive deviations from Raoult’s law. In pure ethanol the molecules are held together by hydrogen bonds; adding acetone molecules comes between them and disrupts that network, so ethanol escapes into the vapour more readily than Raoult’s law would predict.
B. Negative deviations from Raoult’s law
The solutions in which the interactions between solvent and solute molecules are stronger than solute–solute or solvent–solvent interactions exhibit negative deviations.
Here the reverse happens. Mixing creates stronger attractions than existed before, molecules are held more tightly in the liquid, and fewer of them escape. The vapour pressures of such solutions are lower than those of pure components as shown in Fig. 2.4.
Diagram
Fig. 2.4 – Negative deviations from Raoult’s law: vapour pressure versus mole fraction, with dashed straight lines for ideal behaviour and solid curves for the actual solution lying below them; label the lower solid curve “vapour pressure of solution” and mark the axes as in Fig. 2.3.
The solutions of phenol and aniline, and of chloroform and acetone, exhibit negative deviations from Raoult’s law. In the chloroform–acetone mixture a hydrogen bond forms between the hydrogen of chloroform and the oxygen of acetone — an attraction that neither pure liquid possessed.
2.6 Colligative Properties of Nonelectrolyte Solutions
Some properties of a solution depend on what the solute is; others depend only on how much of it is there. The second kind has a name of its own.
The physical properties of solutions that depend on the number of solute particles in solutions and not on their nature are called colligative properties. Whether the dissolved particles are sugar molecules or urea molecules makes no difference — only their number counts.
These are
- vapour pressure lowering
- boiling point elevation
- freezing point depression
- osmotic pressure
A nonelectrolyte is a substance that does not split into ions when it dissolves, so one dissolved molecule contributes exactly one particle to the solution. Sugar and urea behave this way; common salt does not, and that difference is taken up in Section 2.11.
While dealing with colligative properties of nonelectrolyte solutions, the relatively dilute solutions with concentrations 0.2 M or less are considered.
The four properties are not four unrelated facts. All of them follow from a single cause — the lowering of the solvent’s vapour pressure when a solute is present — as set out in FC 2.2, and the chapter now takes them one at a time.
Flowchart
FC 2.2 – Origin of the colligative properties: nonvolatile solute added to solvent → fewer solvent molecules at the surface → vapour pressure of solvent is lowered → branches into four measurable consequences: (1) vapour pressure lowering ΔP, (2) boiling point elevation ΔT_b, (3) freezing point depression ΔT_f, (4) osmotic pressure π. Mark against each branch the section in which it is treated: 2.7, 2.8, 2.9, 2.10.
2.7 Vapour Pressure Lowering
When a liquid in a closed container is in equilibrium with its vapours, the pressure exerted by the vapour on the liquid is its vapour pressure.
- Experiments have shown that when a nonvolatile, nonionizable solid is dissolved in a liquid solvent, the vapour pressure of the solution is lower than that of the pure solvent. In other words, the vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it. When the solute is nonvolatile it does not contribute to the vapour pressure above the solution. Therefore, the vapour pressure of the solution is equal to the vapour pressure of the solvent above the solution.
Two words in that statement carry weight. Nonvolatile means the solute has no measurable tendency to escape into the vapour, so it adds nothing to the pressure above the liquid. Nonionizable means it dissolves as whole molecules without splitting into ions, so the number of particles it contributes is exactly the number of molecules dissolved.
- If \(P_1^{0}\) is the vapour pressure of the pure solvent and \(P_1\) is the vapour pressure of the solvent above the solution, then
\[P_1 < P_1^{0}\]
The vapour pressure lowering is
\[\Delta P = P_1^{0} - P_1 \qquad \text{......(2.5)}\]
Why the vapour pressure falls
- Why is the vapour pressure of a solution containing a nonvolatile solute lower than that of the pure solvent? The vapour pressure of a liquid depends on the ease with which the molecules escape from the surface of the liquid.
When a nonvolatile solute is dissolved in a solvent, some of the surface molecules of the solvent are replaced by nonvolatile solute molecules. These solute molecules do not contribute to the vapour above the solution. The situation at the surface is drawn in Fig. 2.14.
Diagram
Fig. 2.14 – Two magnified liquid surfaces side by side: (a) pure solvent, with solvent molecules across the whole surface and many arrows showing molecules escaping into the vapour; (b) solution, with some surface positions occupied by larger nonvolatile solute particles and correspondingly fewer escape arrows; label “fewer solvent molecules free to vaporize per unit surface area”.
Thus, the number of solvent molecules available for vaporization per unit surface area of the solution is less than the number at the surface of the pure solvent. As a result, the solvent molecules at the surface of the solution vaporize at a slower rate than in the pure solvent. This results in the lowering of vapour pressure.
Notice that this explanation never asks what the solute is — only how much of the surface it occupies. That is precisely why the lowering of vapour pressure turns out to be a colligative property.
2.7.1 Raoult’s Law for Solutions of Nonvolatile Solutes
We saw in section 2.5.1 that Raoult’s law expresses the quantitative relationship between the vapour pressure of a solution and the vapour pressure of the solvent.
In solutions of nonvolatile solutes, the law is applicable only to the volatile solvent. There is no second term for the solute, because a nonvolatile solute contributes no vapour pressure of its own. The law states that the vapour pressure of the solvent over the solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution. Thus
\[P_1 = P_1^{0}x_1\]
A plot of \(P_1\) versus \(x_1\) is a straight line as shown in Fig. 2.5. The line passes through the origin, since a vanishing amount of solvent would give no vapour pressure, and it reaches \(P_1^{0}\) when \(x_1 = 1\), that is, when the solvent is pure.
Diagram
Fig. 2.5 – Variation of vapour pressure of solution with mole fraction of solvent: y-axis “vapour pressure of solvent”, x-axis “mole fraction of solvent” running from x = 0 to x = 1; a straight line through the origin rising to the right, with a few plotted experimental points lying on it.
For a binary solution containing one solute,
\[x_1 = 1 - x_2\]
It therefore follows that
\[P_1 = P_1^{0}x_1 = P_1^{0}(1 - x_2) = P_1^{0} - P_1^{0}x_2\]
or
\[P_1^{0} - P_1 = P_1^{0}x_2\]
The Eq. (2.5) defines \(P_1^{0} - P_1\) as \(\Delta P\), the lowering of vapour pressure. Hence
\[\Delta P = P_1^{0}x_2 \qquad \text{......(2.6)}\]
The Eq. (2.6) shows that \(\Delta P\) depends on \(x_2\), that is, on the number of solute particles. Thus \(\Delta P\), the lowering of vapour pressure, is a colligative property. The identity of the solute has disappeared from the equation altogether; only its mole fraction survives.
2.7.2 Relative Lowering of Vapour Pressure
The lowering \(\Delta P\) by itself still carries the stamp of the particular solvent used, because a solvent with a high vapour pressure will show a larger absolute drop than one with a low vapour pressure. Dividing by the vapour pressure of the pure solvent removes that dependence and leaves a pure number.
The ratio of the vapour pressure lowering of the solvent divided by the vapour pressure of the pure solvent is called the relative lowering of vapour pressure. Thus
\[\text{Relative vapour pressure lowering} = \frac{\Delta P}{P_1^{0}} = \frac{P_1^{0} - P_1}{P_1^{0}} \qquad \text{......(2.7)}\]
The Eq. (2.6) shows that
\[\frac{\Delta P}{P_1^{0}} = x_2 = \frac{P_1^{0} - P_1}{P_1^{0}} \qquad \text{......(2.8)}\]
Thus, the relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution. Therefore, relative vapour pressure lowering is a colligative property.
2.7.3 Molar Mass of Solute from Vapour Pressure Lowering
We studied that the relative lowering of vapour pressure is equal to the mole fraction \(x_2\) of the solute in the solution. From Eq. 2.6, it follows that:
\[\frac{\Delta P}{P_1^{0}} = x_2 \qquad \text{......(2.8)}\]
Recall (Chapter 10, sec 10.5.6) that the mole fraction of a component of a solution is equal to its moles divided by the total moles in the solution. Thus,
\[x_2 = \frac{n_2}{n_1 + n_2}\]
where \(n_1\) and \(n_2\) are the moles of solvent and solute respectively, in the solution.
We are concerned only with dilute solutions, hence \(n_1 \gg n_2\) and \(n_1 + n_2 \approx n_1\). The mole fraction \(x_2\) is then given by
\[x_2 = \frac{n_2}{n_1} \quad \text{and} \quad \frac{\Delta P}{P_1^{0}} = \frac{n_2}{n_1} \qquad \text{......(2.9)}\]
Suppose that we prepare a solution by dissolving \(\mathrm{W_2}\) g of a solute in \(\mathrm{W_1}\) g of solvent. The moles of solute and solvent in the solution are then,
\[n_2 = \frac{\mathrm{W_2}}{M_2} \quad \text{and} \quad n_1 = \frac{\mathrm{W_1}}{M_1} \qquad \text{......(2.10)}\]
where \(M_1\) and \(M_2\) are the molar masses of solvent and solute, respectively. Each of these is just the familiar statement that the number of moles equals mass divided by molar mass.
Substitution of Eq. (2.10) into Eq. (2.9) yields
\[x_2 = \frac{\Delta P}{P_1^{0}} = \frac{\mathrm{W_2}/M_2}{\mathrm{W_1}/M_1}\]
\[\frac{P_1^{0} - P_1}{P_1^{0}} = \frac{\Delta P}{P_1^{0}} = \frac{\mathrm{W_2}M_1}{M_2\mathrm{W_1}} \qquad \text{......(2.11)}\]
Knowing all other quantities, the molar mass of the solute \(M_2\) can be calculated. Everything on the left is measured with a pressure gauge, and everything on the right except \(M_2\) is measured with a balance — which is what makes this an experimental method rather than a piece of algebra.
2.8 Boiling Point Elevation
Recall that the boiling point of a liquid is the temperature at which its vapour pressure equals the applied pressure. For liquids in open containers the applied pressure is atmospheric pressure.
It has been found that the boiling point of a solvent is elevated by dissolving a nonvolatile solute into it. Thus, the solutions containing nonvolatile solutes boil at temperatures higher than the boiling point of a pure solvent.
If \(T_b^{0}\) is the boiling point of a pure solvent and \(T_b\) that of a solution, then
\[T_b > T_b^{0}\]
The difference between the boiling point of the solution and that of the pure solvent at any given constant pressure is called the boiling point elevation.
\[\Delta T_b = T_b - T_b^{0} \qquad \text{......(2.12)}\]
The words at any given constant pressure matter. Boiling point depends on the applied pressure quite apart from anything dissolved in the liquid, so the two boiling points being compared must be measured under the same pressure if the difference is to mean anything.
2.8.1 Boiling Point Elevation as a Consequence of Vapour Pressure Lowering
To understand the elevation of boiling point, let us compare the vapour pressures of solution and those of pure solvent. The vapour pressures of solution and of pure solvent are plotted as a function of temperature as shown in Fig. 2.6.
Diagram
Fig. 2.6 – Vapour pressure–temperature curves of pure solvent and solution: vapour pressure on the y-axis with a horizontal dashed line drawn at 760 mm, temperature on the x-axis; upper curve AB labelled “solvent” and lower curve CD labelled “solution” running parallel below it; mark the intersections of the two curves with the 760 mm line and label the corresponding temperatures on the x-axis as T_b⁰ (solvent) and T_b (solution), showing T_b to the right of T_b⁰.
As stated in section 2.7, at any temperature the vapour pressure of the solution is lower than that of the pure solvent. Hence, the vapour pressure–temperature curve of the solution (CD) lies below that of the solvent (AB). The difference between the two vapour pressures increases as temperature and vapour pressure increase, as predicted by the equation
\[\Delta P = x_2 P_1^{0}\]
The reason for that widening gap is contained in the equation itself. The mole fraction \(x_2\) of the solute is fixed by how the solution was made, but \(P_1^{0}\) rises steeply as the solvent is heated, so the product \(x_2P_1^{0}\) — the size of the gap — rises with it.
The intersection of the curve CD with the line corresponding to 760 mm is the boiling point of the solution. The similar intersection of the curve AB is the boiling point of the pure solvent. It is clear from the figure that the boiling point (\(T_b\)) of the solution is higher than that of the pure solvent (\(T_b^{0}\)).
The argument in words
At the boiling point of a liquid, its vapour pressure is equal to 1 atm. In order to reach the boiling point, the solution and the solvent must be heated to a temperature at which their respective vapour pressures attain 1 atm.
At any given temperature the vapour pressure of the solution is lower than that of the pure solvent. Hence, the vapour pressure of the solution needs a higher temperature to reach 1 atm than that needed for the vapour pressure of the solvent. In other words, the solution must be heated to a higher temperature to cause it to boil than the pure solvent.
Put briefly: the solute holds the solvent’s vapour pressure down, and the extra heating is what makes up the shortfall.
2.8.2 Boiling Point Elevation and Concentration of Solute
The boiling point elevation is directly proportional to the molality of the solution. Thus,
\[\Delta T_b \propto m \quad \text{or} \quad \Delta T_b = K_b m \qquad \text{......(2.13)}\]
where \(m\) is the molality of the solution. The proportionality constant \(K_b\) is called the boiling point elevation constant or molal elevation constant or ebullioscopic constant. All three names refer to the same quantity.
If \(m = 1\), then
\[\Delta T_b = K_b\]
Thus, the ebullioscopic constant is the boiling point elevation produced by a 1 molal solution.
The units of \(K_b\) follow from rearranging Eq. (2.13):
\[K_b = \frac{\Delta T_b}{m} = \frac{\text{K}}{\text{mol kg}^{-1}} = \text{K kg mol}^{-1}\]
Why molality and not molarity
We are dealing with the systems whose temperature is not constant. Therefore, we cannot express the concentration of the solution in molarity, which changes with temperature, whereas molality is temperature independent.
The reason is worth stating plainly. Molarity is moles of solute per litre of solution, and a volume expands when heated; the same solution therefore has a slightly lower molarity when hot than when cold. Molality is moles of solute per kilogram of solvent, and mass does not change on heating. Since the whole experiment consists of raising the temperature to boiling, a concentration unit that shifts as we heat would be useless.
Therefore the concentration of solute is expressed in mol/kg (molality) rather than mol/L (molarity).
2.8.3 Molar Mass of Solute from Boiling Point Elevation
The Eq. (2.13) is
\[\Delta T_b = K_b m\]
Suppose we prepare a solution by dissolving \(\mathrm{W_2}\) g of solute in \(\mathrm{W_1}\) g of solvent.
Moles of solute in \(\mathrm{W_1}\) g of solvent \(= \dfrac{\mathrm{W_2}}{M_2}\), where \(M_2\) is the molar mass of the solute.
Mass of solvent \(= \mathrm{W_1}\) g \(= \dfrac{\mathrm{W_1}\ \text{g}}{1000\ \text{g/kg}} = \dfrac{\mathrm{W_1}}{1000}\) kg. This division by 1000 is simply the conversion from grams to kilograms, which molality demands.
Recall the expression of molality, \(m\).
\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{\mathrm{W_2}/M_2\ \text{mol}}{\mathrm{W_1}/1000\ \text{kg}} = \frac{1000\,\mathrm{W_2}}{M_2\mathrm{W_1}}\ \text{mol kg}^{-1} \qquad \text{......(2.14)}\]
Substitution of this value of \(m\) in Eq. (2.13) gives
\[\Delta T_b = K_b\,\frac{1000\,\mathrm{W_2}}{M_2\mathrm{W_1}}\]
Hence,
\[M_2 = \frac{1000\,K_b\mathrm{W_2}}{\Delta T_b \mathrm{W_1}} \qquad \text{......(2.15)}\]
Every quantity on the right-hand side is measurable: the two masses on a balance, the two boiling points on a thermometer, and \(K_b\) from a table of solvent constants. This is the second route to a molar mass in this chapter, and unlike the vapour pressure method it needs no pressure measurement at all.
2.9 Depression in Freezing Point
Recall that the freezing point of a liquid is the temperature at which liquid and solid are in equilibrium and the two phases have the same vapour pressure.
The general experimental observation is that the freezing point of a solvent is lowered by dissolving a nonvolatile solute into it. Thus, the freezing point of a solution containing a nonvolatile solute is lower than that of the pure solvent.
If \(T_f^{0}\) is the freezing point of the pure solvent and \(T_f\) that of a solution in which a nonvolatile solute is dissolved, then
\[T_f^{0} > T_f\]
The difference between the freezing point of the pure solvent and that of the solution is the depression in freezing point. Thus
\[\Delta T_f = T_f^{0} - T_f \qquad \text{......(2.16)}\]
Compare this with Eq. (2.12) of the previous section. There the solution’s value was written first, because the solution boils higher; here the solvent’s value is written first, because the solution freezes lower. Both conventions are chosen so that the quantity comes out positive.
2.9.1 Freezing Point Depression as a Consequence of Vapour Pressure Lowering
The effect of dissolution of a nonvolatile solute into a solvent on the freezing point of the solvent can be understood in terms of the vapour pressure lowering — the same single cause that produced the elevation of boiling point.
Consider the vapour pressure–temperature diagram as shown in Fig. 2.7. The diagram consists of three curves. AB is the vapour pressure curve of the solid solvent, while CD is the vapour pressure curve of the pure liquid solvent. EF is the vapour pressure curve of the solution, that always lies below the pure solvent.
Diagram
Fig. 2.7 – Variation of vapour pressure with temperature of pure solvent, solid solvent and solution: vapour pressure on the y-axis, temperature on the x-axis; steep curve AB rising from the left labelled “solid solvent”; curve CD labelled “liquid solvent”; curve EF labelled “solution” lying below CD and parallel to it; mark point B where AB and CD intersect and point E where AB and EF intersect, with dotted verticals dropping from B to T⁰ and from E to T on the temperature axis, showing T to the left of T⁰.
It is important to note that the solute does not dissolve in the solid solvent. When a solution freezes, it is pure solid solvent that separates out, leaving the solute behind in the liquid. This is why the curve AB in the figure belongs to the pure solid and is not shifted at all by the presence of the solute — only the liquid curve is pushed down.
The curves AB and CD intersect at point B, where the solid and liquid phases of the pure solvent are in equilibrium. The two phases have the same vapour pressure at B. The temperature corresponding to B is the freezing point of the solvent (\(T_f^{0}\)).
Similarly at E, the point of intersection of EF and AB, the solid solvent and the solution are in equilibrium. They have the same vapour pressure at E. The temperature corresponding to E is the freezing point of the solution, \(T_f\).
It is clear from the figure that the freezing point of the solution \(T_f\) is lower than that of the pure solvent \(T_f^{0}\). It is obvious, because the vapour pressure curve of the solution lies below that of the solvent — a lower curve must meet the rising solid curve further to the left, that is, at a lower temperature.
Why the freezing point falls — the molecular picture
Why is the freezing point of a solvent lowered by dissolving a nonvolatile solute into it? At the freezing point of a pure liquid the attractive forces among molecules are large enough to cause the change of phase from liquid to solid.
In a solution, the solvent molecules are separated from each other because of the solute molecules that have come between them. Thus, the separation of solvent molecules in solution is more than that in the pure solvent. This results in decreasing the attractive forces between solvent molecules.
Consequently, the temperature of the solution is lowered below the freezing point of the solvent to cause the phase change. Since cooling slows the molecules down, extra cooling restores the ability of the weakened attractions to lock the molecules into a solid.
2.9.2 Freezing Point Depression and Concentration of Solute
As verified experimentally for a dilute solution, the freezing point depression (\(\Delta T_f\)) is directly proportional to the molality of the solution. Thus,
\[\Delta T_f \propto m \quad \text{or} \quad \Delta T_f = K_f m \qquad \text{......(2.17)}\]
The proportionality constant \(K_f\) is called the freezing point depression constant or cryoscopic constant.
If \(m = 1\), then \(\Delta T_f = K_f\). The cryoscopic constant thus is the depression in freezing point produced by 1 molal solution of a nonvolatile solute.
Unit of \(K_f\):
\[\frac{\Delta T_f}{m} = \frac{\text{K or }^\circ\text{C}}{\text{mol kg}^{-1}} = \text{K kg mol}^{-1} \ \text{ or } \ ^\circ\text{C kg mol}^{-1}\]
Molality appears here for the same reason it appeared in the previous section: the temperature is being changed during the experiment, and only a mass-based concentration stays fixed while that happens.
2.9.3 Molar Mass of Solute from Freezing Point Depression
Refer to Eq. (2.17),
\[\Delta T_f = K_f m\]
The molality \(m\) of the solution is given by Eq. (2.14) as
\[m = \frac{1000\,\mathrm{W_2}}{M_2\mathrm{W_1}}\]
Substitution of Eq. (2.14) into Eq. (2.17) gives
\[\Delta T_f = K_f\,\frac{1000\,\mathrm{W_2}}{M_2\mathrm{W_1}}\]
Hence,
\[M_2 = \frac{1000\,K_f\mathrm{W_2}}{\Delta T_f\mathrm{W_1}} \qquad \text{......(2.18)}\]
The derivation is the exact counterpart of the one in section 2.8.3, with \(K_f\) in place of \(K_b\) and \(\Delta T_f\) in place of \(\Delta T_b\), because the same expression for molality has been substituted into the same kind of proportionality. In practice the freezing point method is often preferred, since \(K_f\) values are usually larger than \(K_b\) values for the same solvent, giving a bigger temperature change to measure.
2.10 Osmotic Pressure
Besides the boiling point elevation and freezing point depression, the osmotic pressure is associated with vapour pressure lowering and can be used to determine molar masses of dissolved solutes. It is the fourth of the colligative properties listed in Section 2.6, and in several ways the most useful of them.
Semipermeable membrane
The osmotic pressure phenomenon involves the use of a semipermeable membrane.
It is a film such as cellophane which has pores large enough to allow the solvent molecules to pass through them. These pores are small enough not to allow the passage of large solute molecules or ions of high molecular mass through them. The semipermeable membrane selectively allows passage of solvent molecules.
The membrane behaves like a sieve of molecular fineness. Nothing chemical happens at it; the sorting is purely a matter of size.
2.10.1 Osmosis
When a solution and pure solvent, or two solutions of different concentrations, are separated by a semipermeable membrane, the solvent molecules pass through the membrane.
What is the direction of flow of solvent molecules? It is important to understand that the passage of solvent molecules through the semipermeable membrane takes place in both directions, since solvent is on both sides of the membrane.
However, the rate of passage of solvent molecules into the solution, or from the more dilute solution to the more concentrated solution, is found to be greater than the rate in the reverse direction. This is favourable since the vapour pressure of the solvent is greater than that of the solution.
That last sentence ties osmosis back to the common cause running through this whole chapter. The solvent side has the higher escaping tendency, so more molecules leave it per second than return to it, and the traffic is one-way on balance even though it runs both ways in detail.
The net spontaneous flow of solvent molecules into the solution, or from a more dilute solution to a more concentrated solution, through a semipermeable membrane is called osmosis. See Fig. 2.8.
Diagram
Fig. 2.8 – Osmosis, shown as two side-by-side cases: (a) a vessel divided by a semipermeable membrane with pure solvent on the left and solution on the right, arrow through the membrane labelled “net flow of solvent” pointing towards the solution; (b) the same vessel with solution of lower concentration on the left and solution of higher concentration on the right, arrow again pointing towards the more concentrated side; label the membrane in both.
As a result of osmosis, the amount of liquid on the pure solvent side, or on the more dilute solution side, decreases. Consequently, the amount of liquid on the other side increases. This results in a decrease of the concentration of the solution — the incoming solvent dilutes it.
2.10.2 Osmotic Pressure
Osmosis can be demonstrated with the experimental set-up shown in Fig. 2.9. A semipermeable membrane is firmly fastened across the mouth of a thistle tube. The solution of interest is placed inside an inverted thistle tube. This part of the tube and the membrane are then immersed in a container of pure water.
Diagram
Fig. 2.9 – Osmosis and osmotic pressure: an inverted thistle tube with a semipermeable membrane tied across its wide mouth, the bulb filled with sugar solution and dipping into a beaker of water; the solution has risen to a height h up the narrow stem; label the water, the sugar solution, the semipermeable membrane, and mark the risen column as hydrostatic pressure (hρg).
As a result, some of the solvent passes through the membrane into the solution. It causes the liquid level in the tube to rise. The liquid column in the tube creates hydrostatic pressure that pushes the solvent back through the membrane into the container.
The column of liquid in the tube continues to rise and eventually stops rising. At this stage the hydrostatic pressure developed is sufficient to force solvent molecules back through the membrane into the container at the same rate at which they enter the solution.
An equilibrium is thus established where the rates of forward and reverse passages are equal. The height of the liquid column in the tube remains constant. This implies that the hydrostatic pressure has stopped osmosis.
The hydrostatic pressure that stops osmosis is the osmotic pressure (\(\pi\)) of the solution. The hydrostatic pressure is equal to \(h\rho g\), where \(h\) is the height of the liquid column in the tube, \(\rho\) is the density of the solution and \(g\) is the acceleration due to gravity.
2.10.3 Isotonic, Hypertonic and Hypotonic Solutions
i. Isotonic solutions
Two or more solutions having the same osmotic pressure are said to be isotonic solutions.
For example, 0.1 M urea solution and 0.1 M sucrose solution are isotonic because their osmotic pressures are equal. Such solutions have the same molar concentrations but different concentrations in g/L.
The difference in g/L follows from the difference in molar mass: a mole of sucrose weighs far more than a mole of urea, so equal molar concentrations mean very unequal masses per litre. Osmotic pressure, being colligative, counts particles and ignores this entirely.
If these solutions are separated by a semipermeable membrane, there is no flow of solvent in either direction.
ii. Hypertonic and hypotonic solutions
If two solutions have unequal osmotic pressures, the more concentrated solution with the higher osmotic pressure is said to be a hypertonic solution.
The more dilute solution exhibiting lower osmotic pressure is said to be a hypotonic solution.
For example, if the osmotic pressure of a sucrose solution is higher than that of a urea solution, the sucrose solution is hypertonic to the urea solution, and the urea solution is hypotonic to the sucrose solution.
Note that these two words are always comparative. No solution is hypertonic in isolation; it is hypertonic to some other solution named alongside it.
2.10.4 Osmotic Pressure and Concentration of Solution
For very dilute solutions, the osmotic pressure follows the equation
\[\pi = \frac{n_2RT}{V} \qquad \text{......(2.19)}\]
where \(V\) is the volume of a solution in dm³ containing \(n_2\) moles of nonvolatile solute, \(R\) is the gas constant equal to \(0.08206\ \text{dm}^3\ \text{atm K}^{-1}\text{mol}^{-1}\), and \(\pi\) is the osmotic pressure in atm.
The resemblance to the ideal gas equation \(PV = nRT\) is not an accident, and it is worth noticing: the dissolved particles, spread through the solvent, exert an osmotic pressure in much the way that gas particles spread through a container exert a gas pressure.
The term \(n_2/V\) is the concentration in molarity (\(M\)). Eq. (2.19) thus can be written as
\[\pi = MRT \qquad \text{......(2.20)}\]
Note that the solute concentration is expressed in molarity while calculating osmotic pressure rather than molality. The reason is that osmotic pressure measurements are made at a specific constant temperature. It is not necessary to express concentration in a temperature independent unit like molality.
This is the reverse of the situation in Sections 2.8 and 2.9. There, the temperature was being changed during the very measurement, so molality was essential; here the temperature is held fixed throughout, and molarity — which is easier to prepare and to use — does no harm.
2.10.5 Molar Mass of Solute from Osmotic Pressure
Consider Eq. (2.19),
\[\pi = \frac{n_2RT}{V}\]
If the mass of solute in \(V\) litres of solution is \(\mathrm{W_2}\) and its molar mass is \(M_2\), then \(n_2 = \mathrm{W_2}/M_2\). With this value of \(n_2\), Eq. (2.19) becomes
\[\pi = \frac{\mathrm{W_2}RT}{M_2V} \quad \text{or} \quad M_2 = \frac{\mathrm{W_2}RT}{\pi V} \qquad \text{......(2.21)}\]
2.10.6 Reverse Osmosis
As mentioned earlier, osmosis is a flow of solvent through a semipermeable membrane into the solution. The direction of osmosis can be reversed by applying a pressure larger than the osmotic pressure.
Recall what osmotic pressure means: it is the pressure that just stops osmosis. Apply exactly that pressure and the flow halts; apply more than that, and the flow is driven backwards.
The pure solvent then flows from the solution into the pure solvent through the semipermeable membrane. This phenomenon is called reverse osmosis. Fig. 2.10 shows the schematic set-up for reverse osmosis. Fresh water and salty water are separated by a semipermeable membrane. When a pressure larger than the osmotic pressure of the solution is applied to the solution, pure water from the salty water passes into the fresh pure water through the membrane.
Diagram
Fig. 2.10 – Reverse osmosis: a rectangular chamber divided by a vertical semipermeable membrane, fresh water on the left and salt water on the right; a piston pressing down on the salt water marked “pressure > π”; arrow through the membrane pointing from the salt water into the fresh water; label the water outlet on the left.
Flowchart
FC 2.3 – Reverse osmosis water purifier: raw water → sediment and carbon pre-filters (remove suspended matter, chlorine and organics) → high-pressure pump raises pressure above π of the feed water → RO membrane → branch A: purified water (permeate) to storage tank → post-filter → tap; branch B: concentrated reject water carrying the removed salts → drain.
2.11 Colligative Properties of Electrolytes
Solutions of nonelectrolytes in water exhibit colligative properties as described in the preceding sections. These solutions also give, for the dissolved substances, the molar masses expected from their chemical formulae. Sucrose dissolved in water behaves exactly as Sections 2.7 to 2.10 predict, and the molar mass obtained from any of those measurements agrees with the value calculated from its formula.
The study of colligative properties of electrolytes, however, requires a different approach than that used for colligative properties of nonelectrolytes. Following are the experimental observations for the colligative behaviour of electrolytes.
The solutions of electrolytes also exhibit colligative properties, which do not obey the relations of nonelectrolytes.
The colligative properties of the solutions of electrolytes are greater than those to be expected for solutions of nonelectrolytes of the same concentration.
The molar masses of electrolytes in aqueous solutions determined by colligative properties are found to be considerably lower than the formula masses.
Why the discrepancy arises
Why are the colligative properties of electrolyte solutions greater than those for nonelectrolyte solutions of the same concentration? Recall that electrolytes dissociate into two or more ions when dissolved in water, whereas nonelectrolytes do not.
Stated differently, one formula unit of electrolyte dissolved in water produces two or more ions. Consequently the number of particles in solution increases. The colligative properties of electrolyte solutions are thus higher than those of nonelectrolyte solutions.
This also explains observation iii above. A colligative measurement counts particles; from the count it works out the number of moles, and from the mass dissolved it then reports a molar mass. If one formula unit has quietly become two particles, the count is doubled, so the calculated molar mass comes out at roughly half the true formula mass. Nothing is wrong with the measurement — it is answering the question it was asked.
For example, when NaCl is dissolved in water, it produces two ions, \(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\), whereas sucrose does not dissociate. It is expected then that the colligative property of 0.1 m NaCl is twice that of a 0.1 m sucrose solution.
2.11.1 van’t Hoff Factor (i)
To obtain the colligative properties of electrolyte solutions by using relations for nonelectrolytes, van’t Hoff suggested a factor \(i\). It is defined as the ratio of the colligative property of a solution of an electrolyte to the colligative property of a nonelectrolyte solution of the same concentration. Thus
\[i = \frac{\text{colligative property of electrolyte solution}}{\text{colligative property of nonelectrolyte solution of the same concentration}}\]
\[= \frac{(\Delta T_f)}{(\Delta T_f)_0} = \frac{(\Delta T_b)}{(\Delta T_b)_0} = \frac{(\Delta P)}{(\Delta P)_0} = \frac{(\pi)}{(\pi)_0} \qquad \text{......(2.22)}\]
where quantities without subscript refer to electrolytes and those with subscript to nonelectrolytes.
The van’t Hoff factor \(i\) is also defined in an alternative but exactly equivalent manner as
\[i = \frac{\text{actual moles of particles in solution after dissociation}}{\text{moles of formula units dissolved in solution}} \qquad \text{......(2.23)}\]
\[= \frac{\text{formula mass of substance}}{\text{observed molar mass of substance}}\]
\[= \frac{\mathrm{M_{theoretical}}}{\mathrm{M_{observed}}} \qquad \text{......(2.24)}\]
The three expressions say the same thing in three languages — the language of measured properties, the language of particle counting, and the language of molar masses. Whichever data a problem supplies, one of these forms will fit it.
Thus, \(i\) is equal to 1 for a nonelectrolyte, 2 for \(\mathrm{KNO_3}\) and NaCl, 3 for \(\mathrm{Na_2SO_4}\) and \(\mathrm{CaCl_2}\), and so forth. The colligative properties of these electrolytes are, therefore, twice and thrice respectively those of nonelectrolytes of the same concentration.
Counting the ions is worth doing carefully, since examination questions on this topic almost always turn on the count: \(\mathrm{CaCl_2}\) gives one \(\mathrm{Ca^{2+}}\) and two \(\mathrm{Cl^{-}}\), which is three particles, while \(\mathrm{Na_2SO_4}\) gives two \(\mathrm{Na^{+}}\) and one \(\mathrm{SO_4^{2-}}\), again three.
When the ideal count is not reached
The foregoing arguments are valid only for infinitely dilute solutions where the dissociation of electrolytes is complete.
In reality, especially at high concentrations, the colligative properties of strong electrolytes and their \(i\) values are usually smaller than expected. The reason is that the electrostatic forces between oppositely charged ions bring about the formation of ion pairs.
Each ion pair consists of one or more cations and one or more anions held together by electrostatic attractive forces. This results in a decrease in the number of particles in solution, causing a reduction in the expected \(i\) value and colligative properties. Two ions travelling together as a pair are counted by the solution as one particle, not two.
This is why the measured value of \(i\) for a salt such as \(\mathrm{CaCl_2}\) is often quoted as something like 2.47 rather than the ideal 3, as you will see in Solved Example 2.14.
2.11.2 Modification of Expressions of Colligative Properties
The expressions of colligative properties mentioned earlier for nonelectrolytes are to be modified so as to make them applicable to electrolyte solutions. Each modification is the same: multiply by \(i\), since \(i\) is precisely the factor by which the number of particles has been increased. The modified equations are
\[\Delta P = i\,P_1^{0}x_2 = i\,\frac{\mathrm{W_2}M_1}{M_2\mathrm{W_1}} \times P_1^{0}\]
\[\Delta T_b = iK_b m = i\,\frac{1000\,K_b\mathrm{W_2}}{M_2\mathrm{W_1}}\]
\[\Delta T_f = i\,K_f m = i\,\frac{1000\,K_f\mathrm{W_2}}{M_2\mathrm{W_1}}\]
\[\pi = i\,MRT = i\,\frac{\mathrm{W_2}RT}{M_2V}\]
Note that setting \(i = 1\) returns each equation exactly to the form met earlier in the chapter. The nonelectrolyte case has not been replaced; it has been absorbed as the special case in which no dissociation occurs.
2.11.3 van’t Hoff Factor and Degree of Dissociation
A discussion of colligative properties of electrolytes in the preceding sections is based on the fact that the electrolytes are completely dissociated in their aqueous solutions. This is approximately true for solutions of strong electrolytes and not for weak electrolytes that dissociate to a small extent.
The weak electrolytes involve the concept of degree of dissociation (\(\alpha\)), that changes the van’t Hoff factor. The degree of dissociation is simply the fraction of the dissolved formula units that have actually broken into ions — a number between 0, for no dissociation at all, and 1, for complete dissociation.
Relation between van’t Hoff factor and degree of dissociation
Consider an electrolyte \(\mathrm{A_xB_y}\) that dissociates in aqueous solution as
\[\mathrm{A_xB_y} \rightleftharpoons x\,\mathrm{A^{y+}} + y\,\mathrm{B^{x-}} \qquad \text{......(2.25)}\]
| \(\mathrm{A_xB_y}\) | \(x\,\mathrm{A^{y+}}\) | \(y\,\mathrm{B^{x-}}\) | |
|---|---|---|---|
| Initially | 1 mol | 0 | 0 |
| At equilibrium | \((1-\alpha)\) mol | \((x\alpha)\) mol | \((y\alpha)\) mol |
If \(\alpha\) is the degree of dissociation of the electrolyte, then the moles of cations are \(\alpha x\) and those of anions are \(\alpha y\) at equilibrium. We have dissolved just 1 mol of electrolyte initially. \(\alpha\) mol of electrolyte dissociates and \((1-\alpha)\) mol remains undissociated at equilibrium.
Total moles after dissociation
\[= (1-\alpha) + (x\alpha) + (y\alpha)\]
\[= 1 + \alpha(x + y - 1)\]
\[= 1 + \alpha(n-1) \qquad \text{......(2.26)}\]
where \(n = x + y\) = moles of ions obtained from dissociation of 1 mole of electrolyte.
The middle step deserves a second look. Collecting the \(\alpha\) terms gives \(-\alpha + x\alpha + y\alpha\), which is \(\alpha(x + y - 1)\); writing \(n\) for \(x + y\) turns this into \(\alpha(n-1)\).
The van’t Hoff factor given by Eq. (2.23) is
\[i = \frac{\text{actual moles of particles in solution after dissociation}}{\text{moles of formula units dissolved in solution}} = \frac{1 + \alpha(n-1)}{1}\]
Hence
\[i = 1 + \alpha(n-1) \quad \text{or} \quad \alpha = \frac{i-1}{n-1} \qquad \text{......(2.27)}\]
Test the result at its two extremes. If \(\alpha = 0\), nothing dissociates and \(i = 1\), as it should be for a nonelectrolyte. If \(\alpha = 1\), dissociation is complete and \(i = n\), the full number of ions. Every real weak electrolyte lies between these two.
2.12 Summary
Solutions and their types
- A solution is a homogeneous mixture of two or more pure substances. The component that dissolves is the solute; the one that does the dissolving is the solvent. A binary solution has just two components.
- Homogeneous mixtures are classified by particle size as true solutions or colloids. This chapter deals with true solutions.
- Since solute and solvent may each be solid, liquid or gas, there are nine types of solutions (Table 2.1). The chapter concentrates on solids in liquids and gases in liquids, with water as the usual solvent.
Saturation
- Dissolution and crystallization are opposite processes that reach a dynamic equilibrium: \[\text{Solute} + \text{Solvent} \underset{\text{crystallization}}{\overset{\text{dissolution}}{\rightleftharpoons}} \text{Solution}\]
- A saturated solution holds the maximum amount of solute at a given temperature. An unsaturated solution holds less; a supersaturated solution holds more, is unstable, and deposits the excess when a tiny crystal is added.
Solubility and what controls it
- Solubility is the amount of solute per unit volume of saturated solution at a specified temperature, expressed in \(\text{mol L}^{-1}\).
- Nature of solute and solvent: like dissolves like. Substances with similar intermolecular forces dissolve in each other, because solute–solute, solute–solvent and solvent–solvent interactions are then of comparable magnitude.
- Temperature: solids that dissolve endothermically (e.g. KCl) become more soluble on heating; those that dissolve exothermically (e.g. \(\mathrm{CaCl_2}\), \(\mathrm{Li_2SO_4 \cdot H_2O}\)) become less soluble. There is no direct correlation between solubility and the sign of the heat of solution. Gases in water generally become less soluble as temperature rises, because dissolution involves an exothermic condensation.
- Pressure: no effect on solids and liquids, which are incompressible; a large effect on gases in liquids, given by Henry’s law.
Henry’s law
- The solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution: \[S \propto P \quad \text{or} \quad S = K_{\mathrm{H}}P\] with \(S\) in \(\text{mol L}^{-1}\), \(P\) in bar and \(K_{\mathrm{H}}\) in \(\text{mol L}^{-1}\ \text{bar}^{-1}\). When \(P = 1\) bar, \(K_{\mathrm{H}} = S\).
- Gases that react with water, such as \(\mathrm{NH_3}\) and \(\mathrm{CO_2}\), are more soluble than the law predicts and do not obey it.
Raoult’s law and vapour pressure of liquid–liquid solutions
- For a volatile component, the partial vapour pressure equals the vapour pressure of the pure component times its mole fraction: \[P_1 = x_1P_1^{0}, \qquad P_2 = x_2P_2^{0}\]
- By Dalton’s law, the total pressure is \[P = P_1^{0}x_1 + P_2^{0}x_2 = (P_2^{0} - P_1^{0})x_2 + P_1^{0}\] so a plot of \(P\) against \(x_2\) is a straight line.
- The vapour composition follows from \(P_1 = y_1P\) and \(P_2 = y_2P\); the vapour is always richer in the more volatile component.
Ideal and nonideal solutions
- Ideal solutions obey Raoult’s law over the whole concentration range, with \(\Delta_{\text{mix}}H = 0\) and \(\Delta_{\text{mix}}V = 0\), and all molecular interactions comparable. Benzene + toluene is nearly ideal.
- Positive deviation: solute–solvent attractions weaker than the original ones, so vapour pressure is higher than predicted (ethanol + acetone; carbon disulphide + acetone).
- Negative deviation: solute–solvent attractions stronger, so vapour pressure is lower than predicted (phenol + aniline; chloroform + acetone).
Colligative properties
- Colligative properties depend only on the number of solute particles and not on their nature. They are vapour pressure lowering, boiling point elevation, freezing point depression and osmotic pressure, and are treated for dilute solutions of 0.2 M or less.
- All four arise from a single cause, as summarised in FC 2.4.
Flowchart
FC 2.4 – Chain of the colligative properties: nonvolatile solute added → fewer solvent molecules at the surface → vapour pressure of solvent lowered (ΔP = x₂P₁⁰) → three consequences: (a) solution’s vapour pressure curve reaches 1 atm only at a higher temperature → boiling point elevation ΔT_b = K_b m; (b) solution’s curve meets the unchanged solid-solvent curve at a lower temperature → freezing point depression ΔT_f = K_f m; (c) solvent’s escaping tendency higher than the solution’s across a semipermeable membrane → osmosis → osmotic pressure π = MRT. For electrolytes, multiply each result by the van’t Hoff factor i.
Vapour pressure lowering
- \(\Delta P = P_1^{0} - P_1 = P_1^{0}x_2\), so the lowering depends on the mole fraction of solute.
- The relative lowering equals the mole fraction of the solute: \[\frac{\Delta P}{P_1^{0}} = x_2 = \frac{P_1^{0} - P_1}{P_1^{0}}\]
- For a dilute solution this gives the molar mass: \[\frac{\Delta P}{P_1^{0}} = \frac{\mathrm{W_2}M_1}{M_2\mathrm{W_1}}\]
Boiling point elevation
- \(\Delta T_b = T_b - T_b^{0}\), and \(\Delta T_b = K_b m\), where \(K_b\) is the ebullioscopic (molal elevation) constant in \(\text{K kg mol}^{-1}\) — the elevation produced by a 1 molal solution.
- Molality is used, not molarity, because the temperature changes during the measurement and molality is temperature independent.
- \[M_2 = \frac{1000\,K_b\mathrm{W_2}}{\Delta T_b\mathrm{W_1}}\]
Freezing point depression
- \(\Delta T_f = T_f^{0} - T_f\), and \(\Delta T_f = K_f m\), where \(K_f\) is the cryoscopic (molal depression) constant — the depression produced by a 1 molal solution.
- The solute does not dissolve in the solid solvent, so the vapour pressure curve of the solid is unchanged and the intersection point shifts to a lower temperature.
- \[M_2 = \frac{1000\,K_f\mathrm{W_2}}{\Delta T_f\mathrm{W_1}}\]
- \(K_b\) and \(\Delta T_b\), and \(K_f\) and \(\Delta T_f\), are temperature differences, so their values are the same in K and in °C.
Osmosis and osmotic pressure
- A semipermeable membrane lets solvent molecules pass but holds back larger solute particles.
- Osmosis is the net spontaneous flow of solvent through such a membrane into the solution, or from a more dilute to a more concentrated solution.
- The hydrostatic pressure that just stops osmosis is the osmotic pressure \(\pi\), equal to \(h\rho g\) in the thistle-tube experiment. Osmotic pressure is not a pressure produced by the solution itself; it exists only across a semipermeable membrane.
- For very dilute solutions, \[\pi = \frac{n_2RT}{V} = MRT, \qquad M_2 = \frac{\mathrm{W_2}RT}{\pi V}\] with \(R = 0.08206\ \text{dm}^3\ \text{atm K}^{-1}\text{mol}^{-1}\) and \(\pi\) in atm. Molarity is used here because the temperature stays constant.
- Solutions of equal osmotic pressure are isotonic; of the two unequal ones, the more concentrated is hypertonic and the more dilute hypotonic.
- Applying to a solution a pressure greater than its osmotic pressure drives solvent out of it through the membrane — reverse osmosis.
- Osmotic pressure is large and precisely measurable, making it the best of the four for finding the molar masses of costly substances and of those available only in small quantities.
Colligative properties of electrolytes
- Electrolytes dissociate into ions, so one formula unit gives two or more particles. Their colligative properties are therefore greater, and the molar masses obtained from such measurements come out well below the formula masses.
- The van’t Hoff factor corrects for this: \[i = \frac{\text{colligative property of electrolyte solution}}{\text{colligative property of nonelectrolyte solution of same concentration}} = \frac{\text{actual moles of particles after dissociation}}{\text{moles of formula units dissolved}} = \frac{\mathrm{M_{theoretical}}}{\mathrm{M_{observed}}}\]
- \(i\) = 1 for a nonelectrolyte, 2 for NaCl and \(\mathrm{KNO_3}\), 3 for \(\mathrm{Na_2SO_4}\) and \(\mathrm{CaCl_2}\).
- The four expressions become \(\Delta P = i\,P_1^{0}x_2\), \(\Delta T_b = iK_bm\), \(\Delta T_f = iK_fm\) and \(\pi = iMRT\).
- At higher concentrations, oppositely charged ions form ion pairs, reducing the particle count and hence the measured \(i\) below its ideal value.
- For an electrolyte giving \(n\) ions with degree of dissociation \(\alpha\), the total moles after dissociation are \(1 + \alpha(n-1)\), so \[i = 1 + \alpha(n-1) \quad \text{or} \quad \alpha = \frac{i-1}{n-1}\]
2.13 Objective Questions
- The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
- 24 mm Hg
- 32 mm Hg
- 48 mm Hg
- 12 mm Hg
- The colligative property of a solution is
- vapour pressure
- boiling point
- osmotic pressure
- freezing point
- In calculating osmotic pressure the concentration of solute is expressed in
- molarity
- molality
- mole fraction
- mass percent
- Ebullioscopic constant is the boiling point elevation when the concentration of solution is
- 1 m
- 1 M
- 1 mass %
- 1 mole fraction of solute
- Cryoscopic constant depends on
- nature of solvent
- nature of solute
- nature of solution
- number of solvent molecules
- Identify the correct statement
- vapour pressure of solution is higher than that of pure solvent
- boiling point of solvent is lower than that of solution
- osmotic pressure of solution is lower than that of solvent
- osmosis is a colligative property
- A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature?
- 5.08 atm
- 2.54 atm
- 4.92 atm
- 2.46 atm
- The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)
- 5.41 %
- 3.54 %
- 4.53 %
- 53.4 %
- Vapour pressure of a solution is
- directly proportional to the mole fraction of the solute
- inversely proportional to the mole fraction of the solute
- inversely proportional to the mole fraction of the solvent
- directly proportional to the mole fraction of the solvent
- Pressure cooker reduces cooking time for food because
- boiling point of water involved in cooking is increased
- heat is more evenly distributed in the cooking space
- the higher pressure inside the cooker crushes the food material
- cooking involves chemical changes helped by a rise in temperature
- Henry’s law constant for a gas \(\mathrm{CH_3Br}\) is 0.159 mol dm\(^{-3}\) atm at 25 °C. What is the solubility of \(\mathrm{CH_3Br}\) in water at 25 °C and a partial pressure of 0.164 atm?
- 0.0159 mol L\(^{-1}\)
- 0.164 mol L\(^{-1}\)
- 0.026 M
- 0.042 M
- Which of the following statements is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution?
- osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
- urea solution is hypertonic to sucrose solution
- they are isotonic solutions
- sucrose solution is hypotonic to urea solution
2.14 Check Your Concepts
Explain why a saturated solution in contact with undissolved solid is described as being at dynamic equilibrium, even though nothing appears to be happening.
Assertion: A supersaturated solution deposits crystals the moment a tiny seed crystal is dropped into it. Reason: A supersaturated solution holds more solute than the equilibrium amount and is therefore unstable. State whether the assertion and the reason are correct, and whether the reason correctly explains the assertion.
Naphthalene dissolves readily in benzene but hardly at all in water, while common salt behaves in exactly the opposite way. Account for both observations in terms of the intermolecular forces involved.
The dissolution of \(\mathrm{CaCl_2}\) in water is exothermic and that of \(\mathrm{NH_4NO_3}\) is endothermic, yet the solubility of both increases with temperature. What does this tell you about the reliability of predicting solubility trends from the sign of the enthalpy of solution alone?
Ammonia and carbon dioxide are far more soluble in water than Henry’s law predicts. Explain why, writing the relevant equations, and state the condition a gas must satisfy for Henry’s law to apply.
A sealed bottle of a carbonated drink shows no bubbles until it is opened, after which it fizzes vigorously. Explain this using Henry’s law, and state why the dissolved sugar shows no comparable behaviour.
For each pair, predict whether the solution formed will show a positive deviation, a negative deviation, or nearly ideal behaviour from Raoult’s law, giving the reason in each case: (a) chloroform and acetone, (b) ethanol and acetone, (c) benzene and toluene.
For an ideal solution, \(\Delta_{\text{mix}}H = 0\) and \(\Delta_{\text{mix}}V = 0\). Explain how both of these follow from the single statement that solute–solute, solvent–solvent and solute–solvent interactions are of comparable magnitude.
The vapour above a benzene–toluene solution is richer in benzene than the liquid from which it came. Explain why, and state how this fact is put to use in fractional distillation.
Boiling point elevation is written as \(\Delta T_b = T_b - T_b^{0}\), but freezing point depression is written as \(\Delta T_f = T_f^{0} - T_f\). Why are the two differences written in opposite orders?
In the vapour pressure–temperature diagram used to explain freezing point depression, the curve for the solid solvent is drawn unchanged while the curve for the liquid is lowered. What physical fact justifies leaving the solid’s curve untouched, and how does this lead to a lower freezing point?
A student claims that dissolving a solute always makes a liquid harder to boil and easier to freeze. Identify what is right and what is wrong in this statement, and correct it.
Explain why osmotic pressure is not called “the pressure produced by a solution”, and state the condition under which it exists at all.
A red blood cell is placed successively in (a) distilled water, (b) a solution isotonic with its contents, and (c) a strongly hypertonic salt solution. Predict what happens to the cell in each case and explain the direction of water movement.
Of the four colligative properties, osmotic pressure is preferred for finding the molar masses of proteins. Give two reasons.
Assertion: The freezing point depression of a 0.1 m \(\mathrm{Na_2SO_4}\) solution is about three times that of a 0.1 m glucose solution. Reason: Freezing point depression depends on the number of solute particles present and not on their chemical nature. State whether the assertion and the reason are correct, and whether the reason correctly explains the assertion.
The molar masses of electrolytes obtained from colligative property measurements come out considerably lower than their formula masses. Explain why, and state what useful chemical information this apparent error actually conveys.
The measured van’t Hoff factor of \(\mathrm{CaCl_2}\) in a fairly concentrated solution is found to be about 2.5 rather than 3. Give the reason, and state the conditions under which the value would approach 3.
Arrange the following aqueous solutions in order of increasing boiling point, assuming complete dissociation where it applies, and justify the order you choose: 0.1 m glucose, 0.1 m KCl, 0.1 m \(\mathrm{Na_2SO_4}\), 0.05 m \(\mathrm{Al_2(SO_4)_3}\).
Match each item in Column I with the correct item in Column II.
| Column I | Column II |
|---|---|
| a. \(K_b\) | i. equals the mole fraction of the solute |
| b. \(K_f\) | ii. boiling point elevation for a 1 molal solution |
| c. \(K_{\mathrm{H}}\) | iii. equals 1 for a nonelectrolyte |
| d. Relative lowering of vapour pressure | iv. freezing point depression for a 1 molal solution |
| e. van’t Hoff factor \(i\) | v. solubility of a gas at a pressure of 1 bar |
Sea water freezes at a temperature well below 0 °C, while a freshwater lake at the same place freezes at 0 °C. Explain the difference, and state why the dissolved substances in sea water have a larger effect than an equal molar concentration of sugar would have.
What happens to the value of \(i\) for a weak electrolyte as its solution is progressively diluted? Support your answer using the relation between \(i\) and the degree of dissociation.
What is osmotic pressure?
A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why?
Write the equation relating boiling point elevation to the concentration of solution.
What is van’t Hoff factor?
How is van’t Hoff factor related to degree of ionization?
Which of the following solutions will have higher freezing point depression and why?
- 0.1 m NaCl
- 0.05 m \(\mathrm{Al_2(SO_4)_3}\)
State Raoult’s law for a solution containing a nonvolatile solute.
What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm³ of water? Why?
Which of the four colligative properties is most often used for molecular mass determination? Why?
How is vapour pressure lowering related to a rise in the boiling point of a solution?
What are isotonic and hypertonic solutions?
A solvent and its solution containing a nonvolatile solute are separated by a semipermeable membrane. Does the flow of solvent occur in both directions? Comment giving reason.
Explain reverse osmosis.
How is the molar mass of a solute determined by osmotic pressure measurement?
Why is the vapour pressure of a solvent lowered by dissolving a nonvolatile solute into it?
Using Raoult’s law, how will you show that \(\Delta P = P_1^{0}x_2\), where \(x_2\) is the mole fraction of solute in the solution and \(P_1^{0}\) the vapour pressure of pure solvent?
While considering boiling point elevation and freezing point depression, a solution concentration is expressed in molality and not in molarity. Why?
Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor.
What is the effect of temperature on solubility of solids in water? Give examples.
Obtain the relationship between freezing point depression of a solution containing a nonvolatile nonelectrolyte and its molar mass.
Explain with diagram the boiling point elevation in terms of vapour pressure lowering.
2.15 Practice With Numericals
The Henry’s law constant of a gas in water at 298 K is \(4.5 \times 10^{-3}\ \text{mol L}^{-1}\ \text{bar}^{-1}\). Calculate the solubility of the gas in water at 298 K when its partial pressure above the solution is 2.5 bar.
The Henry’s law constant for a gas is \(0.25\ \text{mol L}^{-1}\ \text{bar}^{-1}\) at 298 K. Calculate its solubility in water at 298 K when the pressure of the gas above the solution is 380 mm Hg. (1 atm = 760 mm Hg = 1.013 bar)
At 350 K, the vapour pressures of two pure volatile liquids A and B are 300 mm Hg and 500 mm Hg respectively. A solution is prepared in which the mole fraction of B is 0.40. Calculate (a) the partial vapour pressure of each liquid, (b) the total vapour pressure of the solution, and (c) the mole fraction of B in the vapour above the solution.
The vapour pressure of pure water at 298 K is 23.8 mm Hg. Calculate the vapour pressure of a solution prepared by dissolving 5.4 g of glucose (molar mass 180 g mol\(^{-1}\)) in 90 g of water.
When 10 g of a nonvolatile, nonionizable solute is dissolved in 200 g of water at 293 K, the vapour pressure of water falls from 17.5 mm Hg to 17.2 mm Hg. Calculate (a) the relative lowering of vapour pressure and (b) the molar mass of the solute.
A solution of 5.0 g of a nonvolatile nonelectrolyte in 100 g of water boils at 100.15 °C. Taking \(K_b\) for water as 0.52 K kg mol\(^{-1}\) and the boiling point of pure water as 100.00 °C, calculate the molar mass of the solute.
45 g of ethylene glycol (molar mass 62 g mol\(^{-1}\)) is dissolved in 600 g of water. Taking \(K_f\) for water as 1.86 K kg mol\(^{-1}\) and the freezing point of pure water as 273.15 K, calculate the molality of the solution and the freezing point of the mixture.
3.0 g of a nonvolatile nonelectrolyte dissolved in 150 g of benzene lowers its freezing point by 0.60 K. Calculate the molar mass of the solute, given that \(K_f\) for benzene is 5.12 K kg mol\(^{-1}\).
Calculate the osmotic pressure at 300 K of an aqueous solution containing 36 g of glucose (molar mass 180 g mol\(^{-1}\)) per litre of solution. Take \(R = 0.08206\ \text{dm}^3\ \text{atm K}^{-1}\text{mol}^{-1}\).
A solution prepared by dissolving 1.26 g of a protein in water to make 100 mL of solution has an osmotic pressure of \(2.50 \times 10^{-3}\) atm at 300 K. Calculate the molar mass of the protein. Take \(R = 0.08206\ \text{dm}^3\ \text{atm K}^{-1}\text{mol}^{-1}\).
Calculate the osmotic pressure at 300 K of a 0.020 M aqueous solution of \(\mathrm{CaCl_2}\), assuming complete dissociation. Take \(R = 0.08206\ \text{dm}^3\ \text{atm K}^{-1}\text{mol}^{-1}\).
A 0.10 m aqueous solution of \(\mathrm{K_2SO_4}\) freezes at −0.52 °C. Taking \(K_f\) for water as 1.86 K kg mol\(^{-1}\), calculate (a) the van’t Hoff factor for \(\mathrm{K_2SO_4}\) in this solution and (b) its degree of dissociation, expressed as a percentage.
Calculate the boiling point elevation of (a) a 0.20 m aqueous glucose solution and (b) a 0.20 m aqueous NaCl solution, assuming complete dissociation of the salt. Take \(K_b\) for water as 0.52 K kg mol\(^{-1}\), and state which solution boils at the higher temperature and by how much.
A 0.020 m aqueous solution of a weak monobasic acid HA freezes at −0.042 °C. Taking \(K_f\) for water as 1.86 K kg mol\(^{-1}\) and the freezing point of pure water as 0 °C, calculate the van’t Hoff factor and the degree of dissociation of the acid.
A 0.1 m solution of \(\mathrm{K_2SO_4}\) in water has a freezing point of −4.3 °C. What is the value of the van’t Hoff factor if \(K_f\) for water is 1.86 K kg mol\(^{-1}\)?
The osmotic pressures of \(\mathrm{CaCl_2}\) and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate the van’t Hoff factor for \(\mathrm{CaCl_2}\).
Fish generally need an \(\mathrm{O_2}\) concentration in water of at least 3.8 mg/L for survival. What partial pressure of \(\mathrm{O_2}\) above the water is needed for the survival of fish? Given that the solubility of \(\mathrm{O_2}\) in water at 0 °C and 1 atm partial pressure is \(2.2 \times 10^{-3}\) mol/L.
The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of a solution containing 2.8 g urea in 50 g of water?
A 5 % aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has a freezing point of 271 K. Calculate the freezing point of a 5 % aqueous glucose solution.
A solution of citric acid \(\mathrm{C_6H_8O_7}\) in 50 g of acetic acid has a boiling point elevation of 1.76 K. If \(K_b\) for acetic acid is 3.07 K kg mol\(^{-1}\), what is the molality of the solution?
An aqueous solution of a certain organic compound has a density of 1.063 g mL\(^{-1}\), an osmotic pressure of 12.16 atm at 25 °C and a freezing point of −1.03 °C. What is the molar mass of the compound?
A mixture of benzene and toluene contains 30 % by mass of toluene. At 30 °C, the vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and the partial pressure of each constituent above the solution at 30 °C.
At 25 °C a 0.1 molal solution of \(\mathrm{CH_3COOH}\) is 1.35 % dissociated in aqueous solution. Calculate the freezing point and osmotic pressure of the solution, assuming molality and molarity to be identical.
A 0.15 m aqueous solution of KCl freezes at −0.510 °C. Calculate \(i\) and the osmotic pressure at 0 °C. Assume the volume of the solution to be equal to that of water.