Solutions

Mix chloroform with acetone and the flask gets warm and the volume shrinks. Mix ethanol with hexane and it gets cold and the volume expands. Which of the two mixtures boils at a temperature higher than either pure liquid?

The chloroform-acetone mixture.

The three observations are one observation. Chloroform's slightly acidic hydrogen forms a hydrogen bond with acetone's carbonyl oxygen — an interaction stronger than anything in either pure liquid. Stronger attraction releases energy (warming), pulls the molecules closer (contraction), and holds them in the liquid (lower vapour pressure than Raoult's law predicts). A lower vapour pressure means a higher boiling point, and at $64.5$ mol per cent chloroform the mixture boils at $64.5^{\circ }$C, above both components.

Ethanol and hexane do the reverse. Ethanol's hydrogen-bonded network is broken by the hexane and nothing replaces it, so mixing absorbs heat, the volume expands, escape becomes easier, and the mixture shows a minimum boiling point.

Three observables, one underlying cause. Advanced questions in this chapter almost always supply one of them and ask for the others.

1. Raoult's law and the composition of the vapour

For a solution of two volatile liquids,

$p_A=x_A{p}_A^{\circ },P_{total}=p_A^{\circ }+\left(p_B^{\circ }-p_A^{\circ }\right)x_B$

so the total pressure is linear in composition for an ideal mixture. The vapour composition is not the same as the liquid's:

$y_A=\frac{p_A}{P_{total}}$

and the vapour is always richer in the more volatile component. That difference is precisely what fractional distillation exploits, and its disappearance is what defines an azeotrope.

For a non-volatile solute the law reduces to a statement about lowering:

$\frac{p^{\circ }-p}{p^{\circ }}=x_{solute}$

the relative lowering of vapour pressure being equal to the mole fraction of solute.

Illustration 1

Two liquids have vapour pressures of $450$ and $700$ mm Hg. Find the total pressure and the vapour composition for an equimolar liquid mixture.

$P=0.5\times 450+0.5\times 700=575$ mm Hg

$y_A=\frac{225}{575}=0.391$; $y_B=\frac{350}{575}=0.609$

The vapour is enriched in the more volatile component, from $50\%$ in the liquid to $61\%$ in the vapour. Repeating that enrichment many times is exactly what a fractionating column does.

Illustration 2

The vapour pressure of pure water at $298$ K is $23.8$ mm Hg. Find the vapour pressure of a solution containing $18$ g of glucose in $178.2$ g of water.

$n_{glucose}=\frac{18}{180}=0.1$ mol; $n_{water}=\frac{178.2}{18}=9.9$ mol

$x_{glucose}=\frac{0.1}{10}=0.01$

$p=23.8\left(1-0.01\right)=23.56$ mm Hg

Only the mole fraction matters, not what the solute is. This is the first of the four colligative properties, and it is the one from which the others follow.

Illustration 3

The vapour above a mixture of two liquids of vapour pressures $200$ and $400$ mm Hg contains $60$ mol per cent of the more volatile component. Find the composition of the liquid.

Let $x$ be the mole fraction of the volatile liquid ($400$ mm) in the liquid phase.

$0.60=\frac{400x}{400x+200\left(1-x\right)}=\frac{400x}{200+200x}$

$120+120x=400x\Rightarrow x=0.429$

The liquid is only $43\%$ volatile component while the vapour is $60\%$. Condensing that vapour and re-boiling it enriches it further, which is one theoretical plate of a fractionating column.

2. Non-ideal solutions and azeotropes

An azeotrope is the composition at which liquid and vapour have the same composition, so distillation can go no further. Ethanol and water form one at $95.6\%$ ethanol by mass, boiling at $78.1^{\circ }$C — which is why ordinary distillation cannot produce absolute alcohol. Nitric acid and water form a maximum-boiling azeotrope at $68\%$ acid.

Illustration 4

A mixture of two liquids shows a maximum-boiling azeotrope. Predict the signs of $\Delta H_{mix}$ and $\Delta V_{mix}$, and state whether the mixture obeys Raoult's law.

Maximum boiling means the vapour pressure is lower than Raoult predicts, so this is a negative deviation.

The components attract each other more strongly than they attract themselves, so mixing releases heat: $\Delta H_{mix}<0$.

Stronger attraction pulls molecules closer: $\Delta V_{mix}<0$.

It does not obey Raoult's law at any composition except the pure components.

The chain runs in either direction. Given any one of the five observations, the other four follow without further information.

3. Henry's law

For a gas dissolving in a liquid,

$p=K_{H}x$

so solubility is proportional to partial pressure. A large $K_H$ means low solubility, which is the opposite of what the name suggests and is worth fixing early. $K_H$ rises with temperature, so gases become less soluble in warm liquids — which is why a warm fizzy drink loses its gas faster and why thermal pollution of rivers reduces dissolved oxygen.

The law fails whenever the gas reacts with the solvent, as ammonia, carbon dioxide and hydrogen chloride all do to some extent.

Illustration 5

The Henry constant for oxygen in water at $298$ K is $4.34\times 10^4$ atm. Find the mole fraction of dissolved oxygen in water in contact with air at $1$ atm, taking oxygen as $21\%$ of air.

$p_{{\text{O}}_2}=0.21$ atm

$x=\frac{p}{K_H}=\frac{0.21}{4.34\times 10^4}=4.84\times 10^{-6}$

Under five parts per million by mole. Aquatic life depends on this very small figure, which is why even a modest rise in water temperature, by raising $K_H$, has serious ecological consequences.

4. The four colligative properties

A colligative property depends on the number of solute particles and not at all on what they are:

$\frac{p^{\circ }-p}{p^{\circ }}=x_B,\Delta T_b=K_{b}m,\Delta T_f=K_{f}m,\pi =CRT$

All four descend from the same cause: dissolving a non-volatile solute lowers the solvent's vapour pressure, which moves the liquid-vapour and solid-liquid equilibria in opposite directions.

Osmotic pressure is by far the most sensitive, because it is measured in atmospheres rather than in hundredths of a degree. That is why it, and not freezing point depression, is used for macromolecules.

Illustration 6

A solution of $2.0$ g of a non-volatile solute in $100$ g of benzene raises the boiling point by $0.65$ K. Find the molar mass, with $K_b=2.53$ K kg mol${}^{-1}$.

$m=\frac{\Delta T_b}{K_b}=\frac{0.65}{2.53}=0.257$ mol kg${}^{-1}$

Moles of solute $=0.257\times 0.100=0.0257$

$M=\frac{2.0}{0.0257}=78$ g mol${}^{-1}$

Which is benzene's own molar mass, a coincidence worth noticing but not a check — the method is blind to what the solute is, which is the whole meaning of colligative.

Illustration 7

A solution containing $1.0$ g of a protein in $100$ mL of water has an osmotic pressure of $2.57\times 10^{-3}$ atm at $300$ K. Find the molar mass.

$C=\frac{\pi }{RT}=\frac{2.57\times 10^{-3}}{0.0821\times 300}=1.043\times 10^{-4}$ mol L${}^{-1}$

In $0.1$ L: $1.043\times 10^{-5}$ mol

$M=\frac{1.0}{1.043\times 10^{-5}}=9.6\times 10^4$ g mol${}^{-1}$

The same solution would depress the freezing point by only $0.0002$ K, far below any thermometer's resolution. Osmotic pressure is the only colligative method usable for macromolecules.

Illustration 8

Arrange $0.1$ m aqueous solutions of glucose, sodium chloride and barium chloride by freezing point, assuming complete dissociation.

Glucose does not dissociate: $i=1$, so $\Delta T_f=1.86\times 0.1\times 1=0.186$ K.

Sodium chloride gives two ions: $i=2$, so $\Delta T_f=0.372$ K.

Barium chloride gives three: $i=3$, so $\Delta T_f=0.558$ K.

Freezing points: glucose $-0.186^{\circ }$C, sodium chloride $-0.372^{\circ }$C, barium chloride $-0.558^{\circ }$C.

Identical molality, three different answers. What counts is the number of particles released, which is exactly why these properties are called colligative.

5. Where $K_b$ and $K_f$ come from

The two constants are not empirical accidents; they follow from the thermodynamics of the phase change:

$K_b=\frac{R{T}_b^{2}M}{1000\Delta H_{vap}},K_f=\frac{R{T}_f^{2}M}{1000\Delta H_{fus}}$

with $M$ the solvent's molar mass in grams. Substituting water's values gives $0.512$ and $1.86$ K kg mol${}^{-1}$, matching the tabulated figures.

The expressions explain a useful pattern: $K_f$ exceeds $K_b$ for most solvents because $\Delta H_{fus}$ is much smaller than $\Delta H_{vap}$. Freezing point depression is therefore the more sensitive of the two thermal methods.

Illustration 9

Compute $K_f$ for water from first principles, given $T_f=273$ K and $\Delta H_{fus}=6.01$ kJ mol${}^{-1}$.

$K_f=\frac{8.314\times {\left(273\right)}^2\times 18}{1000\times 6010}$

$=\frac{8.314\times 74529\times 18}{6.01\times 10^6}=\frac{1.115\times 10^7}{6.01\times 10^6}$

$K_f=1.86$ K kg mol${}^{-1}$

Exactly the tabulated value. The same calculation with the enthalpy of vaporisation and the boiling point gives $0.512$, which is why the two constants differ by roughly the ratio of the two enthalpies.

6. The van't Hoff factor

Electrolytes and associating solutes give abnormal colligative results, corrected by a factor counting the actual particles:

$i=\frac{\text{observed colligative effect}}{\text{effect if no association or dissociation}}$

$\text{dissociation into }n:i=1+\left(n-1\right)\alpha ;\text{association of }n:i=1+\left(\frac{1}{n}-1\right)\alpha$

So $i>1$ for dissociation and $i<1$ for association. Every colligative expression then carries the factor: $\Delta T_f=i{K}_{f}m$ and $\pi =iCRT$.

Benzoic acid in benzene dimerises through hydrogen bonding and gives $i$ near $0.5$; sodium chloride in water gives $i$ approaching $2$ but never quite reaching it, because ion pairing removes a fraction of the particles.

Illustration 10

A $0.01$ m aqueous solution of a salt ${\text{AB}}_2$ freezes at $-0.0558^{\circ }$C. Find the degree of dissociation.

$i=\frac{\Delta T_f}{K_{f}m}=\frac{0.0558}{1.86\times 0.01}=3.0$

For ${\text{AB}}_2\to {\text{A}}^{2+}+2{\text{B}}^{-}$, $n=3$:

$3.0=1+\left(3-1\right)\alpha \Rightarrow \alpha =1.0$

Completely dissociated. Any value of $i$ between $1$ and $n$ corresponds to partial dissociation, and $i$ can never exceed $n$ however concentrated the solution.

Illustration 11

Benzoic acid in benzene gives an experimental molar mass of $242$ g mol${}^{-1}$ against a formula mass of $122$. Find the degree of association.

$i=\frac{\text{normal mass}}{\text{observed mass}}=\frac{122}{242}=0.504$

For dimerisation, $n=2$: $i=1+\left(\frac{1}{2}-1\right)\alpha =1-\frac{\alpha }{2}$

$0.504=1-\frac{\alpha }{2}\Rightarrow \alpha =0.99$

Essentially complete dimerisation. In benzene the carboxylic acid pairs form two hydrogen bonds in a closed ring, and water is absent to compete for them.

7. Osmosis and its uses

Osmosis is the passage of solvent through a semipermeable membrane from dilute to concentrated. The pressure required to stop it is

$\pi =iCRT$

Solutions of equal osmotic pressure are isotonic; a more concentrated one is hypertonic and a more dilute one hypotonic. Red blood cells placed in hypotonic water burst and in hypertonic brine shrivel, which is why intravenous fluids must be isotonic with blood at about $0.9\%$ sodium chloride.

Applying a pressure greater than $\pi$ to the concentrated side reverses the flow, and this reverse osmosis is how sea water is desalinated.

Illustration 12

Find the osmotic pressure of a $0.9\%$ mass by volume sodium chloride solution at $310$ K, taking $i=1.86$.

$C=\frac{9{\text{g L}}^{-1}}{58.5}=0.154$ mol L${}^{-1}$

$\pi =1.86\times 0.154\times 0.0821\times 310$

$=7.29$ atm

Which matches the osmotic pressure of blood plasma. The concentration of physiological saline was chosen for exactly this reason, and it is why the value of $i$ used must be the measured one rather than the ideal $2$.

Illustration 13

Sea water contains about $3.5\%$ dissolved salts, with an osmotic pressure near $27$ atm. Explain what pressure a desalination plant must apply and why.

Osmosis would naturally drive fresh water into the sea water, diluting it.

To reverse the flow, the applied pressure must exceed the osmotic pressure of $27$ atm.

Practical plants operate between $50$ and $80$ atm, the excess providing a useful flow rate rather than merely halting osmosis.

Energy cost scales with that pressure, which is why desalination remains expensive and why recovering the pressure from the reject stream is a standard design feature.

Summary

• Warming and contraction on mixing means negative deviation, lower vapour pressure and a maximum-boiling azeotrope. Cooling and expansion means the reverse.
• $p_A=x_A{p}_A^{\circ }$; the vapour is always richer in the more volatile component, which is what makes fractional distillation work.
• $\frac{p^{\circ }-p}{p^{\circ }}=x_{solute}$: relative lowering equals the solute mole fraction.
• An azeotrope is where liquid and vapour compositions coincide, so distillation stops. Ethanol and water azeotrope at $95.6\%$.
• Henry: $p=K_{H}x$, and a large $K_H$ means low solubility. $K_H$ rises with temperature, so gases are less soluble when warm.
• The four colligative properties all descend from vapour pressure lowering, and all count particles rather than identity.
• $K_b=\frac{R{T}_b^{2}M}{1000\Delta H_{vap}}$ and $K_f=\frac{R{T}_f^{2}M}{1000\Delta H_{fus}}$; for water these give $0.512$ and $1.86$.
• $K_f>K_b$ for most solvents because $\Delta H_{fus}\ll \Delta H_{vap}$, so freezing point depression is the more sensitive thermal method.
• Osmotic pressure is the only colligative method usable for macromolecules, being measured in atmospheres rather than millikelvin.
• $i=1+\left(n-1\right)\alpha$ for dissociation and $1+\left(\frac{1}{n}-1\right)\alpha$ for association; $i>1$ and $i<1$ respectively.
• $i$ can never exceed $n$, and falls short of it through ion pairing in concentrated solution.
• $\pi =iCRT$; isotonic solutions share it, and physiological saline at $0.9\%$ matches blood at about $7.3$ atm.
• Reverse osmosis needs an applied pressure above $\pi$, which is why desalination of sea water requires $50$ atm or more.