Which of the following statements about osmosis and osmotic pressure are correct?

Osmometry is preferred to freezing-point depression for a protein's molar mass, because a dilute protein solution has a measurable osmotic pressure but an unmeasurably small ΔTf

Red blood cells placed in distilled water take up water, swell and may burst

In reverse osmosis, a pressure greater than the osmotic pressure applied on the solution side drives solvent out of the solution through the membrane

Osmometry is preferred to freezing-point depression for a protein's molar mass, because a dilute protein solution has a measurable osmotic pressure but an unmeasurably small ΔTf

Solutions of 0.9% (w/v) NaCl and 0.9% (w/v) glucose have nearly the same osmotic pressure at the same temperature

Red blood cells placed in distilled water take up water, swell and may burst

In reverse osmosis, a pressure greater than the osmotic pressure applied on the solution side drives solvent out of the solution through the membrane

At 37 °C the ionic product of water is Kw = 2.4 × 10⁻¹⁴, and pure water at this temperature has pH 6.81. This water is:

neutral, because [H⁺] equals [OH⁻]; the neutral pH itself falls as Kw rises with temperature

acidic, because its pH lies below 7.00, which is the neutral point of water at every temperature

neutral, because [H⁺] equals [OH⁻]; the neutral pH itself falls as Kw rises with temperature

basic, because the rise in Kw on warming produces more OH⁻ ions than H⁺ ions in the water

acidic, because warming the water drives the ionisation forward and leaves H⁺ ions in excess

Aqueous ammonium acetate is prepared at 0.10 M and at 0.010 M at 25 °C. Take pKa(CH₃COOH) = 4.76, pKb(NH₃) = 4.75 and pKw = 14.00. Which statement is correct?

Both have a pH close to 7.0, and the fraction of salt hydrolysed is the same in both, although both ions react with water

Both solutions have a pH close to 7.0, which shows that neither the ammonium ion nor the acetate ion reacts with water

The 0.010 M solution is hydrolysed to a much greater fraction than the 0.10 M one, since dilution always favours the hydrolysis of a salt

Both have a pH close to 7.0, and the fraction of salt hydrolysed is the same in both, although both ions react with water

The 0.010 M solution has a pH much closer to 4.76, because the acetic acid formed by hydrolysis dominates at high dilution

An acetic acid/sodium acetate buffer (pKa of acetic acid 4.76) has its greatest buffer capacity when:

the acetic acid and acetate concentrations are equal, so that the pH equals 4.76

the acetate concentration is ten times that of acetic acid, so that the pH is 5.76

acetic acid is present alone, so that every bit of added base is consumed at once

the buffer is diluted tenfold, so that its pH becomes less sensitive to added acid

Which of the following statements about buffers are correct?

Diluting a buffer tenfold with water leaves its pH almost unchanged but lowers its buffer capacity

An acetic acid/acetate buffer (pKa 4.76) is a poor choice for holding a reaction at pH 7.4

Diluting a buffer tenfold with water leaves its pH almost unchanged but lowers its buffer capacity

A solution of HCl and NaCl is a buffer, because it contains both an acid and its conjugate base

An acetic acid/acetate buffer (pKa 4.76) is a poor choice for holding a reaction at pH 7.4

A Tris buffer adjusted to pH 8.1 at 25 °C will have a lower pH when it is used at 4 °C

A salt MX has Ksp = 4.0 × 10⁻¹⁰ and a salt MX₂ has Ksp = 3.2 × 10⁻¹¹ at the same temperature. Neither ion hydrolyses. Which statement about their molar solubilities in pure water is correct?

MX₂ is about ten times more soluble than MX: 2.0 × 10⁻⁴ M against 2.0 × 10⁻⁵ M

MX is the more soluble, since its Ksp is more than ten times larger than that of MX₂

MX₂ is about ten times more soluble than MX: 2.0 × 10⁻⁴ M against 2.0 × 10⁻⁵ M

Both salts have the same molar solubility, 2.0 × 10⁻⁵ M, once stoichiometry is allowed for

MX is about twelve times more soluble, the same as the ratio of the two Ksp values

Which of the following would increase the molar solubility of calcium oxalate, CaC₂O₄, in water at 25 °C?

Lowering the pH to about 2 with HCl

Adding EDTA, which forms a stable soluble complex with Ca²⁺

Lowering the pH to about 2 with HCl

Adding sodium oxalate to the solution

Adding EDTA, which forms a stable soluble complex with Ca²⁺

Adding more solid calcium oxalate to the saturated solution

The Ksp of CaF₂ is 3.9 × 10⁻¹¹ at 25 °C. Assuming complete dissociation of NaF, no hydrolysis of F⁻, and that the fluoride contributed by CaF₂ is negligible, the molar solubility of CaF₂ in 0.010 M NaF is:

3.9 × 10⁻⁷ M, since s = Ksp/[F⁻]² when the fluoride level is fixed at 0.010 M

3.9 × 10⁻⁹ M, since s = Ksp/[F⁻] when the fluoride level is fixed at 0.010 M

2.1 × 10⁻⁴ M, the same as in pure water, because the Ksp of the salt is unchanged

9.8 × 10⁻⁸ M, since s = Ksp/(2[F⁻])² once the factor of two in CaF₂ is applied

3.9 × 10⁻⁷ M, since s = Ksp/[F⁻]² when the fluoride level is fixed at 0.010 M

The equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is established in a cylinder. The mixture is then compressed to double the total pressure, and equilibrium is re-established at the same temperature. Which statement is correct?

Kp is unchanged, Kx falls to half its earlier value, and the degree of dissociation of PCl₅ decreases

Kp doubles because every partial pressure doubles, while Kx and the degree of dissociation stay as before

Kp and Kx are both unchanged, so the degree of dissociation of PCl₅ is the same as before compression

Kp is unchanged, Kx doubles to compensate for the higher pressure, and the degree of dissociation increases

Chemical Equilibria: Colligative Properties, Ionic Equilibria, Buffers, Solubility Product and Kc, Kp, Kx

This chapter answers Section XL0.5, “Chemical Equilibria”, of XL0 Chemistry, the part of the Life Sciences (XL) paper that every candidate answers: “Osmotic pressure, elevation of boiling point and depression of freezing point, ionic equilibria in solution, solubility product, common ion effect, hydrolysis of salts, pH, buffer and their applications. Equilibrium constants (Kc, Kp, and KX) for homogeneous reactions.” It teaches colligative properties and the van 't Hoff factor, osmotic pressure in cells and in molar-mass work, the pH of acids and bases, the hydrolysis of salts, buffers with the Henderson–Hasselbalch equation and the blood bicarbonate system, the solubility product with the common-ion and pH effects, and the relations among Kc, Kp and Kx. The numericals are molar masses from colligative data, the pH of acids, salts and buffers, precipitation thresholds and equilibrium compositions from Kp.

1. Colligative properties: counting dissolved particles

A colligative property depends on the number of solute particles in a given amount of solvent, not on their identity. The root cause is that a non-volatile solute dilutes the solvent and lowers its escaping tendency. By Raoult's law the solvent's vapour pressure is proportional to its mole fraction, so the relative lowering equals the mole fraction of solute, (p° − p)/p° = x₂. Lower vapour pressure means the solution must be heated further to boil and cooled further to freeze, and that pure solvent flows into it through a semipermeable membrane.

Elevation of boiling point: ΔTb = i Kb m. Depression of freezing point: ΔTf = i Kf m. Here m is the molality (mol solute per kg solvent) and the constants belong to the solvent alone: Kb = R Tb² M₁/ΔvapH, with M₁ the solvent's molar mass in kg mol⁻¹, and Kf likewise with the freezing point and ΔfusH. For water Kf = 1.86 and Kb = 0.512 K kg mol⁻¹. Molality is used because, unlike molarity, it does not change with temperature.

The van 't Hoff factor i is the number of particles actually present per formula unit dissolved. For an electrolyte giving ν ions with degree of dissociation α, i = 1 + (ν − 1)α; for association of n molecules into one, i = 1 − α(1 − 1/n). Complete dissociation gives i = 2 for NaCl and 3 for CaCl₂, while acetic acid, which dimerises through hydrogen bonds in benzene, gives i close to ½. Real strong electrolytes give i a little below ν because of ion pairing.

⚠️ Abnormal molar masses

If i is ignored, a dissociating solute gives too many particles and its apparent molar mass comes out too low; an associating solute gives too few and comes out too high. ΔTb and ΔTf use molality; osmotic pressure uses molarity.

2. Osmotic pressure and its biological uses

Osmosis is the net flow of solvent through a semipermeable membrane from pure solvent, or a dilute solution, into a more concentrated solution. The osmotic pressure π is the extra pressure that must be applied to the solution to stop that flow. For dilute solutions the van 't Hoff equation gives π = iCRT, with C in mol L⁻¹, R = 0.08314 L bar K⁻¹ mol⁻¹ and T in kelvin; the product iC is the osmolarity.

Why osmometry suits macromolecules. A protein of molar mass 50,000 g mol⁻¹ at 10 g L⁻¹ is only 2.0 × 10⁻⁴ M. Its freezing-point depression, about 4 × 10⁻⁴ K, is beyond ordinary thermometry, but at 298 K its osmotic pressure is 2.0 × 10⁻⁴ × 0.08314 × 298 = 4.96 × 10⁻³ bar, about 3.7 mmHg or a 5 cm column of water, easily measured. So M = wRT/(πV), for w grams in volume V, is the standard colligative route to a protein's molar mass (strictly a number-average value).

Cells and tonicity. A solution of the same osmolarity as the cytoplasm is isotonic, and a red cell keeps its shape in it. In a hypotonic solution water enters and the cell swells and may burst (haemolysis); in a hypertonic one it shrinks (crenation), and a plant cell plasmolyses as its membrane pulls away from the wall. Worked example: 0.9% (w/v) NaCl is 9.0/58.44 = 0.154 M, an ideal osmolarity of 0.308 osmol L⁻¹, close to that of plasma; hence “normal saline”. Its ideal π at 310 K is 0.308 × 0.08314 × 310 = 7.9 bar. In reverse osmosis, a pressure greater than π applied to the solution side drives solvent out of it through the membrane, which is how water is desalinated and juices are concentrated.

3. Ionic equilibria in water: the pH of acids and bases

Water self-ionises, H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K, so pH + pOH = 14.00 there (pH = −log[H⁺]). The ionisation is endothermic, so Kw rises with temperature; at 37 °C pure water has a pH near 6.8 and is still neutral, because [H⁺] = [OH⁻]. Strong acids and bases dissociate completely: 0.010 M HCl has pH 2.00.

For a weak acid HA at concentration c with degree of dissociation α, Ka = cα²/(1 − α). When α is small, α ≈ √(Ka/c), which is Ostwald's dilution law (dilution raises α), so [H⁺] = √(Ka c) and pH = ½(pKa − log c). Worked example: 0.10 M acetic acid, pKa 4.76: pH = ½(4.76 + 1.00) = 2.88, and α = 10⁻²·⁸⁸/0.10 = 1.3%, small enough to justify the approximation. For a weak base, pOH = ½(pKb − log c); for a conjugate pair, Ka × Kb = Kw, so pKa + pKb = 14.00 at 298 K. A polyprotic acid such as H₃PO₄ (pKa about 2.1, 7.2 and 12.3) ionises stepwise, and an amphiprotic ion such as H₂PO₄⁻ has pH ≈ ½(pKa₁ + pKa₂), whatever its concentration.

⚠️ An acid cannot make water basic

For 1.0 × 10⁻⁸ M HCl, pH 8 is impossible. When the acid supplies H⁺ comparable to 10⁻⁷ M, include water's own ionisation: [H⁺] = c + Kw/[H⁺], a quadratic whose root gives a pH just below 7. Likewise, the shortcut pH = ½(pKa − log c) fails when α is not small; then solve Ka = x²/(c − x).

4. Hydrolysis of salts

A salt solution is neutral only if neither ion reacts with water. The conjugate base of a weak acid (CH₃COO⁻, CN⁻, CO₃²⁻) takes a proton from water and releases OH⁻; the conjugate acid of a weak base (NH₄⁺) gives a proton to water. The hydrolysis constant is Kh = Kw/Ka for such an anion and Kw/Kb for such a cation, and the degree of hydrolysis is h ≈ √(Kh/c).

The four kinds of salt at 298 K (c is the salt concentration; pKw = 14.00)
Salt of	Example	Ion that hydrolyses	pH
Strong acid + strong base	NaCl, KNO₃	Neither	7
Weak acid + strong base	CH₃COONa	Anion	7 + ½pKa + ½log c (above 7)
Strong acid + weak base	NH₄Cl	Cation	7 − ½pKb − ½log c (below 7)
Weak acid + weak base	CH₃COONH₄	Both	7 + ½(pKa − pKb), independent of c

Worked example: 0.10 M sodium acetate, pKa of acetic acid 4.76: pH = 7 + 2.38 + ½(−1.00) = 8.88. Yet h = √(Kh/c) = √(5.75 × 10⁻¹⁰/0.10) ≈ 7.6 × 10⁻⁵: fewer than one acetate ion in ten thousand reacts, and the pH still moves almost two units. This is why the equivalence point of a weak acid titrated with NaOH lies above pH 7, so phenolphthalein, not methyl orange, is the right indicator. In ammonium acetate the pKa (4.76) and pKb (4.75) nearly cancel and the pH is close to 7, although both ions are hydrolysed; for this type, h does not depend on c.

5. Buffers: the Henderson–Hasselbalch equation and its applications

A buffer keeps the pH nearly constant when small amounts of acid or base are added. It is a weak acid with its conjugate base (CH₃COOH/CH₃COO⁻) or a weak base with its conjugate acid (NH₃/NH₄⁺): added H⁺ is taken up by the base and added OH⁻ by the acid. Rearranging Ka gives the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), valid when both components are far more concentrated than H⁺ and OH⁻.

Range and capacity. A buffer works over about pKa ± 1 (base-to-acid ratio 0.1 to 10). Its buffer capacity, the moles of strong acid or base per litre needed to change the pH by one unit, is greatest at pH = pKa and is proportional to the total buffer concentration. Diluting a buffer leaves the ratio, and hence the pH, almost unchanged but lowers its capacity. Worked example: 1.00 L containing 0.100 mol acetic acid and 0.100 mol sodium acetate (pKa 4.76) receives 0.010 mol HCl. Acetate falls to 0.090 mol and acetic acid rises to 0.110 mol, so pH = 4.76 + log(0.090/0.110) = 4.67; the same acid in a litre of pure water would give pH 2.00.

Blood is held at pH 7.35 to 7.45 mainly by the bicarbonate system, written with dissolved CO₂: pH = 6.1 + log{[HCO₃⁻]/(0.03 × pCO₂)}, with [HCO₃⁻] in mmol L⁻¹, pCO₂ in mmHg and 0.03 mmol L⁻¹ mmHg⁻¹ the solubility of CO₂. With 24 mM bicarbonate and pCO₂ = 40 mmHg, pH = 6.1 + log(24/1.2) = 6.1 + log 20 = 7.40. Although its pKa is 6.1, the system is powerful because it is open: the lungs adjust CO₂ within minutes and the kidneys adjust bicarbonate over days. Inside cells, phosphate (H₂PO₄⁻/HPO₄²⁻, pKa near 7) and the histidine side chains of proteins do most of the buffering.

In the laboratory, choose a buffer whose pKa lies within one unit of the target pH and that does not interfere with the system. Tris (pKa about 8.1 at 25 °C) is common in molecular biology, but its pKa falls by about 0.03 per °C, so a Tris buffer set at room temperature is more alkaline in the cold room. Good's buffers such as HEPES (pKa about 7.5) were designed for physiological pH with little metal binding. Buffers fix the pH of enzyme assays, cell-culture media, electrophoresis and chromatography, and of drug and food formulations.

6. Solubility product, the common-ion effect and pH

For a sparingly soluble salt in equilibrium with its saturated solution, MₓAᵧ(s) ⇌ xMʸ⁺ + yAˣ⁻, the solubility product is Ksp = [Mʸ⁺]ˣ[Aˣ⁻]ʸ; the solid does not appear. In terms of molar solubility s: AB gives Ksp = s², AB₂ or A₂B gives 4s³, AB₃ gives 27s⁴ and A₃B₂ gives 108s⁵. Compare solubilities through s, not Ksp, when the formulas differ: with Ksp(AgCl) = 1.8 × 10⁻¹⁰ and Ksp(Ag₂CrO₄) = 1.1 × 10⁻¹², s is 1.3 × 10⁻⁵ M for AgCl but 6.5 × 10⁻⁵ M for Ag₂CrO₄, so the salt with the smaller Ksp is the more soluble.

Precipitation rule. Compare the ionic product Q (the same expression with actual concentrations) with Ksp: if Q > Ksp a precipitate forms, if Q = Ksp the solution is saturated, and if Q < Ksp more solid can dissolve. Adding Ag⁺ slowly to a solution of Cl⁻ and CrO₄²⁻ precipitates AgCl first because it needs far less Ag⁺; red Ag₂CrO₄ appears only when the chloride is almost used up, the basis of Mohr's titration.

Common-ion effect. A salt is less soluble in a solution that already contains one of its ions. For AgCl in 0.010 M NaCl, [Cl⁻] ≈ 0.010 M, so s = Ksp/0.010 = 1.8 × 10⁻⁸ M, about 750 times less than in pure water. The same effect suppresses the ionisation of a weak acid when its conjugate base is added, which is why a buffer's acid stays almost un-ionised. Conversely, a ligand that complexes the cation (EDTA with Ca²⁺, NH₃ with Ag⁺) raises solubility.

pH and solubility. If the anion is the conjugate base of a weak acid (CO₃²⁻, PO₄³⁻, C₂O₄²⁻, F⁻, OH⁻), acid removes it and pulls more salt into solution. So CaCO₃ dissolves in acid, acids erode tooth enamel (hydroxyapatite, Ca₅(PO₄)₃OH), and fluoride protects teeth by forming the less soluble fluorapatite. Calcium oxalate kidney stones are a biological case of Q exceeding Ksp.

7. Equilibrium constants Kc, Kp and Kx for homogeneous reactions

For a homogeneous gas reaction aA + bB ⇌ cC + dD, the equilibrium constant can be written in molar concentrations, partial pressures or mole fractions: Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ), Kp = p(C)ᶜ p(D)ᵈ/(p(A)ᵃ p(B)ᵇ) and Kx = x(C)ᶜ x(D)ᵈ/(x(A)ᵃ x(B)ᵇ). Since each pᵢ = cᵢRT and pᵢ = xᵢP, with Δn = (c + d) − (a + b) counted over gases only, Kp = Kc(RT)^Δn and Kp = Kx P^Δn. Use R = 0.08314 L bar K⁻¹ mol⁻¹ with pressures in bar (0.08206 L atm K⁻¹ mol⁻¹ with atm). When Δn = 0 all three are equal.

Kc and Kp depend only on temperature (the thermodynamic K divides each pressure by p° = 1 bar and is dimensionless). Kx also depends on the total pressure unless Δn = 0, and this is how Le Chatelier's principle works for pressure: raising P lowers Kx = Kp/P^Δn when Δn > 0, so the mixture shifts towards fewer moles of gas. An inert gas added at constant volume changes no partial pressure and has no effect; added at constant total pressure it dilutes the mixture like a fall in pressure. A catalyst changes neither K nor the equilibrium composition.

Worked example. N₂O₄(g) ⇌ 2NO₂(g) has Kp = 0.148 at 298 K (pressures in bar), with Δn = +1. Then Kc = Kp/(RT) = 0.148/(0.08314 × 298) = 5.97 × 10⁻³ mol L⁻¹. Starting from 1 mol N₂O₄ with degree of dissociation α, the amounts are (1 − α) and 2α, total (1 + α), so Kp = 4α²P/(1 − α²). At P = 1.00 bar, α = √[Kp/(4P + Kp)] = 0.189; at 10.0 bar α falls to 0.061, as Le Chatelier's principle predicts.

⚠️ What changes K and what does not

Only temperature changes Kp and Kc. Pressure, inert gases, catalysts and changes in amounts shift the composition (or not) but leave K alone. Count Δn over gaseous species only, and match the value of R to the pressure unit.

Key takeaways

Colligative properties count particles: ΔTb = iKb m, ΔTf = iKf m, π = iCRT, with i = 1 + (ν − 1)α. Osmometry is the colligative route to protein molar masses; 0.9% NaCl (about 0.31 osmol L⁻¹) is isotonic with plasma.
Kw = 1.0 × 10⁻¹⁴ at 298 K and rises with temperature; neutral means [H⁺] = [OH⁻], not pH 7. For a weak acid, pH = ½(pKa − log c) when α is small; Ka × Kb = Kw.
Salts of a weak acid and strong base are basic (pH = 7 + ½pKa + ½log c), those of a strong acid and weak base acidic (pH = 7 − ½pKb − ½log c), and those of a weak acid and weak base have pH = 7 + ½(pKa − pKb), independent of concentration.
Buffers follow pH = pKa + log([A⁻]/[HA]), work over pKa ± 1 and have most capacity at pH = pKa; dilution keeps the pH but lowers capacity. Blood: pH = 6.1 + log{[HCO₃⁻]/(0.03 pCO₂)} = 7.40 at 24 mM and 40 mmHg.
Ksp: precipitate when Q > Ksp; compare solubilities through s, not Ksp, when formulas differ. A common ion lowers solubility, while acid (for anions of weak acids) and complexing ligands raise it.
Kp = Kc(RT)^Δn = Kx P^Δn with Δn over gases. Only temperature changes Kp and Kc; Kx changes with total pressure unless Δn = 0, which is Le Chatelier's pressure effect.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A 0.100 mol kg⁻¹ aqueous solution of a weak monoprotic acid HA freezes at −0.205 °C. Taking Kf(water) = 1.86 K kg mol⁻¹ and ignoring activity effects, the degree of dissociation of HA is closest to:
1. About 5 per cent
2. About 10 per cent
3. About 21 per cent
4. About 90 per cent
Show answer
Answer: B — About 10 per cent
ΔTf = iKf m gives i = 0.205/(1.86 × 0.100) = 1.102. For HA ⇌ H⁺ + A⁻, ν = 2, so i = 1 + α and α = 0.102, about 10 per cent. Writing i = 1 + 2α (as if three particles formed) gives the tempting 5 per cent; 90 per cent is roughly 1/i, the fraction of formula units not counted as extra particles, and has no meaning here.
A solution made by dissolving 2.00 g of a purified protein in water to a total volume of 100.0 mL shows an osmotic pressure of 7.40 × 10⁻³ bar at 300 K. Using π = CRT with R = 0.08314 L bar K⁻¹ mol⁻¹ and treating the protein as a non-dissociating solute, its molar mass in kg mol⁻¹, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 67.4
C = π/RT = 7.40 × 10⁻³/(0.08314 × 300) = 7.40 × 10⁻³/24.94 = 2.967 × 10⁻⁴ mol L⁻¹. Moles in 0.1000 L = 2.967 × 10⁻⁵ mol. M = 2.00/2.967 × 10⁻⁵ = 6.74 × 10⁴ g mol⁻¹ = 67.4 kg mol⁻¹. Forgetting to convert 100 mL to litres gives a value a thousand times too small.
Which of the following statements about osmosis and osmotic pressure are correct?
1. Osmometry is preferred to freezing-point depression for a protein's molar mass, because a dilute protein solution has a measurable osmotic pressure but an unmeasurably small ΔTf
2. Solutions of 0.9% (w/v) NaCl and 0.9% (w/v) glucose have nearly the same osmotic pressure at the same temperature
3. Red blood cells placed in distilled water take up water, swell and may burst
4. In reverse osmosis, a pressure greater than the osmotic pressure applied on the solution side drives solvent out of the solution through the membrane
Show answer
Answer: A — Osmometry is preferred to freezing-point depression for a protein's molar mass, because a dilute protein solution has a measurable osmotic pressure but an unmeasurably small ΔTf; C — Red blood cells placed in distilled water take up water, swell and may burst; D — In reverse osmosis, a pressure greater than the osmotic pressure applied on the solution side drives solvent out of the solution through the membrane
A protein at a few grams per litre gives π of millibars (centimetres of water) but ΔTf of about 10⁻⁴ K, so osmometry wins. Distilled water is hypotonic, so red cells swell and lyse. Pressure above π reverses the natural flow, which is reverse osmosis. The NaCl–glucose statement is wrong: 0.9% NaCl is 0.154 M giving about 0.31 osmol L⁻¹, while 0.9% glucose is only 0.050 M (0.050 osmol L⁻¹), about six times lower.
At 37 °C the ionic product of water is Kw = 2.4 × 10⁻¹⁴, and pure water at this temperature has pH 6.81. This water is:
1. acidic, because its pH lies below 7.00, which is the neutral point of water at every temperature
2. neutral, because [H⁺] equals [OH⁻]; the neutral pH itself falls as Kw rises with temperature
3. basic, because the rise in Kw on warming produces more OH⁻ ions than H⁺ ions in the water
4. acidic, because warming the water drives the ionisation forward and leaves H⁺ ions in excess
Show answer
Answer: B — neutral, because [H⁺] equals [OH⁻]; the neutral pH itself falls as Kw rises with temperature
In pure water every H₂O that ionises gives one H⁺ and one OH⁻, so [H⁺] = [OH⁻] = √(2.4 × 10⁻¹⁴) = 1.55 × 10⁻⁷ M and pH = 6.81: neutral. pH 7.00 is the neutral point only at 25 °C, so judging acidity against 7.00 is the trap. Ionisation raises both ions equally and can never leave either in excess.
Taking Kw = 1.0 × 10⁻¹⁴ at 25 °C, the pH of a 1.0 × 10⁻⁸ M aqueous solution of HCl is closest to:
1. 8.00
2. 7.00
3. 6.98
4. 6.00
Show answer
Answer: C — 6.98
Charge balance gives [H⁺] = 1.0 × 10⁻⁸ + Kw/[H⁺], i.e. x² − 1.0 × 10⁻⁸x − 1.0 × 10⁻¹⁴ = 0, so x = 1.05 × 10⁻⁷ M and pH = 6.98. Taking −log(10⁻⁸) = 8.00 ignores the water's own H⁺ and absurdly makes an acid solution basic; 7.00 ignores the acid altogether.
Calculate the pH of 0.10 M aqueous NH₄Cl at 25 °C, taking pKb(NH₃) = 4.74 and pKw = 14.00, and assuming that the degree of hydrolysis is small. Give the answer to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 5.13
NH₄⁺ is the conjugate acid of NH₃: pKa = 14.00 − 4.74 = 9.26. [H⁺] = √(Ka c), so pH = ½(pKa − log c) = ½(9.26 + 1.00) = 5.13. Equivalently pH = 7 − ½pKb − ½log c = 7 − 2.37 + 0.50 = 5.13. Using pKb (4.74) in place of pKa gives ½(4.74 + 1.00) = 2.87, far too acidic for the salt of a weak base.
Aqueous ammonium acetate is prepared at 0.10 M and at 0.010 M at 25 °C. Take pKa(CH₃COOH) = 4.76, pKb(NH₃) = 4.75 and pKw = 14.00. Which statement is correct?
1. Both solutions have a pH close to 7.0, which shows that neither the ammonium ion nor the acetate ion reacts with water
2. The 0.010 M solution is hydrolysed to a much greater fraction than the 0.10 M one, since dilution always favours the hydrolysis of a salt
3. Both have a pH close to 7.0, and the fraction of salt hydrolysed is the same in both, although both ions react with water
4. The 0.010 M solution has a pH much closer to 4.76, because the acetic acid formed by hydrolysis dominates at high dilution
Show answer
Answer: C — Both have a pH close to 7.0, and the fraction of salt hydrolysed is the same in both, although both ions react with water
For a salt of a weak acid and a weak base, pH = 7 + ½(pKa − pKb) = 7 + ½(0.01) ≈ 7.0 at any concentration, and Kh = Kw/(KaKb) = h²/(1 − h)² contains no c, so h is the same at both concentrations. The neutral pH arises because the two hydrolyses balance, not because none occurs. The rule that dilution raises h (h = √(Kh/c)) holds only when a single ion hydrolyses, which makes it the tempting wrong choice.
An acetic acid/sodium acetate buffer (pKa of acetic acid 4.76) has its greatest buffer capacity when:
1. the acetic acid and acetate concentrations are equal, so that the pH equals 4.76
2. the acetate concentration is ten times that of acetic acid, so that the pH is 5.76
3. acetic acid is present alone, so that every bit of added base is consumed at once
4. the buffer is diluted tenfold, so that its pH becomes less sensitive to added acid
Show answer
Answer: A — the acetic acid and acetate concentrations are equal, so that the pH equals 4.76
Buffer capacity peaks at pH = pKa, where acid and conjugate base are equal and each can absorb added base or acid. At a 10:1 ratio the buffer is at the edge of its useful range. Acetic acid alone has no conjugate base to absorb added acid, and dilution lowers capacity while leaving the pH nearly unchanged.
What volume, in mL, of 0.100 M NaOH must be added to 100.0 mL of 0.100 M acetic acid (pKa = 4.76) to obtain a buffer of pH 5.76? Use the Henderson–Hasselbalch equation and give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 90.9
pH − pKa = 1.00, so [CH₃COO⁻]/[CH₃COOH] = 10. The acid is 10.0 mmol. Adding x mmol NaOH gives x mmol acetate and leaves (10.0 − x) mmol acid: x/(10.0 − x) = 10, so x = 100/11 = 9.09 mmol. Volume = 9.09 mmol/0.100 M = 90.9 mL. Setting x/10.0 = 10 (forgetting that the acid is consumed) gives an impossible 1000 mL.
A patient's arterial plasma has [HCO₃⁻] = 24 mmol L⁻¹, but hypoventilation has raised pCO₂ to 80 mmHg. Using pH = 6.1 + log{[HCO₃⁻]/(0.03 × pCO₂)}, with pCO₂ in mmHg and concentrations in mmol L⁻¹, the plasma pH is:
1. 6.80
2. 7.40
3. 7.70
4. 7.10
Show answer
Answer: D — 7.10
Dissolved CO₂ = 0.03 × 80 = 2.4 mmol L⁻¹, so the ratio is 24/2.4 = 10 and pH = 6.1 + 1.0 = 7.10, a respiratory acidosis. 7.40 is the normal value at 40 mmHg, and 7.70 would follow if pCO₂ had halved to 20 mmHg (ratio 40), the direction of hyperventilation.
Which of the following statements about buffers are correct?
1. Diluting a buffer tenfold with water leaves its pH almost unchanged but lowers its buffer capacity
2. A solution of HCl and NaCl is a buffer, because it contains both an acid and its conjugate base
3. An acetic acid/acetate buffer (pKa 4.76) is a poor choice for holding a reaction at pH 7.4
4. A Tris buffer adjusted to pH 8.1 at 25 °C will have a lower pH when it is used at 4 °C
Show answer
Answer: A — Diluting a buffer tenfold with water leaves its pH almost unchanged but lowers its buffer capacity; C — An acetic acid/acetate buffer (pKa 4.76) is a poor choice for holding a reaction at pH 7.4
Dilution keeps the base-to-acid ratio, hence the pH, but there are fewer moles to absorb added acid or base. pH 7.4 is 2.6 units above 4.76, a ratio of about 400:1, far outside pKa ± 1. Cl⁻ is the conjugate base of a strong acid and takes up no H⁺, so HCl/NaCl is not a buffer. Tris's pKa rises as it is cooled, so its pH at 4 °C is higher, not lower.
A solution contains 0.010 M Mg²⁺. NaOH is added slowly (neglect the change in volume). Taking Ksp of Mg(OH)₂ = 5.6 × 10⁻¹² and pKw = 14.00 at 25 °C, the pH at which Mg(OH)₂ just begins to precipitate, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 9.37
Precipitation starts when Q = [Mg²⁺][OH⁻]² reaches Ksp: [OH⁻] = √(5.6 × 10⁻¹²/0.010) = √(5.6 × 10⁻¹⁰) = 2.37 × 10⁻⁵ M. pOH = 4.63, so pH = 14.00 − 4.63 = 9.37. Forgetting the square on [OH⁻] gives [OH⁻] = 5.6 × 10⁻¹⁰ M and a pH near 4.7, which is far too acidic for any hydroxide to precipitate.
A salt MX has Ksp = 4.0 × 10⁻¹⁰ and a salt MX₂ has Ksp = 3.2 × 10⁻¹¹ at the same temperature. Neither ion hydrolyses. Which statement about their molar solubilities in pure water is correct?
1. MX is the more soluble, since its Ksp is more than ten times larger than that of MX₂
2. MX₂ is about ten times more soluble than MX: 2.0 × 10⁻⁴ M against 2.0 × 10⁻⁵ M
3. Both salts have the same molar solubility, 2.0 × 10⁻⁵ M, once stoichiometry is allowed for
4. MX is about twelve times more soluble, the same as the ratio of the two Ksp values
Show answer
Answer: B — MX₂ is about ten times more soluble than MX: 2.0 × 10⁻⁴ M against 2.0 × 10⁻⁵ M
For MX, Ksp = s², so s = √(4.0 × 10⁻¹⁰) = 2.0 × 10⁻⁵ M. For MX₂, Ksp = s(2s)² = 4s³, so s³ = 8.0 × 10⁻¹² and s = 2.0 × 10⁻⁴ M. MX₂ is ten times more soluble despite its smaller Ksp; comparing Ksp values directly is valid only for salts of the same formula type, which is the trap in the MX-is-more-soluble choices.
Which of the following would increase the molar solubility of calcium oxalate, CaC₂O₄, in water at 25 °C?
1. Lowering the pH to about 2 with HCl
2. Adding sodium oxalate to the solution
3. Adding EDTA, which forms a stable soluble complex with Ca²⁺
4. Adding more solid calcium oxalate to the saturated solution
Show answer
Answer: A — Lowering the pH to about 2 with HCl; C — Adding EDTA, which forms a stable soluble complex with Ca²⁺
Oxalate is the conjugate base of a weak acid, so H⁺ converts it to HC₂O₄⁻ and H₂C₂O₄; [C₂O₄²⁻] falls, Q drops below Ksp and more solid dissolves. EDTA removes free Ca²⁺ in the same way. Sodium oxalate supplies a common ion and lowers solubility. Extra solid does not change the concentrations in a solution that is already saturated.
The Ksp of CaF₂ is 3.9 × 10⁻¹¹ at 25 °C. Assuming complete dissociation of NaF, no hydrolysis of F⁻, and that the fluoride contributed by CaF₂ is negligible, the molar solubility of CaF₂ in 0.010 M NaF is:
1. 3.9 × 10⁻⁹ M, since s = Ksp/[F⁻] when the fluoride level is fixed at 0.010 M
2. 2.1 × 10⁻⁴ M, the same as in pure water, because the Ksp of the salt is unchanged
3. 9.8 × 10⁻⁸ M, since s = Ksp/(2[F⁻])² once the factor of two in CaF₂ is applied
4. 3.9 × 10⁻⁷ M, since s = Ksp/[F⁻]² when the fluoride level is fixed at 0.010 M
Show answer
Answer: D — 3.9 × 10⁻⁷ M, since s = Ksp/[F⁻]² when the fluoride level is fixed at 0.010 M
Ksp = [Ca²⁺][F⁻]² with [F⁻] = 0.010 M from NaF, so s = [Ca²⁺] = 3.9 × 10⁻¹¹/(0.010)² = 3.9 × 10⁻⁷ M, about 550 times less than in pure water (2.1 × 10⁻⁴ M). The factor of two belongs to the fluoride released by CaF₂ itself, which is negligible here, so squaring 2[F⁻] is wrong; dividing by [F⁻] without squaring ignores the exponent in Ksp.
For the gas-phase reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between Kp (pressures in bar) and Kc (concentrations in mol L⁻¹) is:
1. Kp = Kc(RT)⁻²
2. Kp = Kc(RT)²
3. Kp = Kc
4. Kp = Kc(RT)⁻⁴
Show answer
Answer: A — Kp = Kc(RT)⁻²
Kp = Kc(RT)^Δn with Δn = moles of gaseous products − moles of gaseous reactants = 2 − 4 = −2, so Kp = Kc(RT)⁻². Taking Δn as +2 (reactants minus products) reverses the sign, and −4 comes from counting only the reactant side.
For N₂O₄(g) ⇌ 2NO₂(g) at 298 K, Kp = 0.148 with partial pressures in bar. Pure N₂O₄ is allowed to reach equilibrium at 298 K and a total pressure of 0.50 bar. Treating the gases as ideal, the mole fraction of NO₂ in the equilibrium mixture, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.42
Kx = Kp/P^Δn = 0.148/0.50 = 0.296 (Δn = +1). With x the mole fraction of NO₂, the N₂O₄ fraction is 1 − x, so x²/(1 − x) = 0.296, i.e. x² + 0.296x − 0.296 = 0, giving x = [−0.296 + √(0.0876 + 1.184)]/2 = 0.416, so 0.42. Check via α: α² = Kp/(4P + Kp) = 0.148/2.148, α = 0.262, x = 2α/(1 + α) = 0.416. At 1.00 bar the fraction would be only 0.32, as Le Chatelier's principle predicts.
The equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is established in a cylinder. The mixture is then compressed to double the total pressure, and equilibrium is re-established at the same temperature. Which statement is correct?
1. Kp is unchanged, Kx falls to half its earlier value, and the degree of dissociation of PCl₅ decreases
2. Kp doubles because every partial pressure doubles, while Kx and the degree of dissociation stay as before
3. Kp and Kx are both unchanged, so the degree of dissociation of PCl₅ is the same as before compression
4. Kp is unchanged, Kx doubles to compensate for the higher pressure, and the degree of dissociation increases
Show answer
Answer: A — Kp is unchanged, Kx falls to half its earlier value, and the degree of dissociation of PCl₅ decreases
Kp depends only on temperature. With Δn = +1, Kx = Kp/P, so doubling P halves Kx, and from Kp = α²P/(1 − α²) the degree of dissociation α must fall: the system shifts towards the side with fewer gas moles. Kp cannot double because the system re-equilibrates; Kx would stay constant only if Δn were zero.