CourseCourseQ BankQuestion BankDocsDocuments

25.1 Acids and bases

Syllabus
9701–2028–2029
Topic
25.1
Level
A2

Learning objectives

25.1.1Terms: conjugate acid and conjugate base• Use terms: conjugate acid and conjugate base25.1.2Conjugate acid–base pairs, identifying such• Define conjugate acid–base pairs, identifying such pairs in reactions25.1.3Math definitions: pH, Ka, pKa and Kw and use• Define mathematically: pH, Ka, pKa and Kw and use them in calculations (Kb and the equation Kw = Ka × Kb will not be tested)25.1.4[H+(aq)] and pH values• Calculate [H+(aq)] and pH values for:: (a) strong acids; (b) strong alkalis; (c) weak acids25.1.5Buffer solution• – in controlling pH in blood- (a) define a buffer solution- (b) explain how a buffer solution can be made- (c) explain how buffer solutions control pH; use chemical equations in these explanations- (d) describe and explain the uses of buffer solutions, including the role of HCO 325.1.6Relevant concentrations and the pH of buffer• Calculate relevant concentrations and the pH of buffer solutions, given appropriate data25.1.7Terms: solubility product, K sp• Use terms: solubility product, K sp25.1.8Write an expression for Ksp• Write an expression for Ksp25.1.9Ksp from concentrations and vice versa• Calculate Ksp from concentrations and vice versa25.1.10Common ion effect and solubility• Solution containing a common ion: (a) understand and use the common ion effect to explain the different solubility of a compound in a; (b) perform calculations using Ksp values and concentration of a common ion

Use conjugate acid and conjugate base as proton-transfer terms

Starting species Proton transfer Resulting species
Brønsted–Lowry acid donates H⁺ its conjugate base
Brønsted–Lowry base accepts H⁺ its conjugate acid

A species and its conjugate differ by exactly one H⁺. Losing H⁺ lowers charge by 1; gaining H⁺ raises charge by 1.

NHX3+HX+→NHX4X+HX2O→HX++OHX−\ce{NH3 + H+ -> NH4+}\qquad\ce{H2O -> H+ + OH-}

Track the transferred proton rather than looking only for opposite charges. Water may act as an acid or a base depending on its reaction partner.

Identify conjugate pairs across an acid–base reaction

HA+B⇌AX−+BHX+\ce{HA + B <=> A- + BH+}

Reactant role Product partner Conjugate pair
HA donates H⁺ A⁻ remains HA/A⁻
B accepts H⁺ BH⁺ forms BH⁺/B

Identify the donor and acceptor first. Then pair each reactant with the product that differs from it by one proton; confirm that every other atom is unchanged and the charge difference is one.

In HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻, the pairs are H₂CO₃/HCO₃⁻ and H₂O/OH⁻. Here HCO₃⁻ accepts H⁺, showing that its role depends on the equation.

Do not pair the acid reactant with the conjugate acid product. Members of one pair differ only by H⁺, not by two protons or another atom.

Use the mathematical definitions of pH, Ka, pKa and Kw

Quantity Mathematical definition Meaning
pH pH = −log₁₀[H⁺] logarithmic measure from hydrogen-ion concentration
Ka [H⁺][A⁻]/[HA] for HA ⇌ H⁺ + A⁻ acid dissociation constant
pKa −log₁₀Ka logarithmic acid-strength scale
Kw [H⁺][OH⁻] ionic product of water at the stated temperature

[H+]=10−pHKa=10−pKa[\mathrm{H^+}]=10^{-\mathrm{pH}}\qquad K_a=10^{-\mathrm{p}K_a}

Larger Ka means greater acid dissociation; because of the negative logarithm, this corresponds to smaller pKa. At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

pH and pKa are logarithmic values, not concentrations. Do not introduce Kb or Kw = Ka × Kb: the syllabus explicitly excludes them.

Choose the correct pH model for strong acids, strong alkalis and weak acids

Solution model Route to [H⁺]
strong monoprotic acid complete ionisation: [H⁺] = acid concentration
strong alkali find [OH⁻], then [H⁺] = Kw/[OH⁻]
weak monoprotic acid HA partial ionisation: Ka = [H⁺]²/[HA] when [H⁺] = [A⁻] and dissociation is small

0.0100 mol dm−3 HCl:pH=−log⁡(0.0100)=2.000.0100\ \mathrm{mol\ dm^{-3}\ HCl}:\quad \mathrm{pH}=-\log(0.0100)=2.00

0.0100 mol dm−3 NaOH:[H+]=10−140.0100=10−12;pH=12.000.0100\ \mathrm{mol\ dm^{-3}\ NaOH}:\quad [H^+]=\frac{10^{-14}}{0.0100}=10^{-12};\quad \mathrm{pH}=12.00

0.100 mol dm−3 HA, Ka=1.74×10−5:[H+]≈Kac=1.32×10−3;pH=2.880.100\ \mathrm{mol\ dm^{-3}\ HA},\ K_a=1.74\times10^{-5}:\quad [H^+]\approx\sqrt{K_ac}=1.32\times10^{-3};\quad \mathrm{pH}=2.88

The weak-acid dissociation is 1.32% of 0.100 mol dm⁻³, so using the initial acid concentration in the denominator is reasonable. Apply stoichiometric ion numbers before these routes when the formula releases more than one relevant ion.

Do not assume a weak acid fully ionises or use pH + pOH = 14 without the 298 K Kw condition.

Buffers consume small additions of acid or base

A buffer solution resists a change in pH when a small amount of acid or base is added. It contains appreciable amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Buffer type How to make it
weak acid/conjugate base mix the weak acid with a soluble salt of its conjugate base, or partially neutralise the acid with strong alkali
weak base/conjugate acid mix the weak base with a soluble salt of its conjugate acid, or partially neutralise the base with strong acid

CHX3COOX−(aq)+HX+(aq)→CHX3COOH(aq)\ce{CH3COO-(aq) + H+(aq) -> CH3COOH(aq)}

CHX3COOH(aq)+OHX−(aq)→CHX3COOX−(aq)+HX2O(l)\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}

In blood, the H₂CO₃/HCO₃⁻ pair moderates pH: HCO₃⁻ consumes added H⁺ to form H₂CO₃, while H₂CO₃ consumes added OH⁻ to form HCO₃⁻ and water. Maintaining a narrow pH range supports pH-sensitive biological processes.

A buffer limits rather than prevents pH change. Its capacity is finite and requires both conjugate components; a neutral salt alone is not necessarily a buffer.

Calculate buffer pH from the conjugate-base-to-acid ratio

Ka=[H+][A−][HA]⇒[H+]=Ka[HA][A−]K_a=\frac{[H^+][A^-]}{[HA]}\quad\Rightarrow\quad[H^+]=K_a\frac{[HA]}{[A^-]}

pH=pKa+log⁡10([A−][HA])\mathrm{pH}=\mathrm{p}K_a+\log_{10}\left(\frac{[A^-]}{[HA]}\right)

For an ethanoic acid/ethanoate buffer at 298 K, Ka = 1.74 × 10⁻⁵ mol dm⁻³, [CH₃COOH] = 0.100 mol dm⁻³ and [CH₃COO⁻] = 0.150 mol dm⁻³.

[H+]=(1.74×10−5)0.1000.150=1.16×10−5;pH=4.94[H^+]=(1.74\times10^{-5})\frac{0.100}{0.150}=1.16\times10^{-5};\quad \mathrm{pH}=4.94

If acid or alkali has first reacted with the buffer, use mole stoichiometry to update HA and A⁻ before dividing by the common final volume. If both components share that volume, their mole ratio may be used directly.

Do not substitute initial concentrations after neutralisation or invert the ratio. Equal conjugate-base and acid concentrations give pH = pKa.

Ksp describes saturated dissolution of a sparingly soluble solid

The solubility product, Ksp, is the equilibrium constant for a sparingly soluble ionic solid in equilibrium with its aqueous ions in a saturated solution at a specified temperature.

AXmBXn(s)⇌m AX ∙  ∙  ∙ (aq)+n BX ∙  ∙  ∙ (aq)\ce{A_mB_n(s) <=> mA^{...}(aq) + nB^{...}(aq)}

Ksp uses equilibrium aqueous-ion concentrations; the pure solid is omitted. Its value is constant at fixed temperature even when a common ion changes the molar solubility.

A smaller Ksp often suggests lower solubility only when dissolution stoichiometries are comparable. Ksp is not itself the molar solubility.

Write Ksp from the balanced dissolution equilibrium

Dissolution equilibrium Ksp expression
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) [Ag⁺][Cl⁻]
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq) [Pb²⁺][I⁻]²
Al(OH)₃(s) ⇌ Al³⁺(aq) + 3OH⁻(aq) [Al³⁺][OH⁻]³

Balance the dissolution equation, omit the pure solid, multiply the aqueous-ion concentrations and use each stoichiometric coefficient as its power.

Powers come from equation coefficients, not ionic charges. Do not include the concentration of the undissolved solid.

Convert between molar solubility and Ksp

Let molar solubility be s, translate the balanced dissolution coefficients into equilibrium ion concentrations, then substitute those multiples of s into Ksp.

PbBrX2(s)⇌PbX2+(aq)+2 BrX−(aq)⇒Ksp=s(2s)2=4s3\ce{PbBr2(s) <=> Pb^{2+}(aq) + 2Br-(aq)}\quad\Rightarrow\quad K_{sp}=s(2s)^2=4s^3

For saturated PbBr₂ with s = 1.39 × 10⁻³ mol dm⁻³, [Pb²⁺] = 1.39 × 10⁻³ and [Br⁻] = 2.78 × 10⁻³ mol dm⁻³.

Ksp=(1.39×10−3)(2.78×10−3)2=1.07×10−8 mol3 dm−9K_{sp}=(1.39\times10^{-3})(2.78\times10^{-3})^2=1.07\times10^{-8}\ \mathrm{mol^3\ dm^{-9}}

For a supplied Ksp, form the same equation in s and solve the required square or cube root. Check that the resulting ion concentrations retain their stoichiometric ratios.

Do not set every ion concentration equal to s or use s² for every salt; the balanced dissolution stoichiometry determines the powers and numerical factors.

A common ion reduces solubility without changing Ksp

Adding an aqueous ion already present in a dissolution equilibrium raises that ion concentration. The equilibrium shifts toward the solid, so less salt dissolves; Ksp itself remains constant at fixed temperature.

CaFX2(s)⇌CaX2+(aq)+2 FX−(aq)Ksp=[Ca2+][F−]2\ce{CaF2(s) <=> Ca^{2+}(aq) + 2F-(aq)}\qquad K_{sp}=[Ca^{2+}][F^-]^2

Take Ksp(CaF₂) = 3.2 × 10⁻¹¹ mol³ dm⁻⁹. In pure water, Ksp = 4s³ gives s = 2.0 × 10⁻⁴ mol dm⁻³.

in 0.0100 mol dm−3 NaF:s=3.2×10−11(0.0100)2=3.2×10−7 mol dm−3\text{in }0.0100\ \mathrm{mol\ dm^{-3}\ NaF}:\quad s=\frac{3.2\times10^{-11}}{(0.0100)^2}=3.2\times10^{-7}\ \mathrm{mol\ dm^{-3}}

The added 0.0100 mol dm⁻³ F⁻ greatly exceeds the extra 2s, so [F⁻] ≈ 0.0100 is self-consistent. The common ion lowers the calculated solubility by more than two orders of magnitude.

Include dilution before testing concentrations. A precipitate forms when the relevant ion product exceeds Ksp; adding a common ion does not change the numerical Ksp.