Ionic Equilibrium — pH, Buffers & Solubility Product

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Ionic equilibrium is chemical equilibrium applied to ions in water. Almost every question comes down to one of five formulas: pH of a weak acid, pH of a salt, pH of a buffer, Ksp from solubility, and whether a precipitate forms. Learn which situation calls for which formula and the numericals become routine.

What you will be able to do

• Use Arrhenius, Brønsted–Lowry and Lewis definitions and identify conjugate pairs
• Calculate pH for strong and weak acids and bases, including very dilute solutions
• Find the pH of salt solutions and buffers with the hydrolysis and Henderson equations
• Relate Ksp to solubility, apply the common-ion effect and predict precipitation
• Choose the right indicator for a titration

The big picture

Pick the situation, then the formula.

In one line: Strong electrolytes ionise fully; weak ones set up an equilibrium described by Ka, Kb or Ksp. pH, hydrolysis, buffers and precipitation are all that equilibrium seen from different angles.

Situation	Formula
Strong acid, concentration C	[H⁺] = C (add water’s 10⁻⁷ only if C < 10⁻⁶ M)
Weak acid	[H⁺] = √(Ka C); α = √(Ka / C)
Salt of weak acid + strong base	pH = 7 + ½pKa + ½log C
Salt of strong acid + weak base	pH = 7 − ½pKb − ½log C
Salt of weak acid + weak base	pH = 7 + ½pKa − ½pKb
Acidic buffer	pH = pKa + log([salt]/[acid])
Sparingly soluble salt	Ksp from s; precipitate if ionic product > Ksp

Think of it like this: A weak acid in water is like a crowd at a gate where only a few people pass through at a time. Add more people at the gate (a common ion) and even fewer get through. Add more water (dilution) and a larger fraction passes, even though the total number per litre falls.

Acid–base definitions

Three definitions, each wider than the last.

Theory	Acid	Base	Example
Arrhenius	Gives H⁺ in water	Gives OH⁻ in water	HCl, NaOH
Brønsted–Lowry	Proton donor	Proton acceptor	NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Lewis	Electron-pair acceptor	Electron-pair donor	BF₃ (acid), NH₃ (base)

• Conjugate pairs differ by one H⁺: HCl/Cl⁻, NH₄⁺/NH₃, H₂O/OH⁻, H₃O⁺/H₂O.
• Strong acid → very weak conjugate base, and vice versa.
• Ka × Kb = Kw for a conjugate pair; so pKa + pKb = 14 at 298 K.
• Amphiprotic species can both give and accept H⁺: H₂O, HCO₃⁻, HSO₄⁻, H₂PO₄⁻.
• Lewis acids include electron-deficient molecules (BF₃, AlCl₃), cations (Ag⁺, Fe³⁺) and molecules with polar multiple bonds (CO₂, SO₂).

Relative strength of acids and bases

What makes one acid stronger than another.

• Down a group, bond strength decides: HF < HCl < HBr < HI in acid strength. The H–I bond is longest and weakest, so it releases H⁺ most easily, even though F is the most electronegative.
• Across a period, electronegativity decides: CH₄ < NH₃ < H₂O < HF. The more electronegative atom stabilises the negative charge left behind.
• Oxyacids of one element: more oxygen atoms → stronger acid, because the negative charge on the anion spreads over more O atoms. HClO < HClO₂ < HClO₃ < HClO₄.
• Oxyacids with the same structure: more electronegative central atom → stronger. HOCl > HOBr > HOI; H₂SO₄ > H₂SeO₄.
• Levelling effect: in water, HCl, HBr, HI, HNO₃, H₂SO₄ and HClO₄ all appear equally strong — each converts completely to H₃O⁺, the strongest acid that can exist in water. In a less basic solvent such as acetic acid their differences show: HClO₄ is the strongest.

Acid	Ka (298 K)	pKa
HF	3.5 × 10⁻⁴	3.45
HNO₂	4.5 × 10⁻⁴	3.35
HCOOH	1.8 × 10⁻⁴	3.74
CH₃COOH	1.8 × 10⁻⁵	4.74
HCN	4.9 × 10⁻¹⁰	9.31
Phenol	1.3 × 10⁻¹⁰	9.89

Base	Kb (298 K)	pKb
NH₃	1.8 × 10⁻⁵	4.74
CH₃NH₂	4.4 × 10⁻⁴	3.36
(CH₃)₂NH	5.4 × 10⁻⁴	3.27
C₆H₅NH₂	4.3 × 10⁻¹⁰	9.37
Pyridine	1.8 × 10⁻⁹	8.75

Check yourself

• C1. Arrange HClO, HClO₃ and HClO₄ in increasing acid strength.
• C2. Why is HF a weaker acid than HCl in water?

Answers at the end.

Kw and the pH scale

Water ionises a little, and that sets the scale.

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K
pH = −log[H⁺] pOH = −log[OH⁻] pH + pOH = 14

pKw changes with temperature

Key idea: Kw increases with temperature (ionisation of water is endothermic). At higher temperature, neutral water has pH below 7 — but it is still neutral because [H⁺] = [OH⁻].

Solved example 01 — A very dilute strong acid

Find the pH of 1.0 × 10⁻⁸ M HCl.

1. Writing pH = 8 would make an acid basic — wrong. Water’s own H⁺ cannot be ignored.
2. Let [H⁺] = 10⁻⁸ + x and [OH⁻] = x, with (10⁻⁸ + x)x = 10⁻¹⁴.
3. Solving: x ≈ 0.95 × 10⁻⁷, so [H⁺] ≈ 1.05 × 10⁻⁷ M.

Answer: pH ≈ 6.98 — slightly acidic, as it must be.

Solved example 02 — Mixing strong acid and base

50 mL of 0.10 M HCl is mixed with 50 mL of 0.08 M NaOH. Find the pH.

1. Moles H⁺ = 5.0 mmol; moles OH⁻ = 4.0 mmol; excess H⁺ = 1.0 mmol.
2. Total volume = 100 mL → [H⁺] = 0.010 M.

Answer: pH = 2.

pH of mixtures

• Two strong acids: add the moles of H⁺ and divide by the total volume.
• Strong acid + weak acid: the strong acid fixes [H⁺]; the weak acid’s ionisation is suppressed (common ion) and can usually be ignored.
• Strong acid + strong base: find the excess moles of H⁺ or OH⁻ and divide by the total volume.
• Weak acid + strong base (not in excess): a buffer forms — use the Henderson equation (see Buffers).

Solved example 03 — Strong and weak acid together

Find the pH of a solution 0.10 M in HCl and 0.10 M in CH₃COOH, and the degree of ionisation of acetic acid in it. Ka = 1.8 × 10⁻⁵.

1. HCl is fully ionised: [H⁺] ≈ 0.10 M → pH = 1.
2. For acetic acid: Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] → [CH₃COO⁻] = 1.8 × 10⁻⁵ × 0.10 / 0.10 = 1.8 × 10⁻⁵ M.
3. α = 1.8 × 10⁻⁵ / 0.10 = 1.8 × 10⁻⁴.

Answer: pH = 1; α of acetic acid falls from 1.3% (alone) to 0.018%.

Weak acids and bases

Partial ionisation, governed by Ka or Kb.

HA ⇌ H⁺ + A⁻ Ka = Cα² / (1 − α) ≈ Cα²
α = √(Ka / C) [H⁺] = Cα = √(Ka C)

Ostwald’s dilution law; the approximation holds when α < about 5%

• Dilution increases α but decreases [H⁺]: diluting 100 times raises α 10 times while [H⁺] falls 10 times.
• For a weak base: [OH⁻] = √(Kb C), then pH = 14 − pOH.
• Polyprotic acids (H₂S, H₃PO₄, H₂CO₃): Ka₁ ≫ Ka₂ ≫ Ka₃; the pH comes almost entirely from the first step.
• Common-ion effect: adding a salt with a shared ion (CH₃COONa to CH₃COOH) suppresses ionisation of the weak electrolyte.

Solved example 04 — pH of a weak acid

Find the pH and degree of ionisation of 0.10 M acetic acid. Ka = 1.8 × 10⁻⁵.

1. α = √(1.8 × 10⁻⁵ / 0.10) = √(1.8 × 10⁻⁴) = 1.34 × 10⁻² (1.34%).
2. [H⁺] = Cα = 1.34 × 10⁻³ M.
3. pH = −log(1.34 × 10⁻³) = 3 − 0.127.

Answer: pH ≈ 2.87; α ≈ 1.3%.

Polyprotic acids in detail

H₂S ⇌ H⁺ + HS⁻ Ka₁ = 9.1 × 10⁻⁸
HS⁻ ⇌ H⁺ + S²⁻ Ka₂ = 1.2 × 10⁻¹³
Ka₁ × Ka₂ = [H⁺]² [S²⁻] / [H₂S]

The combined expression is what group analysis uses

• Each step is weaker than the one before: removing H⁺ from a negative ion is harder.
• In a solution of H₂S alone, [S²⁻] ≈ Ka₂ — independent of concentration.
• Adding a strong acid raises [H⁺] and drives [S²⁻] down by the square of [H⁺].

Solved example 05 — Sulphide ion in acid

Saturated H₂S is 0.10 M. Find [S²⁻] when the solution also contains 0.10 M HCl.

1. [H⁺] = 0.10 M (from HCl).
2. [S²⁻] = Ka₁Ka₂[H₂S] / [H⁺]² = (9.1 × 10⁻⁸)(1.2 × 10⁻¹³)(0.10) / (0.10)².
3. = 1.09 × 10⁻²¹ / 0.010.

Answer: [S²⁻] ≈ 1.1 × 10⁻¹⁹ M — tiny, so only sulphides with very small Ksp (group II) precipitate.

Check yourself

• C3. What is the conjugate base of HSO₄⁻?
• C4. A weak acid is diluted 4 times. By what factor does α change?

Answers at the end.

Salt hydrolysis

Ions of weak partners react with water.

Salt from	Example	Ion that hydrolyses	Solution	pH
Strong acid + strong base	NaCl, KNO₃	None	Neutral	7
Weak acid + strong base	CH₃COONa, Na₂CO₃	Anion (gives OH⁻)	Basic	7 + ½pKa + ½log C
Strong acid + weak base	NH₄Cl, CuSO₄	Cation (gives H⁺)	Acidic	7 − ½pKb − ½log C
Weak acid + weak base	CH₃COONH₄	Both	Depends on Ka vs Kb	7 + ½pKa − ½pKb

Key idea: For a weak acid–weak base salt the pH does not depend on concentration. CH₃COONH₄ has pKa ≈ pKb ≈ 4.75, so its solution is almost exactly neutral — yet it is hydrolysed.

Solved example 06 — pH of sodium acetate

Find the pH of 0.10 M CH₃COONa. pKa (CH₃COOH) = 4.74.

1. Weak acid + strong base → pH = 7 + ½pKa + ½log C.
2. = 7 + 2.37 + ½(−1) = 7 + 2.37 − 0.50.

Answer: pH ≈ 8.87 (basic).

Degree of hydrolysis

Salt type	Hydrolysis constant Kh	Degree of hydrolysis h
Weak acid + strong base	Kw / Ka	√(Kh / C)
Strong acid + weak base	Kw / Kb	√(Kh / C)
Weak acid + weak base	Kw / (Ka × Kb)	√Kh (independent of C)

Solved example 07 — Hydrolysis of ammonium chloride

Find the hydrolysis constant, degree of hydrolysis and pH of 0.10 M NH₄Cl. Kb (NH₃) = 1.8 × 10⁻⁵.

1. Kh = Kw / Kb = 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.
2. h = √(5.6 × 10⁻¹⁰ / 0.10) = 7.5 × 10⁻⁵.
3. [H⁺] = Ch = 7.5 × 10⁻⁶ M → pH = 5.13.

Answer: Kh = 5.6 × 10⁻¹⁰; h = 7.5 × 10⁻⁵; pH ≈ 5.13 (the formula 7 − ½pKb − ½log C gives the same).

Buffers

Solutions that resist a change in pH.

Acidic buffer: pH = pKa + log ([salt] / [acid])
Basic buffer: pOH = pKb + log ([salt] / [base])

Henderson–Hasselbalch equation

• Acidic buffer: weak acid + its salt with a strong base (CH₃COOH + CH₃COONa). Basic buffer: weak base + its salt with a strong acid (NH₄OH + NH₄Cl).
• Added H⁺ is taken up by the conjugate base; added OH⁻ by the weak acid.
• Maximum buffer capacity when [salt] = [acid], i.e. pH = pKa. The useful range is pKa ± 1.
• A buffer is also made by partly neutralising a weak acid with a strong base: half-neutralisation gives pH = pKa.
• Blood is buffered near pH 7.4 mainly by H₂CO₃/HCO₃⁻.

Solved example 08 — Buffer pH

Find the pH of a solution 0.10 M in CH₃COOH and 0.20 M in CH₃COONa. pKa = 4.74.

1. pH = 4.74 + log(0.20 / 0.10) = 4.74 + log 2.
2. = 4.74 + 0.30.

Answer: pH ≈ 5.04.

Solved example 09 — Adding acid to a buffer

To 1.0 L of buffer containing 0.10 mol CH₃COOH and 0.10 mol CH₃COONa, 0.010 mol HCl is added. Find the new pH (pKa = 4.74).

1. H⁺ reacts with acetate: CH₃COO⁻ + H⁺ → CH₃COOH.
2. Acid: 0.10 + 0.010 = 0.11 mol; salt: 0.10 − 0.010 = 0.09 mol.
3. pH = 4.74 + log(0.09 / 0.11) = 4.74 − 0.087.

Answer: pH ≈ 4.65. The same HCl in 1 L of pure water would give pH 2 — a drop of 5 units instead of 0.09.

Solved example 10 — Designing a buffer

In what ratio must sodium acetate and acetic acid be mixed to get a buffer of pH 5.74? (pKa = 4.74)

1. 5.74 = 4.74 + log([salt]/[acid]).
2. log([salt]/[acid]) = 1.

Answer: [salt] : [acid] = 10 : 1.

Check yourself

• C5. What pH does a buffer have when [salt] = [acid]?
• C6. Is a mixture of HCl and NaCl a buffer?

Answers at the end.

Titration curves

How pH changes as base is added to acid.

Point in the titration	Weak acid (HA) + NaOH	What to calculate
Start	Weak acid alone	[H⁺] = √(Ka C)
Before equivalence	Buffer of HA + A⁻	Henderson equation
Half-way to equivalence	[HA] = [A⁻]	pH = pKa
Equivalence point	Salt NaA only	Hydrolysis: pH = 7 + ½pKa + ½log C
After equivalence	Excess OH⁻	[OH⁻] from excess base

Solved example 11 — Points on a titration curve

25 mL of 0.10 M CH₃COOH is titrated with 0.10 M NaOH (pKa = 4.74). Find the pH (a) at the start, (b) after 12.5 mL, (c) at 25 mL.

1. (a) [H⁺] = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ → pH 2.87.
2. (b) Half-neutralised: [salt] = [acid] → pH = pKa = 4.74.
3. (c) All acid is now 2.5 mmol of acetate in 50 mL → C = 0.050 M; pH = 7 + 2.37 + ½log 0.050 = 9.37 − 0.65.

Answer: (a) 2.87 (b) 4.74 (c) 8.72 — so phenolphthalein is the right indicator.

Key idea: Half-equivalence point gives pKa directly. This is how Ka of a weak acid is measured in the lab: titrate, find the volume at equivalence, read the pH at half that volume.

Solubility product

Equilibrium between a solid and its ions in a saturated solution.

AₓBᵧ(s) ⇌ xA^y+ + yB^x− Ksp = (xs)^x (ys)^y = x^x y^y s^x+y

s = molar solubility (mol L⁻¹)

Type	Example	Ksp in terms of s	s in terms of Ksp
AB	AgCl, BaSO₄	s²	√Ksp
AB₂ or A₂B	PbCl₂, Ag₂CrO₄	4s³	∛(Ksp/4)
AB₃	Fe(OH)₃, Al(OH)₃	27s⁴	(Ksp/27) to the ¼
A₃B₂	Ca₃(PO₄)₂	108s⁵	(Ksp/108) to the ⅕

Key idea: Compare solubilities only through s, not Ksp, unless the salts are of the same type. Ag₂CrO₄ (Ksp ≈ 1.1 × 10⁻¹²) is more soluble than AgCl (Ksp ≈ 1.8 × 10⁻¹⁰) because its formula type is different.

• Precipitation rule: if the ionic product Q > Ksp, a precipitate forms; if Q < Ksp, more solid can dissolve.
• Common-ion effect lowers solubility: AgCl is far less soluble in NaCl solution than in water.
• Group analysis: group II sulphides (CuS, PbS) have very small Ksp, so they precipitate with H₂S in acid, where HCl (common H⁺) keeps [S²⁻] low; group IV sulphides (ZnS, MnS) need ammoniacal solution, where [S²⁻] is higher.
• Purification of common salt: passing HCl gas into a saturated NaCl solution precipitates pure NaCl (common Cl⁻).

Solved example 12 — Solubility and common ion

Ksp of AgCl = 1.8 × 10⁻¹⁰. Find its solubility (a) in water, (b) in 0.10 M NaCl.

1. (a) Ksp = s² → s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M.
2. (b) [Cl⁻] ≈ 0.10 M from NaCl; s × 0.10 = 1.8 × 10⁻¹⁰.

Answer: (a) 1.34 × 10⁻⁵ M. (b) 1.8 × 10⁻⁹ M — about 7400 times less.

Solved example 13 — Will it precipitate?

Equal volumes of 2 × 10⁻⁴ M AgNO₃ and 2 × 10⁻⁴ M NaCl are mixed. Does AgCl precipitate? (Ksp = 1.8 × 10⁻¹⁰)

1. Mixing halves each concentration: [Ag⁺] = [Cl⁻] = 1 × 10⁻⁴ M.
2. Q = (1 × 10⁻⁴)² = 1 × 10⁻⁸.
3. Q (10⁻⁸) > Ksp (1.8 × 10⁻¹⁰).

Answer: Yes, AgCl precipitates.

Solved example 14 — Ksp from grams per litre

The solubility of BaSO₄ is 2.45 × 10⁻³ g L⁻¹ at 298 K. Find Ksp. (Molar mass 233 g mol⁻¹)

1. s = 2.45 × 10⁻³ / 233 = 1.05 × 10⁻⁵ mol L⁻¹.
2. Ksp = s² = (1.05 × 10⁻⁵)².

Answer: Ksp ≈ 1.1 × 10⁻¹⁰.

Solved example 15 — At what pH does a hydroxide precipitate?

A solution is 0.010 M in Mg²⁺. Above what pH does Mg(OH)₂ begin to precipitate? (Ksp = 1.8 × 10⁻¹¹)

1. Ksp = [Mg²⁺][OH⁻]² → [OH⁻]² = 1.8 × 10⁻¹¹ / 0.010 = 1.8 × 10⁻⁹.
2. [OH⁻] = 4.2 × 10⁻⁵ M → pOH = 4.37.

Answer: Above pH ≈ 9.63.

• Why NH₄Cl is added with NH₄OH in group III: the common NH₄⁺ ion lowers [OH⁻], so only the very insoluble hydroxides (Fe³⁺, Al³⁺, Cr³⁺) precipitate; Mg(OH)₂ and the group IV–VI hydroxides stay in solution.

Check yourself

• C7. How does the solubility of AgCl change in 0.1 M AgNO₃ compared with water?
• C8. A salt AB₂ has solubility s. Write its Ksp.

Answers at the end.

Indicators and titrations

Choose an indicator that changes colour at the equivalence-point pH.

Indicator	pH range	Colour change (acid → base)
Methyl orange	3.1 – 4.4	Red → yellow
Methyl red	4.2 – 6.3	Red → yellow
Phenolphthalein	8.2 – 10.0	Colourless → pink

HIn ⇌ H⁺ + In⁻ pH = pKIn + log([In⁻] / [HIn])

The colour changes over about pKIn ± 1, where one form is ten times the other

An indicator is itself a weak acid (or base) whose two forms have different colours. Phenolphthalein’s HIn form is colourless and its In⁻ form pink; in acid it stays as HIn, in base it changes to In⁻.

Titration	pH at equivalence	Suitable indicator
Strong acid – strong base	7	Either (sharp jump from about 4 to 10)
Weak acid – strong base	> 7	Phenolphthalein
Strong acid – weak base	< 7	Methyl orange
Weak acid – weak base	≈ 7, no sharp jump	No simple indicator works well

Exam trap gallery

Where marks are lost in ionic equilibrium.

#	What students write	What is actually correct	Why
1	pH of 10⁻⁸ M HCl is 8.	About 6.98.	Water’s own H⁺ (10⁻⁷ M) is larger than the acid’s.
2	Dilution increases [H⁺] of a weak acid because α increases.	α increases but [H⁺] = √(Ka C) decreases.	The concentration falls faster than α rises.
3	Comparing Ksp directly tells which salt is more soluble.	Only for salts of the same formula type; otherwise compare s.	Ag₂CrO₄ is more soluble than AgCl despite a smaller Ksp.
4	CH₃COONH₄ is not hydrolysed because its solution is neutral.	It is hydrolysed; the acid and base effects cancel.	pKa ≈ pKb.
5	HCl + NaCl is a buffer.	No: a buffer needs a weak acid (or base) and its conjugate.	A strong acid has no reserve to absorb added OH⁻.
6	Pure water at 350 K with pH 6.6 is acidic.	It is neutral: [H⁺] = [OH⁻].	Kw rises with temperature.
7	Phenolphthalein is suitable for a strong acid – weak base titration.	Use methyl orange (equivalence pH < 7).	Phenolphthalein changes colour above pH 8.
8	Adding NaCl increases the solubility of AgCl.	It decreases it (common ion).	Q rises above Ksp until AgCl precipitates.
9	HI is a weaker acid than HF because I is less electronegative.	HI is the strongest hydrohalic acid.	Bond strength, not electronegativity, decides down a group.
10	At the equivalence point of any acid–base titration, pH = 7.	Only for strong acid – strong base.	Salt hydrolysis shifts the pH.
11	At half-equivalence, pH = 7.	At half-equivalence, pH = pKa.	[salt] = [acid] in the Henderson equation.
12	Adding NH₄Cl to NH₄OH increases [OH⁻].	It decreases [OH⁻].	Common-ion effect of NH₄⁺.

Memory hook: “Weak acid: root of Ka C.” · “Seven plus half pKa plus half log C” — salt of weak acid, strong base. · “Equal salt and acid → pH = pKa.” · “Same type, compare Ksp; different type, compare s.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The conjugate base of H₂O is: (Easy)

• (a) H₃O⁺
• (b) OH⁻
• (c) O²⁻
• (d) H₂

Q02. BF₃ is a: (Easy)

• (a) Brønsted acid
• (b) Lewis acid
• (c) Lewis base
• (d) Arrhenius base

Q03. pH of 0.001 M NaOH is: (Easy)

• (a) 3
• (b) 11
• (c) 10
• (d) 14

Q04. An aqueous solution of NH₄Cl is: (Easy)

• (a) neutral
• (b) acidic
• (c) basic
• (d) amphoteric

Q05. Ksp of a salt AB₂ with solubility s is: (Easy)

• (a) s²
• (b) 2s³
• (c) 4s³
• (d) s³

Q06. pH of 10⁻⁸ M HCl is closest to: (Moderate)

• (a) 8.0
• (b) 7.0
• (c) 6.98
• (d) 6.0

Q07. A buffer of CH₃COOH and CH₃COONa has pH = pKa when: (Moderate)

• (a) [salt] = 2[acid]
• (b) [salt] = [acid]
• (c) [acid] = 0
• (d) [salt] = 0

Q08. pH of 0.1 M CH₃COONa (pKa = 4.74) is about: (Moderate)

• (a) 8.87
• (b) 5.13
• (c) 9.87
• (d) 7.00

Q09. Indicator for titrating NH₄OH with HCl: (Moderate)

• (a) phenolphthalein
• (b) methyl orange
• (c) either
• (d) none

Q10. Solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) in 0.01 M NaCl is: (Moderate)

• (a) 1.8 × 10⁻⁸ M
• (b) 1.34 × 10⁻⁵ M
• (c) 1.8 × 10⁻¹² M
• (d) 1.8 × 10⁻⁶ M

Q11. Group II sulphides are precipitated in acidic medium because: (Moderate)

• (a) acid increases [S²⁻]
• (b) acid lowers [S²⁻] so only low-Ksp sulphides precipitate
• (c) acid dissolves group IV cations
• (d) H₂S is a strong acid

Q12. Which salt solution has pH independent of concentration? (Moderate)

• (a) NH₄Cl
• (b) CH₃COONa
• (c) CH₃COONH₄
• (d) Na₂CO₃

Q13. A weak acid HA (0.01 M) is 1% ionised. Its Ka is about: (Tough)

• (a) 10⁻⁴
• (b) 10⁻⁶
• (c) 10⁻⁸
• (d) 10⁻²

Q14. Which is the most soluble in water (molar solubility)? AgCl (Ksp 1.8 × 10⁻¹⁰), AgBr (5 × 10⁻¹³), Ag₂CrO₄ (1.1 × 10⁻¹²) (Tough)

• (a) AgCl
• (b) AgBr
• (c) Ag₂CrO₄
• (d) all equal

Q15. 50 mL of 0.2 M CH₃COOH is mixed with 50 mL of 0.1 M NaOH (pKa = 4.74). The pH is: (Tough)

• (a) 4.74
• (b) 2.87
• (c) 8.72
• (d) 7.00

Q16. Strongest acid in water among these: (Easy)

• (a) HF
• (b) HCl
• (c) HBr
• (d) HI

Q17. Strongest oxyacid: (Easy)

• (a) HClO
• (b) HClO₂
• (c) HClO₃
• (d) HClO₄

Q18. pH of a solution with [OH⁻] = 10⁻⁵ M is: (Easy)

• (a) 5
• (b) 9
• (c) 10
• (d) 4

Q19. At the half-equivalence point of CH₃COOH vs NaOH, pH equals: (Moderate)

• (a) 7
• (b) pKa
• (c) pKb
• (d) 14 − pKa

Q20. Kh for CH₃COONa equals: (Moderate)

• (a) Ka/Kw
• (b) Kw/Ka
• (c) Kw/Kb
• (d) Ka × Kb

Q21. The degree of hydrolysis of CH₃COONH₄ depends on: (Moderate)

• (a) concentration only
• (b) Ka and Kb only
• (c) concentration and Ka
• (d) nothing

Q22. NH₄Cl is added before NH₄OH in group III analysis to: (Moderate)

• (a) increase [OH⁻]
• (b) decrease [OH⁻]
• (c) increase [NH₃]
• (d) dissolve Fe(OH)₃

Q23. Ksp of Ca₃(PO₄)₂ in terms of solubility s is: (Moderate)

• (a) 108s⁵
• (b) 27s⁴
• (c) 6s⁵
• (d) 36s⁵

Q24. Solubility of Ag₂CrO₄ (Ksp = 1.1 × 10⁻¹²) in 0.10 M AgNO₃ is: (Tough)

• (a) 1.1 × 10⁻¹⁰ M
• (b) 1.1 × 10⁻¹¹ M
• (c) 6.5 × 10⁻⁵ M
• (d) 1.1 × 10⁻⁶ M

Q25. pH at the equivalence point when 0.10 M CH₃COOH is titrated with 0.10 M NaOH (pKa 4.74): (Tough)

• (a) 7.00
• (b) 8.72
• (c) 4.74
• (d) 10.0

Answer key with explanations

1. (b) OH⁻ — Remove one H⁺.
2. (b) Lewis acid — Accepts an electron pair.
3. (b) 11 — pOH = 3.
4. (b) acidic — Strong acid + weak base salt.
5. (c) 4s³ — (s)(2s)².
6. (c) 6.98 — Include water’s H⁺.
7. (b) [salt] = [acid] — log 1 = 0.
8. (a) 8.87 — 7 + 2.37 − 0.5.
9. (b) methyl orange — Equivalence pH < 7.
10. (a) 1.8 × 10⁻⁸ M — Ksp / 0.01.
11. (b) acid lowers [S²⁻] so only low-Ksp sulphides precipitate — Common-ion effect of H⁺.
12. (c) CH₃COONH₄ — Weak acid – weak base salt.
13. (b) 10⁻⁶ — Ka = Cα² = 0.01 × 10⁻⁴.
14. (c) Ag₂CrO₄ — s = ∛(Ksp/4) ≈ 6.5 × 10⁻⁵ M.
15. (a) 4.74 — Half-neutralised → equal acid and salt.
16. (d) HI — Weakest H–X bond.
17. (d) HClO₄ — Most O atoms to spread the charge.
18. (b) 9 — pOH 5 → pH 9.
19. (b) pKa — Equal acid and salt.
20. (b) Kw/Ka — Anion of a weak acid hydrolyses.
21. (b) Ka and Kb only — h = √(Kw/KaKb).
22. (b) decrease [OH⁻] — Common ion suppresses NH₄OH.
23. (a) 108s⁵ — (3s)³(2s)².
24. (a) 1.1 × 10⁻¹⁰ M — Ksp = (0.10)² s.
25. (b) 8.72 — 0.050 M acetate.

Advanced problems

Each needs two ideas from the chapter.

Solved example 16 — Ka from pH

A 0.050 M solution of a weak monobasic acid has pH 3.0. Find Ka and α.

1. [H⁺] = 1.0 × 10⁻³ M; α = 10⁻³ / 0.050 = 0.02 (2%).
2. Ka ≈ Cα² = 0.050 × (0.02)² = 2.0 × 10⁻⁵.

Answer: Ka ≈ 2.0 × 10⁻⁵; α = 2%.

Solved example 17 — Selective precipitation

A solution is 0.01 M in both Cl⁻ and CrO₄²⁻. AgNO₃ is added slowly. Which precipitates first? (Ksp: AgCl 1.8 × 10⁻¹⁰; Ag₂CrO₄ 1.1 × 10⁻¹²)

1. [Ag⁺] needed for AgCl: 1.8 × 10⁻¹⁰ / 0.01 = 1.8 × 10⁻⁸ M.
2. [Ag⁺] needed for Ag₂CrO₄: √(1.1 × 10⁻¹² / 0.01) = √(1.1 × 10⁻¹⁰) ≈ 1.05 × 10⁻⁵ M.
3. The smaller [Ag⁺] is reached first.

Answer: AgCl precipitates first — the basis of Mohr’s titration, where red Ag₂CrO₄ appears only after the chloride is used up.

Solved example 18 — Common ion with a 2:1 salt

Find the solubility of Ag₂CrO₄ in 0.10 M AgNO₃. Ksp = 1.1 × 10⁻¹².

1. [Ag⁺] ≈ 0.10 M from AgNO₃; [CrO₄²⁻] = s.
2. Ksp = (0.10)² × s → s = 1.1 × 10⁻¹² / 0.010.

Answer: s = 1.1 × 10⁻¹⁰ M — about 600 000 times lower than in water (6.5 × 10⁻⁵ M).

Solved example 19 — Basic buffer

Find the pH of a solution 0.20 M in NH₃ and 0.10 M in NH₄Cl. pKb = 4.74.

1. pOH = pKb + log([salt]/[base]) = 4.74 + log(0.10/0.20) = 4.74 − 0.30 = 4.44.
2. pH = 14 − 4.44.

Answer: pH ≈ 9.56.

Check-yourself answers

• C1. HClO < HClO₃ < HClO₄.
• C2. The H–F bond is very strong (and F⁻ is strongly hydrated by H bonding to HF), so HF releases H⁺ far less readily.
• C3. SO₄²⁻.
• C4. α = √(Ka/C): C becomes C/4, so α doubles.
• C5. pH = pKa.
• C6. No — HCl is a strong acid; there is no weak acid/conjugate base pair.
• C7. It decreases (common Ag⁺ ion): s = Ksp/[Ag⁺] = 1.8 × 10⁻⁹ M in 0.1 M AgNO₃.
• C8. Ksp = (s)(2s)² = 4s³.

One-page revision

Read this page the night before the exam.

Definitions

• Arrhenius, Brønsted (H⁺), Lewis (e⁻ pair)
• Conjugate pair differs by one H⁺
• Ka × Kb = Kw; pKa + pKb = 14
• Amphiprotic: H₂O, HCO₃⁻, HSO₄⁻

pH basics

• Kw = 10⁻¹⁴ at 298 K; rises with T
• pH + pOH = 14
• 10⁻⁸ M HCl: pH ≈ 6.98
• Mix: find excess moles ÷ total volume

Weak electrolytes

• α = √(Ka/C); [H⁺] = √(Ka C)
• Dilution: α ↑, [H⁺] ↓
• Polyprotic: first step dominates
• Common ion suppresses ionisation

Hydrolysis and buffers

• WA–SB: 7 + ½pKa + ½log C
• SA–WB: 7 − ½pKb − ½log C
• WA–WB: 7 + ½pKa − ½pKb
• pH = pKa + log(salt/acid)

Solubility product

• AB: s²; AB₂: 4s³; AB₃: 27s⁴; A₃B₂: 108s⁵
• Q > Ksp → precipitate
• Common ion lowers s
• Group II: H₂S in acid; group IV: in NH₃

Indicators

• Methyl orange 3.1–4.4
• Phenolphthalein 8.2–10
• WA–SB: phenolphthalein
• SA–WB: methyl orange