Ionic Equilibrium

Water is never just water. In every aqueous solution a set of reversible ionic reactions sits at equilibrium — the water splitting into H⁺ and OH⁻, a weak acid half-ionised, a sparingly soluble salt dissolving as fast as it re-precipitates. This unit gives you the equilibrium constants that describe those processes (K_w, K_a, K_b, K_{sp}, K_h) and the arithmetic — almost all of it logarithms — that turns them into a pH or a solubility. It is the longest unit in the physical-chemistry area and the most heavily examined.

2.1 Limitation of Arrhenius concept

Svante Arrhenius (1887) defined an acid as a substance that gives H⁺ ions in aqueous solution and a base as one that gives OH⁻ ions. The definition works — HCl → H⁺ + Cl⁻ and NaOH → Na⁺ + OH⁻ — but it fails in five ways that the board asks about directly.

1. It is restricted to water. Acid-base behaviour in liquid ammonia, in benzene or in the gas phase is outside it. NH₃(g) + HCl(g) → NH₄Cl(s) is clearly a neutralisation, yet no water and no OH⁻ are involved.
2. Free H⁺ does not exist in solution. A bare proton is only about 10⁻¹⁵ m across and would have an enormous charge density; in water it is always hydrated as the hydronium ion, H₃O⁺.
3. It cannot explain acids without hydrogen. CO₂, SO₂, AlCl₃ and BF₃ all behave as acids, yet none of them contains an ionisable H.
4. It cannot explain bases without OH. NH₃, Na₂CO₃, CH₃NH₂ and NaHCO₃ all turn red litmus blue although no OH group is present in the formula.
5. It ignores the role of the solvent. Dry HCl gas does not turn dry blue litmus red, and HCl in benzene is not acidic at all. The solvent is an active partner, not a spectator.

2.2 Bronsted-Lowry definition; conjugate acid-base pairs

Because a proton must be given to something, every Bronsted acid-base reaction has two acids and two bases:

CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻

Here CH₃COOH gives a proton to H₂O in the forward direction, and H₃O⁺ gives a proton back to CH₃COO⁻ in the reverse direction.

Acid	Its conjugate base	Base	Its conjugate acid
HCl	Cl⁻	H₂O	H₃O⁺
CH₃COOH	CH₃COO⁻	NH₃	NH₄⁺
H₂SO₄	HSO₄⁻	OH⁻	H₂O
HSO₄⁻	SO₄²⁻	CO₃²⁻	HCO₃⁻
H₂CO₃	HCO₃⁻	HCO₃⁻	H₂CO₃
NH₄⁺	NH₃	CH₃COO⁻	CH₃COOH
H₃O⁺	H₂O	H₂PO₄⁻	H₃PO₄

Species such as H₂O, HCO₃⁻, HSO₄⁻ and H₂PO₄⁻ appear in both columns: they can donate a proton to a stronger base and accept one from a stronger acid. They are called amphiprotic (or amphoteric) species.

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

The Bronsted definition still has a limitation of its own: it needs a proton. Reactions such as BF₃ + NH₃ → F₃B←NH₃, which show every chemical feature of neutralisation, lie outside it.

2.3 Relative strength of acids and bases

The strength of a Bronsted acid is the extent to which it transfers its proton to water. For a weak acid HA,

HA + H₂O ⇌ H₃O⁺ + A⁻

A larger K_a (smaller pK_a) means a stronger acid. The same logic with K_b measures base strength.

Acid	K_a (25 °C)	pK_a	Conjugate base	K_b of that base
HCl	very large	< 0	Cl⁻	negligible
HSO₄⁻	1.2\times10^{-2}	1.92	SO₄²⁻	8.3\times10^{-13}
HF	6.8\times10^{-4}	3.17	F⁻	1.5\times10^{-11}
HCOOH	1.8\times10^{-4}	3.75	HCOO⁻	5.6\times10^{-11}
CH₃COOH	1.8\times10^{-5}	4.74	CH₃COO⁻	5.6\times10^{-10}
H₂CO₃	4.5\times10^{-7}	6.35	HCO₃⁻	2.2\times10^{-8}
NH₄⁺	5.6\times10^{-10}	9.25	NH₃	1.8\times10^{-5}
HCN	4.9\times10^{-10}	9.31	CN⁻	2.0\times10^{-5}

If two weak acids of the same concentration C are compared, then since \alpha = \sqrt{K_a/C} (next section),

So acetic acid (1.8\times10^{-5}) is about 192 times stronger than hydrocyanic acid at the same concentration, because \sqrt{1.8\times10^{-5}/4.9\times10^{-10}} = \sqrt{3.67\times10^{4}} = 192.

Levelling effect. HCl, HBr, HI, HNO₃ and HClO₄ all appear equally strong in water, because each of them transfers its proton to water completely; the strongest acid that can exist in water is H₃O⁺ itself. Water is said to level them. To rank them you must use a weaker base as solvent, such as glacial acetic acid, which shows the true order HClO₄ > HI > HBr > HCl > HNO₃.

2.4 Lewis definition

BF₃ + :NH₃ → F₃B←NH₃

Lewis acids (electron-pair acceptors)	Lewis bases (electron-pair donors)
Molecules with an incomplete octet: BF₃, AlCl₃, BeCl₂	Molecules with a lone pair: NH₃, H₂O, R–OH, R–NH₂, R₂S
Simple cations: H⁺, Ag⁺, Cu²⁺, Fe³⁺	Anions: OH⁻, Cl⁻, CN⁻, F⁻
Molecules with expandable octets: SiF₄, PF₅, SF₄	Molecules with π electrons: C₂H₄, C₆H₆
Molecules with a multiple bond to an electronegative atom: CO₂, SO₃, SO₂	Carbanions and alkoxides

Ag⁺ + 2NH₃ → [Ag(NH₃)₂]⁺ and Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ are Lewis neutralisations; this is why complex formation belongs in the same chapter of ideas as acidity.

	Arrhenius	Bronsted-Lowry	Lewis
Acid	gives H⁺ in water	proton donor	electron-pair acceptor
Base	gives OH⁻ in water	proton acceptor	electron-pair donor
Solvent	must be water	any solvent, or none	any solvent, or none
Scope	narrowest	wider	widest
Weakness	no non-aqueous acids/bases	cannot handle BF₃, AlCl₃	too wide; says nothing about relative strength; protonic acids such as HCl are not strictly Lewis acids

2.5 Ionization of weak electrolyte (Ostwald's dilution law)

A weak electrolyte is only partly ionised, and the un-ionised molecules are in equilibrium with the ions.

The law works well for weak electrolytes but fails completely for strong electrolytes such as NaCl or HCl, which are fully ionised in the solid state already; for them the observed "degree of ionisation" changes with concentration only because of interionic attraction, not because of any equilibrium.

2.6 Ionic product of water (Kw)

Pure water conducts electricity very slightly, so it must ionise to a small extent:

H₂O + H₂O ⇌ H₃O⁺ + OH⁻, usually written H₂O ⇌ H⁺ + OH⁻

Because only about two water molecules in 10⁹ are ionised, [\text{H}_2\text{O}] is effectively constant (55.5 mol L⁻¹), and it is absorbed into the constant:

Taking negative logarithms of K_w = [\text{H}^{+}][\text{OH}^{-}],

Ionisation of water is endothermic, so by Le Chatelier's principle K_w rises with temperature:

Temperature	K_w	pK_w	pH of pure water
0 °C	0.11\times10^{-14}	14.94	7.47
25 °C	1.00\times10^{-14}	14.00	7.00
50 °C	5.48\times10^{-14}	13.26	6.63
100 °C	51.3\times10^{-14}	12.29	6.14

2.7 Dissociation constant of acid and base (Ka and Kb); concept of pKa and pKb

For the weak acid HA and the weak base BOH:

K_a and K_b are true equilibrium constants: they depend on temperature but not on concentration. Combining with Ostwald's law, K_a = C\alpha^{2}.

Because these constants span many powers of ten, they are quoted logarithmically:

A small pK_a means a strong acid. Acetic acid, K_a = 1.8\times10^{-5}, has pK_a = 4.74.

2.8 pH value: pH of strong and weak acids; pH of strong and weak bases

Hydrogen-ion concentrations run from about 10 to 10⁻¹⁵ mol L⁻¹. Sørensen (1909) replaced them by a logarithmic scale.

Strong acid. A strong monobasic acid is completely ionised, so [\text{H}^{+}] = C and

For a dibasic strong acid such as H₂SO₄, [\text{H}^{+}] = 2C.

Weak acid. Only a fraction \alpha ionises, and [\text{H}^{+}] = C\alpha. Substituting \alpha = \sqrt{K_a/C},

Strong base. [\text{OH}^{-}] = C \times (acidity), pOH = -\log[\text{OH}^{-}] and pH = 14 - pOH.

Weak base. [\text{OH}^{-}] = \sqrt{K_bC}, so

2.9 Solubility and solubility product principle

In a saturated solution of AgCl an equilibrium exists between the undissolved solid and its ions:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

The concentration of a pure solid is constant, so it is absorbed into the constant, giving K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}].

Salt type	Example	K_{sp} in terms of S	S in terms of K_{sp}
AB	AgCl, BaSO₄	S^{2}	\sqrt{K_{sp}}
AB₂ or A₂B	PbI₂, Ag₂CrO₄	4S^{3}	\left(K_{sp}/4\right)^{1/3}
AB₃	Fe(OH)₃	27S^{4}	\left(K_{sp}/27\right)^{1/4}
A₃B₂	Ca₃(PO₄)₂	108S^{5}	\left(K_{sp}/108\right)^{1/5}

K_{sp} depends only on temperature. Solubility, by contrast, depends on what else is in the solution — which is the subject of the next two sections.

2.10 Common ion effect

It is Le Chatelier's principle applied to an ionic equilibrium. Add acetate ions to acetic acid and the equilibrium

CH₃COOH ⇌ CH₃COO⁻ + H⁺

shifts to the left, so [\text{H}^{+}] falls and the pH rises — even though more "acid material" has been added. In the same way NH₄Cl suppresses the ionisation of NH₄OH, and HCl suppresses that of H₂S.

System	Added common ion	Effect
CH₃COOH + CH₃COONa	CH₃COO⁻	ionisation of the acid suppressed; pH rises
NH₄OH + NH₄Cl	NH₄⁺	ionisation of the base suppressed; [OH⁻] falls
H₂S + HCl	H⁺	[S²⁻] falls sharply — basis of Group II analysis
AgCl + NaCl	Cl⁻	solubility of AgCl falls
Saturated NaCl + HCl gas	Cl⁻	pure NaCl crystallises out (salt purification)

2.11 Application of solubility product principle and common ion effect in precipitation reactions

Write Q (the ionic product) for the product of the ion concentrations at any moment, calculated exactly like K_{sp} but for a solution that need not be saturated. Then:

Condition	Meaning
Q < K_{sp}	unsaturated — no precipitate; more solid will dissolve
Q = K_{sp}	exactly saturated — solid and solution in equilibrium
Q > K_{sp}	supersaturated — precipitation occurs until Q falls to K_{sp}

Qualitative analysis. The whole group separation scheme rests on these two ideas.

• Group II (Cu²⁺, Pb²⁺, Cd²⁺, Bi³⁺, …) is precipitated as sulphides by passing H₂S in the presence of dilute HCl. The H⁺ from HCl is a common ion for H₂S ⇌ 2H⁺ + S²⁻, so it pushes that equilibrium far to the left and keeps [\text{S}^{2-}] extremely low. Only the sulphides with the very smallest K_{sp} (CuS, PbS, CdS) then satisfy Q > K_{sp}; ZnS, MnS and NiS do not precipitate.
• Group III A (Fe³⁺, Al³⁺, Cr³⁺) is precipitated as hydroxides by NH₄OH in the presence of NH₄Cl. The NH₄⁺ common ion suppresses the ionisation of NH₄OH, keeping [\text{OH}^{-}] low, so only the hydroxides of very low K_{sp} come down and Mg(OH)₂ stays in solution.
• Group IV sulphides (ZnS, MnS, CoS, NiS) are then precipitated by H₂S in alkaline medium, where [\text{S}^{2-}] is much higher.

Purification of common salt. Saturated brine made from crude salt is treated with HCl gas. The large excess of the common ion Cl⁻ makes Q = [\text{Na}^{+}][\text{Cl}^{-}] exceed K_{sp} for NaCl, and pure sodium chloride crystallises out while the more soluble impurities (MgCl₂, CaCl₂) stay in solution.

Salting out of soap. Adding solid NaCl to the soap solution after saponification raises [\text{Na}^{+}], exceeds the solubility product of sodium stearate, and the soap separates as a curd.

2.12 Buffer solution and its application

Buffer solution

A buffer solution is one that resists a change in its pH when a small amount of a strong acid or a strong base is added to it, or when it is diluted.

Type	Made from	Example	pH
Acidic buffer	weak acid + its salt with a strong base	CH₃COOH + CH₃COONa	< 7
Basic buffer	weak base + its salt with a strong acid	NH₄OH + NH₄Cl	> 7
Salt-only buffer	salt of a weak acid and a weak base	CH₃COONH₄	≈ 7

How a buffer works. Take CH₃COOH + CH₃COONa. The mixture contains a large reserve of un-ionised acid and a large reserve of acetate ions.

• Add H⁺: it is mopped up by the acetate reserve, CH₃COO⁻ + H⁺ → CH₃COOH. A weak, barely ionised acid replaces the strong acid, so the pH hardly moves.
• Add OH⁻: it is neutralised by the acid reserve, CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.

Only the ratio [salt]/[acid] changes, and pH depends on its logarithm, so a large change in the ratio gives a small change in pH.

Applications of buffers.

• Blood is buffered at pH 7.4 by H₂CO₃/HCO₃⁻ (with phosphate and protein buffers). A drift of 0.4 pH unit either way is fatal.
• Intracellular fluid is buffered by H₂PO₄⁻/HPO₄²⁻ at about pH 7.2.
• Agriculture: soil must be buffered near pH 6–7; acidic hill soils in Nepal are limed to raise and hold the pH.
• Industry: electroplating baths, fermentation of beer and curd, dyeing of textiles, sugar refining and paper manufacture all need fixed pH.
• Pharmacy: eye drops and injections are buffered to the pH of body fluid to avoid irritation.
• Analysis: complexometric EDTA titrations are run in an NH₄Cl/NH₄OH buffer at pH 10.

2.13 Indicators and selection of indicators in acid-base titration

Ostwald's theory. Let the indicator be a weak acid HIn:

HIn ⇌ H⁺ + In⁻ (colour A) → (colour B)

The eye sees colour A alone when [HIn] is about ten times [In⁻], and colour B alone when the ratio is reversed. Putting the ratio equal to 1/10 and 10/1 gives the working range of the indicator:

In an acidic solution the high [\text{H}^{+}] pushes the equilibrium left, so phenolphthalein (whose ionised form is pink) is colourless in acid; in alkali the H⁺ is removed and the pink In⁻ appears.

Indicator	Range (pH)	Colour in acid	Colour in alkali
Methyl orange	3.1 – 4.4	red	yellow
Methyl red	4.2 – 6.3	red	yellow
Litmus	5.0 – 8.0	red	blue
Bromothymol blue	6.0 – 7.6	yellow	blue
Phenol red	6.8 – 8.4	yellow	red
Phenolphthalein	8.3 – 10.0	colourless	pink
Thymolphthalein	9.3 – 10.5	colourless	blue

Titration	pH at equivalence	Why	Indicator
Strong acid vs strong base	7	salt not hydrolysed	methyl orange or phenolphthalein
Weak acid vs strong base	> 7 (≈ 8.7)	anionic hydrolysis	phenolphthalein
Strong acid vs weak base	< 7 (≈ 5.3)	cationic hydrolysis	methyl orange
Weak acid vs weak base	≈ 7	both ions hydrolysed	no sharp jump — no indicator works; use a pH meter

2.14 Types of salts: acidic, basic, simple, complex

A salt is the product of the reaction of an acid with a base. Salts are classified in two independent ways.

By composition:

Class	Formed by	Examples
Normal (neutral) salt	complete replacement of all H⁺ of the acid	NaCl, Na₂SO₄, K₃PO₄
Acid salt	partial replacement of the replaceable H⁺	NaHSO₄, NaHCO₃, NaH₂PO₄
Basic salt	partial replacement of the OH⁻ of a polyacidic base	Pb(OH)Cl, Zn(OH)Cl, Mg(OH)Cl
Double salt	two simple salts crystallised together	Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O; potash alum K₂SO₄·Al₂(SO₄)₃·24H₂O
Mixed salt	one acid with two different bases, or one base with two acids	CaOCl₂ (bleaching powder), NaKSO₄
Complex salt	contains a complex ion that survives in solution	K₄[Fe(CN)₆], [Cu(NH₃)₄]SO₄, K₃[Fe(CN)₆]

By the pH of their solution (the result of hydrolysis, §2.15):

Salt of	Nature of solution	Example
Strong acid + strong base	neutral, pH = 7	NaCl, KNO₃, Na₂SO₄
Weak acid + strong base	basic, pH > 7	CH₃COONa, Na₂CO₃, KCN
Strong acid + weak base	acidic, pH < 7	NH₄Cl, CuSO₄, FeCl₃
Weak acid + weak base	depends on K_a vs K_b	CH₃COONH₄ (≈ 7)

Note that an acid salt is not the same thing as an acidic salt: NaHCO₃ is an acid salt by composition but its solution is alkaline, because the carbonic acid it comes from is much weaker than the base NaOH.

2.15 Hydrolysis of salts

Case 1 — salt of a strong acid and a strong base (NaCl).

NaCl → Na⁺ + Cl⁻. Neither ion reacts with water: Na⁺ would give NaOH, a strong base which is completely ionised again, and Cl⁻ would give HCl, likewise. There is no hydrolysis; the solution is neutral, pH 7.

Case 2 — salt of a weak acid and a strong base (CH₃COONa): anionic hydrolysis.

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

The acetate ion takes a proton from water, leaving OH⁻ behind; the solution is alkaline.

Case 3 — salt of a strong acid and a weak base (NH₄Cl): cationic hydrolysis.

NH₄⁺ + H₂O ⇌ NH₄OH + H⁺

The solution is acidic. By exactly the same argument,

Case 4 — salt of a weak acid and a weak base (CH₃COONH₄). Both ions hydrolyse and

Notice that here the degree of hydrolysis and the pH are independent of concentration — a favourite one-mark point. Dilution does not change the pH of an ammonium acetate solution.

Salt of	Ion hydrolysed	K_h	Nature
SA + SB	none	—	neutral
WA + SB	anion	K_w/K_a	alkaline
SA + WB	cation	K_w/K_b	acidic
WA + WB	both	K_w/(K_aK_b)	depends on K_a vs K_b

Chapter summary

• Arrhenius acids/bases work only in water; Bronsted-Lowry defines acid = proton donor, base = proton acceptor, giving conjugate pairs that differ by one H⁺; Lewis defines acid = electron-pair acceptor, base = electron-pair donor.
• For a conjugate pair K_a \times K_b = K_w and pK_a + pK_b = 14; the stronger the acid, the weaker its conjugate base.
• Ostwald's dilution law: K = C\alpha^{2}/(1-\alpha) \approx C\alpha^{2}, so \alpha = \sqrt{K/C} = \sqrt{KV}. It fails for strong electrolytes.
• K_w = [\text{H}^{+}][\text{OH}^{-}] = 1.0\times10^{-14} at 25 °C; pH + pOH = 14. K_w rises with temperature, so neutral pH falls below 7 on heating.
• pH master formulas: strong acid pH = -\log C; weak acid pH = \frac{1}{2}(pK_a - \log C); strong base pH = 14 + \log C; weak base pOH = \frac{1}{2}(pK_b - \log C).
• K_{sp} = x^{x}y^{y}S^{(x+y)} for A_xB_y. Precipitation occurs when the ionic product Q > K_{sp}; the common ion effect lowers solubility and suppresses the ionisation of weak electrolytes.
• Buffers: pH = pK_a + \log([\text{salt}]/[\text{acid}]) and pOH = pK_b + \log([\text{salt}]/[\text{base}]); capacity is greatest at pH = pK_a and useful over pK_a \pm 1.
• Choose an indicator whose range lies inside the vertical part of the titration curve: phenolphthalein (8.3–10.0) for weak acid vs strong base, methyl orange (3.1–4.4) for strong acid vs weak base, either for strong vs strong.
• Hydrolysis: WA+SB gives pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C (alkaline); SA+WB gives pH = 7 - \frac{1}{2}pK_b - \frac{1}{2}\log C (acidic); WA+WB gives pH = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b, independent of concentration.

Practice questions

Group A — Multiple choice (1 mark each)

1. The conjugate base of HSO₄⁻ is [1]aH₂SO₄bSO₄²⁻cH₂OdOH⁻
2. Which of the following is a Lewis acid but not a Bronsted acid? [1]aHClbNH₄⁺cBF₃dCH₃COOH
3. The pH of a 0.001 M solution of NaOH at 25 °C is [1]a3b7c11d14
4. On dilution, the degree of ionisation of a weak electrolyte [1]adecreasesbincreasescremains the samedbecomes zero
5. An aqueous solution of ammonium chloride is [1]aneutralbacidiccalkalinedamphoteric
6. The indicator most suitable for the titration of acetic acid against sodium hydroxide is [1]amethyl orangebmethyl redcphenolphthaleindany of them
7. If the solubility of a salt AB₂ is S mol L⁻¹, its solubility product is [1]aS^{2}b2S^{3}c4S^{3}dS^{3}

Answers to Group A

1. (b) — removing one proton from HSO₄⁻ leaves SO₄²⁻.

2. (c) — BF₃ has no proton to donate but an empty p orbital to accept a lone pair.

3. (c) — pOH = -\log 10^{-3} = 3, so pH = 14 - 3 = 11.

4. (b) — Ostwald's law, \alpha = \sqrt{K/C}, and C falls on dilution.

5. (b) — NH₄⁺ hydrolyses to give H⁺ (salt of strong acid + weak base).

6. (c) — the equivalence pH is about 8.7, inside the phenolphthalein range 8.3–10.0.

7. (c) — AB₂ ⇌ A²⁺ + 2B⁻, so K_{sp} = (S)(2S)^{2} = 4S^{3}.

Group B — Short answer (5 marks each)

1. State the limitations of the Arrhenius concept of acids and bases, and show how the Bronsted-Lowry concept removes them. Identify the conjugate acid-base pairs in HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻. [5]
2. State and derive Ostwald's dilution law. Why does it fail for strong electrolytes? [5]
3. Calculate the pH and the degree of ionisation of a 0.020 M solution of a weak monobasic acid whose K_a is 2.0\times10^{-5}. [5]
4. What is a buffer solution? Derive the Henderson equation for an acidic buffer and calculate the pH of a solution that is 0.20 M in CH₃COOH and 0.10 M in CH₃COONa. (pK_a = 4.74) [5]
5. Define solubility product. The solubility of Mg(OH)₂ in water is 1.65\times10^{-4} mol L⁻¹. Calculate its solubility product. [5]
6. What is meant by hydrolysis of a salt? Explain, with equations, why an aqueous solution of sodium carbonate is alkaline while that of ferric chloride is acidic. [5]

Answers to Group B

1. Limitations: restricted to water; free H⁺ cannot exist; cannot explain acids without H (CO₂, BF₃) or bases without OH (NH₃); ignores the solvent. The Bronsted definition needs only a proton transfer, so it works in any solvent and classifies NH₃ as a base and NH₄⁺ as an acid. In the given equation the pairs are HCO₃⁻ / H₂CO₃ (base / its conjugate acid) and H₂O / OH⁻ (acid / its conjugate base).

3. Using [\text{H}^{+}] = \sqrt{K_aC},

[\text{H}^{+}] = \sqrt{2.0\times10^{-5}\times 0.020} = \sqrt{4.0\times10^{-7}} = 6.32\times10^{-4}\ \text{mol L}^{-1}

so pH = 4 - \log 6.32 = 4 - 0.801 = 3.20. And \alpha = \sqrt{K_a/C} = \sqrt{1.0\times10^{-3}} = 3.16\times10^{-2}, i.e. 3.16 per cent.

4. pH = pK_a + \log([\text{salt}]/[\text{acid}]), so pH = 4.74 + \log(0.10/0.20) = 4.74 - 0.301 = 4.44.

5. Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻, so K_{sp} = (S)(2S)^{2} = 4S^{3} and K_{sp} = 4(1.65\times10^{-4})^{3} = 4 \times 4.49\times10^{-12} = 1.80\times10^{-11}.

6. Hydrolysis is the reaction of the ions of a salt with water to regenerate the parent acid or base. Na₂CO₃ is the salt of the weak acid H₂CO₃ and the strong base NaOH, so the anion hydrolyses: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻, releasing OH⁻ and making the solution alkaline. FeCl₃ is the salt of the strong acid HCl and the weak base Fe(OH)₃, so the cation hydrolyses: Fe³⁺ + 3H₂O ⇌ Fe(OH)₃ + 3H⁺, releasing H⁺ and making the solution acidic.

Group C — Long answer (8 marks each)

1. (a) Define pH and derive the expression for the pH of a weak monobasic acid. [3](b) Calculate the pH of 0.10 M ammonium hydroxide (K_b = 1.8\times10^{-5}), and the pH of the solution obtained when 0.10 mol of NH₄Cl is dissolved in 1.00 L of that ammonia solution. [5]
2. (a) What is the common ion effect? Explain how it and the solubility product principle are used to separate Group II from Group IV cations in qualitative analysis. [4](b) The solubility product of AgCl is 1.8\times10^{-10}. Calculate its solubility in pure water and in 0.10 M AgNO₃ solution. [4]
3. (a) State Ostwald's theory of indicators and derive the working range of an indicator. [4](b) Sketch the titration curve for a weak acid against a strong base, mark the equivalence point, and justify the choice of phenolphthalein rather than methyl orange. [4]

Answers to Group C

1.(b) For the ammonia alone, [\text{OH}^{-}] = \sqrt{K_bC} = \sqrt{1.8\times10^{-6}} = 1.34\times10^{-3} mol L⁻¹, so pOH = 2.87 and pH = 11.13.

Adding NH₄Cl creates a basic buffer with [base] = [salt] = 0.10 M:

pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]} = 4.74 + \log 1 = 4.74

so pH = 14 - 4.74 = 9.26. The common ion NH₄⁺ has suppressed the ionisation of the base and lowered the pH by nearly two units.

2.(a) The common ion effect is the suppression of the ionisation of a weak electrolyte by adding a strong electrolyte with an ion in common. H₂S is a weak acid; in the presence of dilute HCl the common ion H⁺ drives H₂S ⇌ 2H⁺ + S²⁻ to the left, so [\text{S}^{2-}] is very small. Only the Group II sulphides, whose K_{sp} values are extremely small, then reach Q > K_{sp} and precipitate. In Group IV the solution is made alkaline with NH₄OH; the OH⁻ removes H⁺, [\text{S}^{2-}] rises by many powers of ten, and the more soluble sulphides ZnS, MnS, CoS and NiS now satisfy Q > K_{sp}.

2.(b) In pure water S = \sqrt{K_{sp}} = 1.34\times10^{-5} mol L⁻¹.

In 0.10 M AgNO₃, [\text{Ag}^{+}] \approx 0.10, so

S' = \frac{K_{sp}}{[\text{Ag}^{+}]} = \frac{1.8\times10^{-10}}{0.10} = 1.8\times10^{-9}\ \text{mol L}^{-1}

about 7450 times smaller.

3.(b) The curve starts near pH 2.9, rises quickly, flattens into a buffer region around pH = pK_a = 4.74 at half-neutralisation, then jumps steeply to the equivalence point at pH ≈ 8.7 (alkaline because of anionic hydrolysis of the acetate formed) and levels off near pH 12–13. The vertical portion runs only from about pH 7 to pH 11. Phenolphthalein (8.3–10.0) lies inside that jump, so its colour changes within one drop of the equivalence point. Methyl orange (3.1–4.4) lies far below it, in the buffer region, and would change colour long before the acid was neutralised.