Equilibrium expression:
BiO(OH)(s) ⇌ BiO⁺(aq) + OH⁻(aq)
K = [BiO⁺][OH⁻] = 4 × 10⁻¹⁰
(Pure solid BiO(OH) is not included in K expression)
Let solubility = s mol/L:
[BiO⁺] = s and [OH⁻] = s
s² = 4 × 10⁻¹⁰
s = √(4 × 10⁻¹⁰) = 2 × 10⁻⁵ mol/L
Calculate pOH:
[OH⁻] = 2 × 10⁻⁵
pOH = −log(2 × 10⁻⁵) = −log 2 − log 10⁻⁵ = −0·3010 + 5 = 4·699
Calculate pH:
pH + pOH = 14
pH = 14 − 4·699 = 9·301
s = 2×10⁻⁵ M → [OH⁻] = 2×10⁻⁵
pOH = 5 − log2 = 5 − 0·301 = 4·699
pH = 14 − 4·699 = 9·301
For a sparingly soluble salt AxBy dissolving in water: AxBy(s) ⇌ xAʸ⁺(aq) + yBˣ⁻(aq). Ksp = [Aʸ⁺]ˣ[Bˣ⁻]ʸ. The solid phase is not included in Ksp (activity of pure solid = 1). In this problem, BiO(OH) is a sparingly soluble compound — the equilibrium involves its dissolution into BiO⁺ and OH⁻ ions. K = [BiO⁺][OH⁻] = Ksp = 4×10⁻¹⁰. Lower Ksp → less soluble → less ions in solution.
📌 AB type (1:1): Ksp = s² → s = √Ksp
📌 AB₂ or A₂B type (1:2 or 2:1): Ksp = 4s³ → s = (Ksp/4)^(1/3)
📌 A₂B₃ or A₃B₂ type (2:3 or 3:2): Ksp = 108s⁵ → s = (Ksp/108)^(1/5)
📌 This problem: BiO(OH) → BiO⁺ + OH⁻ (1:1 type) → Ksp = s²
📌 s = √(4×10⁻¹⁰) = 2×10⁻⁵ mol/L
pH = −log[H⁺] and pOH = −log[OH⁻]. At 25°C (298K): Kw = [H⁺][OH⁻] = 10⁻¹⁴. Taking −log: pH + pOH = 14. This relationship holds at 25°C only — at higher temperatures Kw increases so pH + pOH < 14. Neutral solution at 25°C: pH = 7 (neither acidic nor basic). Acidic: pH < 7. Basic: pH > 7. In this problem, [OH⁻] = 2×10⁻⁵ → basic solution → pH > 7 → pH = 9·301 ✓ (consistent with basic BiO(OH) dissolution).
Adding a common ion to a solution of sparingly soluble salt decreases its solubility (Le Chatelier's principle). Example: Adding NaOH (provides OH⁻) to BiO(OH) solution shifts equilibrium left → less BiO(OH) dissolves → lower solubility. Adding BiONO₃ (provides BiO⁺) also decreases solubility. The common ion effect is widely used in qualitative analysis to selectively precipitate ions. In this problem, if the solution already contained OH⁻ from another source, the solubility of BiO(OH) would be even lower.
A buffer resists change in pH on addition of small amounts of acid or base. Consists of a weak acid + its conjugate base (or weak base + conjugate acid). Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). For basic buffer: pOH = pKb + log([Salt]/[Base]). Buffer capacity is maximum when [A⁻] = [HA], i.e., pH = pKa (or [Salt] = [Base], pOH = pKb). Biological buffers: blood pH 7·35–7·45 maintained by H₂CO₃/HCO₃⁻ buffer system. Phosphate buffer: H₂PO₄⁻/HPO₄²⁻.
Bi³⁺ belongs to Group II of qualitative analysis (precipitated as sulphides in the presence of H₂S at low pH). However, this question involves detection via BiO(OH) precipitation in alkaline conditions. Qualitative analysis uses differences in Ksp values to selectively precipitate ions. By controlling pH and the precipitating agent, different groups of cations are separated. Group I (HCl): Pb²⁺, Ag⁺, Hg₂²⁺ precipitate as chlorides. Group II (H₂S, acidic): Cu²⁺, Bi³⁺, Hg²⁺ as sulphides. Group III (NH₃): Fe³⁺, Al³⁺, Cr³⁺ as hydroxides.
Salts of weak acid + strong base hydrolyse to give basic solution. Salts of strong acid + weak base hydrolyse to give acidic solution. Hydrolysis constant Kh = Kw/Ka (for salt of weak acid + strong base). Degree of hydrolysis h = √(Kw/Ka·C) for very dilute solutions or weak acids. pH of salt solution: for CH₃COONa: pH = 7 + ½(pKa + log C); for NH₄Cl: pH = 7 − ½(pKb + log C). These formulas are important for NEET numerical problems.
Kw = [H⁺][OH⁻]. At 25°C: Kw = 10⁻¹⁴, pH of neutral water = 7. At 37°C (body temperature): Kw ≈ 2·4×10⁻¹⁴, neutral pH ≈ 6·8 (still neutral, but not 7). At 100°C: Kw ≈ 10⁻¹², neutral pH = 6. Dissolution of water is endothermic → increasing temperature increases Kw (more dissociation). This is a highly tested NEET concept: neutral solution always has [H⁺] = [OH⁻], but the neutral pH is 7 ONLY at 25°C.