Write Ksp expression:
BiO(OH)(s) ⇌ BiO⁺(aq) + OH⁻(aq)
Solid is not included in equilibrium expression:
Let solubility = s mol/L:
From stoichiometry: [BiO⁺] = s and [OH⁻] = s
s² = 4 × 10⁻¹⁰
s = 2 × 10⁻⁵ mol/L
Find pH:
[OH⁻] = s = 2 × 10⁻⁵ mol/L
pH = 14 − pOH = 14 − 4.7 = 9.3 ≈ 9 ✓
When a sparingly soluble ionic compound dissolves in water, it establishes an equilibrium between the undissolved solid and the dissolved ions. The solubility product Ksp is the equilibrium constant for this dissolution. Pure solids and liquids have activity = 1 and are not included in the expression.
For AxBy(s) ⇌ xA^(y+)(aq) + yB^(x−)(aq)
Ksp = [A^(y+)]^x × [B^(x−)]^y
Ksp is temperature dependent — it increases with temperature for most salts (endothermic dissolution). A smaller Ksp means a less soluble compound. Ksp is used to predict whether a precipitate will form: if the ionic product Q > Ksp, precipitation occurs; if Q < Ksp, no precipitate.
📌 AB type (1:1): Ksp = s² → s = √Ksp
📌 AB₂ or A₂B type: Ksp = 4s³ → s = (Ksp/4)^(1/3)
📌 AB₃ or A₃B type: Ksp = 27s⁴ → s = (Ksp/27)^(1/4)
📌 A₂B₃ type: Ksp = 108s⁵ → s = (Ksp/108)^(1/5)
📌 Here BiO(OH): 1:1 type → Ksp = s² → s = √(4×10⁻¹⁰) = 2×10⁻⁵
pH = −log[H⁺]
pOH = −log[OH⁻]
pH + pOH = 14 (at 25°C)
Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C
Since [OH⁻] = 2×10⁻⁵: pOH = −log(2×10⁻⁵) = 5 − log2 = 5 − 0.301 ≈ 4.7. Therefore pH = 14 − 4.7 = 9.3. The solution is basic (pH > 7) because OH⁻ ions are released — this makes sense since BiO(OH) is a base-like compound releasing OH⁻.
Adding a common ion to a solution of a sparingly soluble salt decreases its solubility (Le Chatelier's Principle). For example, adding NaOH to BiO(OH) solution increases [OH⁻], shifting the equilibrium left, reducing solubility of BiO(OH). This principle is used in qualitative analysis to selectively precipitate ions.
A buffer solution resists changes in pH when small amounts of acid or base are added. It contains a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer). Buffer pH = pKa + log([A⁻]/[HA]) — Henderson-Hasselbalch equation. Maximum buffer capacity is at pH = pKa (equal concentrations of acid and conjugate base).
A pH indicator changes colour at its pKIn. For a titration, choose an indicator whose transition range overlaps with the steep portion of the titration curve. Strong acid + strong base: use phenolphthalein (pH 8.3–10) or methyl orange (pH 3.1–4.4). Weak acid + strong base: only phenolphthalein (endpoint at pH 8–9). Strong acid + weak base: only methyl orange. Weak acid + weak base: no suitable indicator (no sharp endpoint).
Salts of weak acids/bases undergo hydrolysis in water. For a salt of weak acid (Ka) and strong base: pH = 7 + ½(pKa + log C). For salt of strong acid and weak base: pH = 7 − ½(pKb + log C). For salt of weak acid and weak base: pH = 7 + ½(pKa − pKb). The degree of hydrolysis h ≈ √(Kh/C) where Kh = Kw/Ka (or Kw/Kb).
Solubility of metal hydroxides increases in acidic solutions — added H⁺ reacts with OH⁻ (common ion removed), shifting equilibrium right. Solubility of CaF₂ increases in acid (F⁻ + H⁺ → HF). Solubility of AgCl is independent of pH (Cl⁻ doesn't react with H⁺ to form a weak acid significantly). This pH dependence of solubility is extensively used in qualitative analysis and separation schemes.