K_sp of a sparingly soluble salt AB at 25 °C is 4 × 10⁻¹⁰. Molar solubility s of AB is:
A2 × 10⁻¹⁰
B2 × 10⁻⁵
C4 × 10⁻⁵
D1 × 10⁻⁵
Answer & Solution
Correct answer: B. 2 × 10⁻⁵
1. For AB(s) ⇌ A⁺ + B⁻, K_sp = [A⁺][B⁻] = s · s = s².
2. s = √K_sp = √(4 × 10⁻¹⁰) = 2 × 10⁻⁵ M.
3. For A₂B-type the formula is K_sp = 4s³ (s = (K_sp/4)^(1/3)).
_Source: NCERT Class 11 Chem Ch 7 §7.13 Solubility Equilibria_
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