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: syllabus Chem Ch 7 §7.13 Solubility Equilibria_
Keep practising Chemical Thermodynamics
Browse all Chemical Thermodynamics questions for AP Intermediate 1st Year, or explore more Chemistry topics.
Related questions
A process whose enthalpy and entropy changes share the same arithmetic sign will show:At minus 10.00 degrees Celsius the entropy change of the universe for melting is negative,Melting is described as spontaneous at 10.00 degrees Celsius because at that temperature:A buffer solution is most resistant to pH change when:K_w (ionic product of water at 25 °C) is:The pH of a 0.01 M HCl solution is:For an exothermic reaction at equilibrium, raising the temperature:Adding an inert gas at constant volume to a gaseous equilibrium: