1. 0.5×10⁻¹⁵ 2. 0.25×10⁻¹⁰ 3. 0.125×10⁻¹⁵ 4. 0.5×10⁻¹⁰

Answer: 1. 0.5×10⁻¹⁵ Use pH to get [OH^-] , then write K_sp . pH = 9 so pOH = 5 and [OH^-] = 10⁻⁵ M. mathrmCa(OH)₂ ⇌ Ca²⁺ + 2OH^- , so [mathrmCa²⁺] = 1/2[OH^-] = 0.5×10⁻⁵ M. K_sp = [mathrmCa²⁺][OH^-]² = 0.5×10⁻⁵×(10⁻⁵)² = 0.5×10⁻¹⁵ . Option (1).

Topic: Solubility product· Equilibrium

pH of a saturated solution of Ca(OH)_2 is 9. The solubility product (K_sp) of Ca(OH)_2 is :

1. \(0.5\times10^{-15}\)

2. \(0.25\times10^{-10}\)

3. \(0.125\times10^{-15}\)

4. \(0.5\times10^{-10}\)

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Answer: 1. \(0.5\times10^{-15}\)

Use pH to get \([\mathrm{OH^-}]\), then write \(K_{sp}\).

pH = 9 so pOH = 5 and \([\mathrm{OH^-}] = 10^{-5}\) M.

\(\mathrm{Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-}\), so \([\mathrm{Ca^{2+}}] = \tfrac{1}{2}[\mathrm{OH^-}] = 0.5\times10^{-5}\) M.

\(K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2 = 0.5\times10^{-5}\times(10^{-5})^2 = 0.5\times10^{-15}\). Option (1).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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pH of a saturated solution of Ca(OH)_2 is 9. The solubility product (K_sp) of Ca(OH)_2 is :

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