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Find out the solubility product KspK_{sp} of Ca(OH)2Ca(OH)_2 if the pH of its saturated solution is 9:

A

0.5×10−100.5 \times 10^{-10}

B

0.5×10−150.5 \times 10^{-15}

C

0.25×10−100.25 \times 10^{-10}

D

0.125×10−150.125 \times 10^{-15}

Step-by-Step Solution

Given, pH of the saturated solution = 9 We know that at 298K, pH+pOH=14pH + pOH = 14. pOH=14−9=5pOH = 14 - 9 = 5 Therefore, the concentration of hydroxyl ions is [OH−]=10−5 M[OH^-] = 10^{-5} \text{ M}. The dissociation equilibrium of Ca(OH)2Ca(OH)_2 is: Ca(OH)2(s)⇌Ca2+(aq)+2OH−(aq)Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq) If the molar solubility of Ca(OH)2Ca(OH)_2 is SS, then [Ca2+]=S[Ca^{2+}] = S and [OH−]=2S[OH^-] = 2S. Given [OH−]=10−5 M[OH^-] = 10^{-5} \text{ M}, we have: 2S=10−5⇒S=0.5×10−5 M2S = 10^{-5} \Rightarrow S = 0.5 \times 10^{-5} \text{ M} Now, the solubility product constant KspK_{sp} is given by: Ksp=[Ca2+][OH−]2=(S)(2S)2K_{sp} = [Ca^{2+}][OH^-]^2 = (S)(2S)^2 Ksp=(0.5×10−5)(10−5)2K_{sp} = (0.5 \times 10^{-5})(10^{-5})^2 Ksp=0.5×10−5×10−10=0.5×10−15K_{sp} = 0.5 \times 10^{-5} \times 10^{-10} = 0.5 \times 10^{-15}

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