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Chemistry Question on Equilibrium

pHpH of a saturated solution of Ba(OH)2Ba(OH)_2 is 1212 . The value of solubility product K(sp) K_({sp} ) of Ba(OH)2Ba(OH)_2 is

A

3.3×10−73.3 \times 10^{-7}

B

5.0×10−75.0 \times 10^{-7}

C

4.0×10−64.0 \times 10^{-6}

D

5.0×10−65.0 \times 10^{-6}

Answer

5.0×10−75.0 \times 10^{-7}

Explanation

Solution

Given, pHpH of Ba(OH)2=12Ba(OH)_2 = 12 ∴pOH=14−pH\therefore pOH = 14 - p^H =14−12=2= 14 - 12 = 2 We know that, pOH=−log⁡[OH−]pOH = - \log [OH^-] 2=−log[OH−]2 = - log [OH^-]   [OH−]=antilog(−2)\, \, [OH^-] = antilog(-2)   [OH−]=1×10−2\, \, [OH^-] = 1 \times 10^{-2} Ba(OH)2Ba(OH)_2 dissolves in water as      SmolL−1Ba(OH)2⇌  SBa2++  2S2OH−\, \, \, _{\, \, S mol L^{-1}}^{Ba(OH)_2} \rightleftharpoons _{\, \, S}^{Ba^{2+}} + _{\, \, 2S}^{2OH^-} ∴  [OH−]=2S=1×10−2\therefore \, \, [OH^-] = 2S = 1 \times 10^{-2} S=[OH−]2     [Ba2+=S]S = \frac{[OH^-]}{2} \, \, \, \, \, [Ba^{2+} = S] [Ba2+]=[OH−]2a=1×10−22[Ba^{2+} ] = \frac{[OH^-]}{2a} = \frac{1 \times 10^{-2}}{2} Ksp=[Ba2+][OH−]K_{sp} = [Ba^{2+} ] [OH^-] =(1×10−22)(1×10−2)2= \bigg (\frac{1 \times 10^{-2}}{2} \bigg ) (1 \times 10^{-2} )^2 =0.5×10−6=5×10−7= 0.5 \times 10^{-6} = 5 \times 10^{-7}