A) \[1.2\times {{10}^{-\,3}}M\]
B) \[1.8\times {{10}^{-\,5}}M\]
C) \[3.6\times {{10}^{-\,3}}M\]
D) \[4.1\times {{10}^{-\,4}}M\]
Correct Answer: A
Solution :
| In water, |
| \[Pb{{(OH)}_{2}}P{{\underset{S}{\mathop{b}}\,}^{2}}+2{{\underset{2S}{\mathop{OH}}\,}^{-}}\] |
| \[{{K}_{sp}}=4.{{S}^{3}}=4{{(6.7\times {{10}^{-\,6}})}^{3}}\] |
| \[=1.2\times {{10}^{-\,15}}\] |
| In buffer of \[pH=8\] |
| \[pOH=6,\] \[[O{{H}^{-}}]={{10}^{-\,6}}\] |
| \[{{K}_{sp}}=S\,{{[O{{H}^{-}}]}^{2}}\] |
| \[S=\frac{1.2\times {{10}^{-\,15}}}{{{10}^{^{-\,12}}}}=1.2\times {{10}^{\,-3}}M\] |
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