A) \[1.1\times {{10}^{-9}}\]
B) \[1.1\times {{10}^{-10}}\]
C) \[1.1\times {{10}^{-11}}\]
D) \[{{10}^{-12}}\]
Correct Answer: C
Solution :
\[\underset{x}{\mathop{Ag}}\,(l)+\underset{y}{\mathop{{{H}_{2}}O(\ell )}}\,\rightleftharpoons HA\underset{y}{\mathop{(aq)}}\,+O{{H}^{-}}(aq)\]\[{{K}_{sp}}=x(x-y)\] \[{{K}_{b}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{y}^{2}}}{(x-y)};\,\,\frac{{{10}^{-14}}}{{{10}^{-10}}}=\frac{{{({{10}^{-5}})}^{2}}}{(x-y)}\] \[x-y={{10}^{-6}}\] \[x={{10}^{-5}}+{{10}^{-6}}\] \[x=1.1\times {{10}^{-5}}\] \[{{K}_{sp}}=1.1\times {{10}^{-5}}\times {{10}^{-6}}\] \[{{K}_{sp}}=1.1\times {{10}^{-11}}\]You need to login to perform this action.
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