Equilibrium Question 836
Question: The degree of hydrolysis in hydrolytic equilibrium $ {A^{-}}~+H_2O\rightarrow HA+O{H^{-}} $ at salt concentration of 0.001 M is : $ (K_{a}=1\times {10^{-5}}) $
Options:
A) $ 1\times {10^{-3}} $
B) $ 1\times {10^{-4}} $
C) $ 5\times {10^{-4}} $
D) $ 1\times {10^{-6}} $
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Answer:
Correct Answer: A
Solution:
$ \begin{matrix} \underset{(Ateq.)}{\mathop{\underset{(Initial)}{\mathop{{}}},}}, & \underset{\begin{matrix} ~0.001M \\ ~(0.001-R) \\ \end{matrix}}{\mathop{{A^{-}}}},~+H_2O\rightarrow \underset{(0.001,\times ,h)}{\mathop{\underset{0}{\mathop{HA}},}},+\underset{(0.001,\times ,h)}{\mathop{\underset{0}{\mathop{HA}},}}, \\ \end{matrix} $
$ K_{h}=\frac{(0.001\times h)(0.001\times h)}{(0.001-h)} $ or, $ K_{h}=0.001\times h^{2}(as,0.001-h\approx 0.001) $
$ {10^{-9}}=0.001\times h^{2} $
$ ( K_{h}=\frac{K_{w}}{K_{a}}=\frac{{10^{-14}}}{{10^{-5}}}={10^{-9}} ) $
$ h^{2}={10^{-6}}; $
$ \therefore h={10^{-3}} $
