A) \[1\times {{10}^{-3}}\]
B) \[1\times {{10}^{-4}}\]
C) \[5\times {{10}^{-4}}\]
D) \[1\times {{10}^{-6}}\]
Correct Answer: A
Solution :
[a] \[\begin{matrix} \underset{(At\,\,eq.)}{\mathop{\underset{(Initial)}{\mathop{{}}}\,}}\, & \underset{\begin{matrix} \text{ }\!\!~\!\!\text{ }0.001M \\ \text{ }\!\!~\!\!\text{ }(0.001-R) \\ \end{matrix}}{\mathop{{{A}^{-}}}}\,~+{{H}_{2}}O\rightleftharpoons \underset{(0.001\,\times \,h)}{\mathop{\underset{0}{\mathop{HA}}\,}}\,+\underset{(0.001\,\times \,h)}{\mathop{\underset{0}{\mathop{HA}}\,}}\, \\ \end{matrix}\] \[{{K}_{h}}=\frac{(0.001\times h)(0.001\times h)}{(0.001-h)}\] or, \[{{K}_{h}}=0.001\times {{h}^{2}}(as,0.001-h\approx 0.001)\] \[{{10}^{-9}}=0.001\times {{h}^{2}}\] \[\left( {{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{{{10}^{-5}}}={{10}^{-9}} \right)\] \[{{h}^{2}}={{10}^{-6}};\] \[\therefore h={{10}^{-3}}\]
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