A) \[2\times {{10}^{-3}}\]M and \[5\times {{10}^{-12}}\]M
B) \[1\times {{10}^{3}}\ M\ \text{and}\ 3\times {{10}^{-11}}M\]
C) \[0.02\times {{10}^{-3}}\ M\ \text{and}\ 5\times {{10}^{-11}}M\]
D) \[3\times {{10}^{-2}}\ M\ \text{and}\ 4\times {{10}^{-13}}M\]
Correct Answer: A
Solution :
Given that Concentration of solution =.1 Degree of ionisation \[=2%=\frac{2}{100}=.02\] Ionic product of water \[=1\times {{10}^{-14}}\] Concentration of \[[{{H}^{+}}]\]= Concentration of solution X degree of ionisation \[=.1\times .02=2\times {{10}^{-3}}M\] Concentration of \[[O{{H}^{-}}]=\frac{\text{Ionic product of water}}{[{{H}^{+}}]}\] \[=\frac{1\times {{10}^{-14}}}{2\times {{10}^{-3}}}=0.5\times {{10}^{-11}}=5\times {{10}^{-12}}M\].
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