A) \[0.02\times {{10}^{-3}}\,M\] and \[5\times {{10}^{-11}}\,M\]
B) \[1\times {{10}^{-3}}\,M\] and \[3\times {{10}^{-11}}\,M\]
C) \[2\times {{10}^{-3}}\,M\] and \[5\times {{10}^{-12}}\,M\]
D) \[3\times {{10}^{-2}}\,M\] and \[4\times {{10}^{-13}}\,M\]
Correct Answer: C
Solution :
\[[{{H}^{+}}]\] in monobasic acid = molarity \[\times \] degree of ionisation \[=0.1\times \frac{2}{100}=2\times {{10}^{-3}}\,m\] \[[O{{H}^{-}}]=\frac{{{K}_{w}}}{[{{H}^{+}}]}=\frac{1\times {{10}^{-14}}}{2\times {{10}^{-3}}}=5\times {{10}^{-12}}\,\,M\]
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