A) \[3.1\times {{10}^{10}}\]
B) \[6.2\times {{10}^{5}}\]
C) \[6.2\times {{10}^{-10}}\]
D) \[3.2\times {{10}^{-9}}\]
Correct Answer: D
Solution :
\[pH=-\log {{K}_{a}}+\log \frac{[NaCN]}{HCN}\] \[=-\log 6.2\times {{10}^{-10}}+\log \frac{0.2}{0.1}\] \[=-log\text{ }6.2+10\text{ }log\text{ }10+log\text{ }2-log\,1\] \[pH=-\text{ }0.7924+10+0.3010-0\] \[=9.5086\] \[\because \] \[pH=-\log [{{H}^{+}}]\] \[0.5086=-\log [{{H}^{+}}]\] \[[{{H}^{+}}]=3.225\times {{10}^{-9}}M\]
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