A) \[11.80\]
B) \[10.69\]
C) \[7.22\]
D) \[12.24\]
Correct Answer: B
Solution :
\[[O{{H}^{-}}]=C.\,\alpha \] \[=0.01\times \frac{5}{100}=5\times {{10}^{-4}}\] \[pOH=-\log [O{{H}^{-}}]\] \[=-[log\,5\times {{10}^{-4}}]\] \[=4\log 10-\log 5\] \[=4-0.06989\] \[=3.3010\] \[pH=14-pOH\] \[=14-3.3010=10.6990\]
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