A) 4
B) 11
C) 10.5
D) 10
Correct Answer: D
Solution :
\[\text{1}{{\text{0}}^{-4}}\,M\,KOH={{10}^{-4}}M[O{{H}^{-}}]\] We know that \[[{{H}^{+}}][O{{H}^{-}}]=1\times {{10}^{-14}}\] \[[{{H}^{+}}]=\frac{1\times {{10}^{-14}}}{1\times {{10}^{-4}}}=1\times {{10}^{-10}}\,M\] \[\therefore \] \[pH=-\log [{{H}^{+}}]=-\log (1\times {{10}^{-10}})\] \[=10\]
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