A) \[{{10}^{-4}}\]M KOH
B) \[{{10}^{-10}}M\text{ }KOH\]
C) \[{{10}^{-10}}\]M \[\text{HCl}\]
D) \[{{10}^{-4}}M\text{ HCl}\]
Correct Answer: A
Solution :
\[[O{{H}^{-}}]\]in\[{{10}^{-4}}M\text{ }KOH\]solution \[={{10}^{-4}}M\] (KOH is a strong electrolyte) \[[{{H}^{+}}]=\frac{{{K}_{w}}}{[O{{H}^{-}}]}\] \[=1\times \frac{{{10}^{-14}}}{{{10}^{-4}}}=1\times {{10}^{-10}}M\] or \[pH=10\]
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