A) \[{{10}^{-7}}\]
B) \[{{10}^{-6}}\]
C) \[{{10}^{-8}}\]
D) \[{{10}^{-4}}\]
Correct Answer: A
Solution :
\[pH=4\] \[[{{H}^{+}}]={{10}^{-4}}\] \[[{{H}^{+}}]=N.\alpha \] \[{{10}^{-4}}=\frac{1}{10}.\alpha \] \[\alpha ={{10}^{-3}}\] For weak electrolyte \[K={{\alpha }^{2}}.C\] \[K={{\left( {{10}^{-3}} \right)}^{2}}\times \frac{1}{10}={{10}^{-7}}\]
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