• question_answer What is the$[{{H}^{+}}]$in a 0.40 M solution of $HOCl,{{K}_{a}}=3.5\times {{10}^{-8}}\,M$ A) $1.4\times {{10}^{-8}}M$             B) $1.2\times {{10}^{-4}}M$C) $1.9\times {{10}^{-4}}M$             D) $3.7\times {{10}^{-4}}M$

$\alpha =\sqrt{{{K}_{a}}V}$ ${{K}_{a}}=3.5\times {{10}^{-8}},V=\frac{1}{0.40}L=2.5L.$ $\alpha =\sqrt{3.5\times {{10}^{-8}}\times 2.5}$ $=2.95\times {{10}^{-4}}$ $[{{H}^{+}}]=\frac{\alpha }{V}=\frac{2.95\times {{10}^{-4}}}{2.5}=1.18\times {{10}^{-14}}$ $\approx 1.2\times {{10}^{-4}}M$ Hence, the correct option is [b].