Question

What is the $\left[\right.H^{+}\left]\right.$ in a 0.40 M solution of HOCl, $K_{a}=3.5\times 10^{- 8}$ ?

• A. $1.4\times 10^{- 8 \, }M$
• B. $1.2\times 10^{- 4 \, }M$
• C. $1.9\times 10^{- 4 \, }M$
• D. $3.7\times 10^{- 4 \, }M$

Answer 1

Answer: (B)

$\text{HOCl}\overset{}{\rightleftharpoons}\text{} \text{H}^{\text{+}}+\text{OCl}^{-}$
$C\left(1 - \alpha \right) \, \text{}C\alpha \, C\alpha $
$\text{K}_{\text{a}} = \text{Cα}^{2}$
$\text{H}^{\oplus} = \text{Cα} = \sqrt{\text{K}_{\text{a}} \times \text{C}}$
$=\sqrt{3.5 \times 10^{- 8} \times 0.40}$
$\left[\text{H}^{\text{+}}\right]=1.183\times 10^{- 4}$