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 $

$k\alpha =C\alpha ^{2}$

$\text{H}^{\oplus}=C\alpha =\sqrt{k_{a} \times C}$

$=\sqrt{3.5 \times 10^{- 8} \times 0.40}$

$\left[\text{H}^{\text{+}}\right]=1.83\times 10^{- 14}$