Question

Chlorine gas used for disinfection combines with water to form hypochlorous acid (HOCI). The HOCl ionizes to form hypochlorite $\left(OC{l}^{-}\right)$ in a reversible reaction: $HOCl\leftrightarrow H^{+}+OC{L}^{-}(k=2.7\times {10}^{-8}at{20}^{\circ }C)$ the equilibrium of which is governed by pH. The sum of HOCl and $OC{l}^{-}$ is known as free chlorine residual and HOCl is the more effective disinfectant. The 90%, fraction of HOCl in the free chlorine residual is available at a pH value

• A. $7.5$
• B. $6.6$
• C. $9.4$
• D. $4.8$

Answer 1

The correct option is B $6.6$

$\underset{1-x}{HOCl}\rightleftharpoons H^{+}+\underset{x}{OC{l}^{-}}$

1 equilibrium 90% fraction of HOCl in the free chlorine is available, means 10% dissociates

$\left[OC{l}^{-}\right]=x=10\%$

$=0.1$ $\left[HOCl\right]=0.9$

$k=\frac{\left[H^{+}\right]\left[OC{l}^{-}\right]}{\left[HOC{l}^{-}\right]}$

$\Rightarrow 2.7\times {10}^{-8}=\frac{\left[H^{+}\right]\times 0.1}{0.9}$

$\Rightarrow \left[H^{+}\right]=24.3\times {10}^{-8}mol/L$

We know, $pH=-log\left[H^{+}\right]$

= $-log[24.3\times {10}^{-8}]$

$=6.6$