Question

Chlorination with $C{l}_2$ produces hypoch-lorous acid (HOCL), which may further dissociate as hypochlorite ion $\left(OC{l}^{-}\right)$ depending upon pH of the water, the reaction is represented as:



$(k=2.5\times {10}^{-8}moles/Lat{20}^{\circ }C)$

What is the fraction of HOCl in the water at pH 7.0?

• A. 0.05
• B. 0.80
• C. 0.95
• D. 0.20

Answer 1

The correct option is B 0.80

(b) For pH = 7

$\text{Concentration of}\left[H^{+}\right]={10}^{-pH}moles/L$

Ionization constant

$k=\frac{\left[H^{+}\right]\left[OC{l}^{-}\right]}{\left[HOCl\right]}=2.5\times {10}^{-8}moles/L$

The total quantity of HOCl and $OC{l}^{-}$ present in water is called free available chlorine. The relative distribution of these two species is very important because the killing efficiency of HOCl is about 40 to 80 times that of $OC{L}^{-}$. The percent distribution of HOCl

$\frac{\left[HOCl\right]}{\left[HOCl\right]+\left[OC{l}^{-}\right]}=\frac{1}{1+k{10}^{pH}}$

$=\frac{1}{1+2.5\times {10}^{-8}\times {10}^7}=\frac{1}{1.25}=0.8$