Calculate the $pH$ of $0.08\, M$ solution of hypochlorous acid, $HOCl$. The ionization constant of the acid is $2.5 \times 10^{-5}$ Determine the percent dissociation of $HOCl.$

 $HOCl(aq) + {H_2}O(l) \rightleftharpoons {H_3}{O^ + }(aq) + Cl{O^ - }(aq)$

Initial concentration $(M)$

           $0.08$                                                  $0$            $0$                  

Change to reach equilibrium concentration $(M)$

           $-x$                                                 $+x$           $+x$

equilibrium concentartion $( M )$

           $0.08-x$                                            $x$                $x$

$K_{ a }=\left\{\left[ H _{3} O ^{+}\right]\left[ ClO ^{-}\right] /[ HOCl ]\right\}$

$=x^{2} /(0.08-x)$ 

As $x\,<\,<\,0.08,$ therefore $0.08 - x \simeq 0.08$

$x^{2} / 0.08=2.5 \times 10^{-5}$

$x^{2}=2.0 \times 10^{-6},$ thus, $x=1.41 \times 10^{-3}$

$\left[ H ^{+}\right]=1.41 \times 10^{-3} \,M$

Therefore

Percent dissociation

$ = \{ {[HOCl]_{{\rm{dissociated }}}}/{[HOCl]_{{\rm{initial }}}}] \times 100$

$=1.41 \times 10^{-3} \times 10^{2} / 0.08=1.76 \%$

$pH =-\log \left(1.41 \times 10^{-3}\right)=2.85$