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$
$=1.41 \times 10^{-3} \times 10^{2} / 0.08=1.76 \%$
$pH =-\log \left(1.41 \times 10^{-3}\right)=2.85$
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