Question

One litre solution contains 1M HOCl [$K_a={10}^{-8}$] and 1M NaOH. What is the pH of the solution?

Answer 1

$\begin{array}{ccc}HOCl& \rightleftharpoons & H^{+}+{\alpha }^{-}\\ 1M& & \\ 1-a& a& a\end{array}$

$H_a=\frac{H^{+}OC{l}^{-}}{HOCl}=\frac{a-\alpha }{(1-\alpha )}=\frac{{\alpha }^2}{1-\alpha }={\alpha }^2={10}^{-8}$

$\therefore =\alpha ={10}^{-9}$

$=\left[H^{+}\right]={10}^{-4}M={10}^{-4}mal[\because volumwis1L]$

$\begin{array}{ccccc}NaOH& \to & {Na}^{+}& +& {OH}^{-}\\ 1M& & & & \\ -& & 1M& & 1M\\ & & ||& & ||\\ & & 1mol& & 1mol\end{array}$

$\begin{array}{ccccc}{H}^{+}& +& {OH}^{-}& \to & H_{2}O\\ {10}^{-4}& & 1.& & -\\ -& & 1-{10}^{-4}& & {10}^{-4}\end{array}$

$\therefore [O{H}^{-}{]}_{final}=0.9999mol$

$[O{H}^{-}{]}_{final}=\frac{0.9999}{2}m$ [volume because $2L$]

$=\frac{1}{2}M$

$\therefore pOH=-\log \left[O{H}^{-}\right]=\log 2=0.301$

$\therefore pH=14-pOH=14-0.301$

$pH=13.699$