Question

What is the pH of a $1.0M$ solution of acetic acid? To what volume of one litre of this solution be diluted so that the pH of the resulting solution will be twice the original value? Given, $K_a=1.8\times {10}^{-5}$

Answer 1

We know that degree of dissociation

$\alpha =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{1.8\times {10}^{-5}}{1}}=4.2426\times {10}^{-3}$

$\left[H^{+}\right]=C\alpha =1\times 4.2426\times {10}^{-3}=4.2426\times {10}^{-3}mol{L}^{-1}$

$pH=\log \left[H^{+}\right]=-\log 4.2426\times {10}^{-3}=2.3724$

so, pH of the acetic acid solution after dilution$=2\times 2.3724$

$=4.7448$

New $\left[H^{+}\right]={10}^{-4.7448}=1.8\times {10}^{-5}$

Let the new concentration be $C_0$

${CH}_{3}COOH\rightleftharpoons H^{+}+{CH}_{3}CO{O}^{-}$

At equilibrium $C_0-1.8\times {10}^{-5}$ $1.8\times {10}^{-5}$ $1.8\times {10}^{-5}$

$K_a=\frac{\left[H^{+}\right]\left[{CH}_{3}CO{O}^{-}\right]}{\left[{CH}_{3}COOH\right]}=\frac{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}{(C_0-1.8\times {10}^{-5})}=1.8\times {10}^{-5}$

so, $C_0=3.6\times {10}^{-5}$

Let the new volume be $V$ litre

$1\times 1=3.6\times {10}^{-5}\times V$

$V=\frac{1}{3.6\times {10}^{-5}}=2.78\times {10}^{4}l$