Question

What is the pH of a 1.0 M solution of acetic acid? To what volume of one litre of this solution be diluted so that the pH of resulting solution will be twice the original value? Given $K_a$=1.8 $\times {10}^{-5},lo{g}_{10}4.2426\times {10}^{-3}=2.3724$

• A. $2.78\times {10}^4\text{litre}$
• B. $2.78\times {10}^2\text{litre}$
• C. $3.68\times {10}^5\text{litre}$
• D. $3.68\times {10}^2\text{litre}$

Answer 1

The correct option is A $2.78\times {10}^4\text{litre}$

As degree of dissociation, $\alpha =\sqrt{\frac{\text{Ka}}{\text{C}}}$
$\alpha =\sqrt{\frac{1.8\times {10}^{-5}}{1}}$
$\alpha =4.2426\times {10}^{-3}$
$\left[{\text{H}}^{+}\right]=\text{C}\times \alpha$
$\left[{\text{H}}^{+}\right]=1\times 4.2426\times {10}^{-3}$
$\left[{\text{H}}^{+}\right]=4.2426\times {10}^{-3}{\text{mol L}}^{-1}$
$\text{pH}=-{\text{log}}_{10}\left[{\text{H}}^{+}\right]$
$\text{pH}=-{\text{log}}_{10}4.2426\times {10}^{-3}$
pH=2.3724

$\text{So pH of the acetic acid solution after dilution}$
$=2\times 2.3724=4.7448$

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

${\text{Suppose the new concentration is C}}_0$

${\text{CH}}_3\text{COOH}={\text{H}}^{+}+{\text{CH}}_3{\text{COO}}^{-}$

${\text{At. eq. C}}_0-1.8\times {10}^{-5}1.8\times {10}^{-5}1.8\times {10}^{-5}$

$\text{As Ka}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$

$=\frac{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}{({\text{C}}_0-1.8\times {10}^{-5})}=1.8\times {10}^{-5}$ ${\text{So C}}_0=3.6\times {10}^{-5}$
${\text{So C}}_0=3.6\times {10}^{-5}$
$\text{Suppose the new volume is V litre.}$
$1\times 1=3.6\times {10}^{-5}\times \text{V}$

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