Question

Calculate change in pH upon ten-fold dilution of the following solutions:
(a) $0.1$ $HCl$
(b) $0.1M$ acetic acid
(c) $0.1M$ ${NH}_{4}Cl$
$OH=1.8\times {10}^{-5},K_b\left({NH}_3\right)=1.8\times {10}^{-5}$

Answer 1

$\left(a\right)$

$HCl$ is a strong acid. It is completely ionised in solution.

$HCl\rightleftharpoons H^{+}+{Cl}^{-}$

$\left[H^{+}\right]=0.1M={10}^{-1}M$

$pH=-\log \left[H^{+}\right]=-\log \left({10}^{-1}\right)=1$

After dilution, $\left[H^{+}\right]=0.1/10={10}^{-2}M$

$pH=-\log \left[H^{+}\right]=-\log \left({10}^{-2}\right)=2$

pH changes from 1 to 2

$\left(b\right)$ ${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+H^{+}$

$(0.1-x)$ $x$ $x$

(${CH}_{3}COOH$ is a weak acid)

$0.1-x\sim 0.1$.So

$\frac{x\times x}{0.1-x}=\frac{x^2}{0.1}=1.8\times {10}^{-5}$

$x=1.34\times {10}^{-3}$

$pH=-\log x=-\log (1.34\times {10}^{-3})=2.87$

After dilution

$\frac{{x_1}^2}{0.01}=1.8\times {10}^{-5}$

$x_1=4.24\times {10}^{-4}M$

$pH=-\log x=-\log (4.24\times {10}^{-4})=3.37$

pH change from $2.87$ to $3.37$

(c) ${NH}_{4}Cl$ is a salt of weak base and strong acid.

${NH}_4^{+}+H_{2}O\rightleftharpoons {NH}_{4}OH+H^{+}$

$(0.1-h)$ $h$ $h$

$\frac{h^2}{0.1}=K_h$ or $h^2=0.1\times K_h$

$[K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}]$

$=0.1\times 5.5\times {10}^{-10}$

$h=7.45\times {10}^{-6}=\left[H^{+}\right]$

$pH=-\log (7.45\times {10}^{-6})=5.128$

After dilution

$h^2=0.01\times K_h$

$h=2.35\times {10}^{-6}$

$pH=-\log (2.35\times {10}^{-6})=5.627$

pH change from $5.128$ to $5.627$