# 7.64     The ionization constant of chloroacetic acid is 1.35 × 10-3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Answers (1)
M manish

We have,
Ionisation constant of chloroacetic acid($K_a$) is $1.35\times 10^{-3}$
The concentration of acid = 0.1 M
Ionisation if acid, =
$ClCH_2COOH\rightleftharpoons ClCH_2COO^-+H^+$

We know that,
$\Rightarrow K_a =\frac{[ClCH_2COO^-][H^+]}{[ClCH_2COOH]}$....................(i)
As it completely ionised
$[ClCH_2COO^-]=[H^+]$

Putting the values in eq (i)
$1.35\times 10^{-3} = \frac{[H^+]^2}{0.02}$
$[H^+] = \sqrt{1.35\times 10^{-3}\times 0.02}=1.16\times 10^{-2}$
Therefore, $pH$ of the solution = $-\log (1.16\times 10^{-2})$
= $2-\log(1.16)$
= $1.94$

Now,

0.1 M  $\dpi{100} ClCH_2COONa$ (sod. chloroacetate) is basic due to hydrolysis-

$\dpi{100} ClCH_2COO^-+ \: H_2O\rightleftharpoons CH_2ClCOOH+OH^-$

For a salt of strong base+strong acid

$\dpi{100} \\pH=7+\frac{pK_a+logC}{2}=7+\frac{2.87+log0.1}{2}\\pH=7.94$

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