Question

A weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $NaOH$ solution. The $pH$ at the point of one-third neutralisation of the acid will be

• A. $5+\log 2-\log 3$
• B. $5-\log 2$
• C. $5-\log 3$
• D. $5-\log 6$

Answer 1

Answer: (B)

$K_a=10^{-5}\Rightarrow p{K}_a=-\log K_a=-\log 10^{-5}=5$

(Assumed weak acid to be monoprotic, since only one dissociation constant value is provided.)
Final solution acts as an acidic buffer.
$pH=p{K}_a+\log \frac{[\text{ salt }]}{[\text{ acid] }}$
or $pH=5+\log \frac{\frac{1}{3}}{\frac{2}{3}}$
$=5+\log \frac{1}{2}$
$\therefore pH=5-\log 2$