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. $5+log2-log3$
• B. $5-log2$
• C. $5-log3$
• D. $5-log6$

Answer 1

The correct option is B $5-log2$

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

Initial $\underset{1mole}{HA}+\underset{0}{NaOH}\to \underset{0}{NaA}+\underset{0}{H)2O}$

Final $\left(\begin{array}{c}1-\frac{1}{3}\end{array}\right)mol\frac{1}{3}mol\frac{1}{3}mol$

$=\frac{2}{3}mol\frac{1}{3}mol$

(Assumed weak acid to be monoprotic, since only one dissociation constant value is provided).

Final solution acts as an acidic buffer.

$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$

or $pH=5+log\frac{\frac{1}{3}}{\frac{2}{3}}$

$=5+log\frac{1}{2}$

$\therefore pH=5-log2$