Question

When 0.1 M NaOH is titrated with 0,1 M, 20 mL HA till the end point, $K_{\alpha }\left(HA\right)=6\times {10}^{-6}$ and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point?

• A. 6.23
• B. 9.22
• C. 7.21
• D. 8.95

Answer 1

The correct option is D 8.95

$\underset{2}{NaOH}+\underset{2}{HA}⟶\underset{-}{NaA}+\underset{-}{H_{2}O}$
At end point = 0.1 x 20 = 2
$\because$ 20 mL of NaOH is required for the complete neutralisation of HA.
NaA is a salt of strong base and weak acid.
Thus, will undergo bydrolysis and solution will become basic.
$C=\left[NaA\right]=\frac{2}{20+20}=0.05M$
and $p{K}_a=-\log (6\times {10}^{-6})=5.2$
pH at the end point $=7+\frac{1}{2}(p{K}_a+\log C)$
$7+\frac{1}{2}(5.2+\log 0.05)=8.95$