Question

$0.1 \,M \,NaOH$ is titrated with $0.1 \,M , 20\, ml \,HA$ till the end point, $K _{ a }( HA )=6 \times 10^{-6}$ 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

Answer: (D)

$\underset{2}{ NaOH }+\underset{2}{ HA } \rightarrow NaA + H _{2} O$
At end point $\equiv 0.1 \times 20=2$
$\because 20\, mL\,NaOH$ is required for the complete neutralization of $HA$
$NaA$ is a salt of strong base and weak acid
Thus, will undergo hydrolysis and solution will becomes basic
$C =[ NaA ]=\frac{2}{20+20}=0.05\, M$
And $pK _{ a }=-\log \left(6 \times 10^{-6}\right)=5.2$
$pH$ at the end point $=7+\frac{1}{2}\left( pK _{ a }+\log C \right)$
$7+\frac{1}{2}(5.2+\log 0.05)=8.95$