Question

What is the difference in $pH$ for $1/3$ and $2/3$ stages of neutralisation of $0.1MC{H}_{3}COOH$ with $0.1MNaOH$ ?

• A. $-2\log 3$
• B. $2\log (1/4)$
• C. $2\log (2/3)$
• D. $-2\log 2$

Answer 1

Answer: (D)

$pH=pKa+\log \frac{[\text{ salt }]}{[\text{ acid }]}$
$p{H}_1=pKa+\log \frac{1/3}{2/3}\dots (i)$
at 2/3 neutralization
$p{H}_2=pKa+\log \frac{2/3}{1/3}\dots (2)$
$p{H}_1-p{H}_2=\log \frac{1}{2}-\log 2$
$=\log \frac{1}{4}=-2\log 2$