Question

What is the difference in $pH$ for $1 / 3$ and $2 / 3$ stages of neutralisation of $0.1 \,M\, CH _{3} COOH$ with $0.1\, M\, NaOH $ ?

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

Answer 1

Answer: (D)

$pH =p K a+\log \frac{[\text { salt }]}{[\text { acid }]}$
$pH _{1}=p K a+\log \frac{1 / 3}{2 / 3}\, \dots(i)$
at 2/3 neutralization
$pH _{2}=p K a+\log \frac{2 / 3}{1 / 3}\, \dots(2)$
$pH _{1}- pH _{2}=\log \frac{1}{2}-\log 2$
$=\log \frac{1}{4}=-2 \log 2$