Question

$0.1\text{M}C{H}_{3}COOH$ is titrated with $0.1MNaOH$ solution. What would be the difference in pH between $\frac{1}{4^{th}}\text{and}\frac{3}{4^{th}}$ stages of neutralization of the acid?

• A. $2\text{log}\left(\frac{1}{4}\right)$
• B. $2\text{log 3}$
• C. $0.47$
• D. $\text{log 2}$

Answer 1

The correct option is B $2\text{log 3}$

In the given solution these two processes are happening $C{H}_{3}COOH+NaOH\to C{H}_{3}COONa+H_{2}O$
$C{H}_{3}CO{O}^{-}+H_{2}O\leftrightharpoons C{H}_{3}COOH+O{H}^{-}$
This will form an acidic buffer system for which $pH$ will be :
$pH=p{K}_a+log\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$
$p{H}_{1/4}-p{H}_{3/4}=log\frac{1/4}{3/4}-log\frac{3/4}{1/4}=-2log3$
so difference in $pH=2log3$