The 100ml of 0.1M $C{H}_{3}COOH(K_a={10}^{-5})$ is titrated with 0.1 M $NaOH$ and $pH$ is observed at three states at 298K.

$1^{st} step\Rightarrow$50ml of 0.1M NaOH added

$2^{nd}step\Rightarrow$100ml of 0.1 M NaOH added

$3^{rd}step\Rightarrow$120ml of 0.1 M NaOH added

Calculate the difference in pH between stages (I & II) and (II & III) stages

At Stage I:

Left $\underset{5meq}{C{H}_{3}COOH}+\underset{0}{NaOH}\to \underset{5meq}{C{H}_{3}COONa}$

So, its is a buffer solutions , [salt]=[Acid]

$pH=p{K}_a\Rightarrow p{H}_1=5$

At Stage II:

Left $\underset{0}{C{H}_{3}COOH}+\underset{0}{NaOH}\to \underset{10meqin200ml}{C{H}_{3}COONa}$

So, c = 0.05

So, anionic hydrolysis takes place and pH is:

& $pH=7+\frac{1}{2}(5+\log 0.05)=8.84$

At Stage III:

Left $\underset{0}{C{H}_{3}COOH}+\underset{0}{NaOH}\to \underset{10meqin200ml}{C{H}_{3}COONa}$

So, pH will be according to NaOH

$\left[NaOH\right]=[a\times {10}^{-3}]$

So, $\left[O{H}^{-}\right]=a\times {10}^{-3}$

$pOH=2.040, p{H}_{III}=11.95$