20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The P^H of the resulting solution is ________ x 10^-2(Nearest integer) 
(Given :  ${\mathrm{pK}}_{\mathrm{a}}\left({\mathrm{CH}}_3\mathrm{COOH}\right)=4.76;\log 2=0.30;\log 3=0.48)$
 

$\begin{array}{l}\mathrm{NaOH}+{\mathrm{CH}}_3\mathrm{COOH}\to {\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}\\ 0.1\mathrm{M},20\mathrm{ml}0.1\mathrm{M},50\mathrm{ml}\end{array}$
Millimole $=0.1\times 20=2,0.1\times 50=5$
$$\begin{gathered} \mathrm{L}.\mathrm{R} = \mathrm{NaOH} \\ 0 5 - 2 = 3 \end{gathered}$$
So resultant solution is Acidic buffer solution
$$\begin{gathered} {\mathrm{P}}^{\mathrm{H}}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{{\displaystyle \mathrm{salt}}}{{\displaystyle \mathrm{acid}}}\right) \\ {\mathrm{P}}^{\mathrm{H}}=4.76+\log \frac{2}{3} \\ {\mathrm{P}}^{\mathrm{H}}=4.76+(-0.18)=4.58=458\times {10}^{-2} \end{gathered}$$