A 10L vessel contain 1L of ${\text{10}}^{\text{-3}}\text{M}$ ${\text{CH}}_{\text{3}}\text{COOH}$ $\left[{\text{P}}^{\text{ka}}\text{=4}\text{.8}\right]$ to this 40 mg of NaOH is added [ignore the change in volume]. Then the change in ${\text{P}}^{\text{H}}$ observed is     

$C{H}_{3}COOH+NaOH\to C{H}_{3}COONa+H_{2}O$ 

                   60g           40g

${10}^{-3}mole\stackrel{}{\to }{10}^{-3}mole\stackrel{}{\to }{10}^{-3}mole$ 

$\therefore$ conc. of $C{H}_{3}COONa$ formed = ${10}^{-3}$ molar

$P^H=\frac{P^{kw}+P^{ka}+\log C}{2}=\frac{14+4.8-3}{2}=7.9$ 

Initial P^H of $C{H}_{3}COOH=\frac{1}{2}\left[P^{ka}-\log C\right]$ 

                         $=\frac{4.8-\left(-3\right)}{2}=3.9$ 

$\therefore \Delta P^H$ = 4 units (increase)