`0.1` M `CH_(3)COOH` solution is titrated against ` 0.05` M NaOh . Calculate pH at `1//4^(th)` and `3//4^(th)` stage of neutralization of acid , the pH for ` 0.1 ` M `CH_(3)COOH` is 3 .

Explanation

Calculation of dissociation constant of the acid
`CH_(3)CO OH hArr CH_(3)CO O^(-) + H^(+)`
As `pH = 3, :. [H^(+)] = 10^(-3)M, [CH_(3)CO O^(-)]=[H^(+)]=10^(-3)M`
`K_(a) = ([CH_(3)CO O^(-)][H^(+)])/([CH_(3)CO OH])=(10^(-3)xx10^(-3))/(0.1) = 10^(-5)`
(i) When 1/4th of the acid has been neutralized
`{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa ,+,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 1/4th neutralization",0.1xx(3)/(4),,,,0.1xx(1)/(4),,),(,=0.075 M ,,,,=0.025 M ,,):}`
`:. pH = pK_(a) + log. (["Salt"])/(["Acid"])=-log 10^(-5) + log. (0.025)/(0.075) = 5 - 0.4771 = 4.5229`
(ii) When 3/4th of the acid has been neutralized
`{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa , +,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 3/4th",0.1xx1/4M,,,,0.1xx3/4M,,),("neutralization",=0.025 M,,,,=0.075 M,,):}`
`:. pH = - log 10^(-5) + log. (0.075)/(0.025) = 5 + 0.4771=5.4771`