`500mL` of `0.2M` aqueous solution of acetic acid is mixed with `500mL` of `0.2HCI` at `25^(@)C`.
a. Calculate the degree of dissociation of acetic acid in the resulting solution and `pH` of the folution.
b. If `6g` of `NaOH` is added to the above solution determine the final `pH.[K_(a)` of `CH_(3)COOH =2 xx 10^(-5)`.

Explanation

The volume being doubled by mixing two solutions , the molarity of each component will be halved i.e
` [CH_(3)COOH] = 0.1 M , [HCl] = 0.1 M `
HCl being a strong acid will remain completely ionised and hence `H^(+)` ion concentration furnished by it will be `0.1 ` M . This would exert common ion effect on the dissociation of acetic acid , a weak acid .
`{:(CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),(C,,0,,0),(C(1-alpha),,Calpha,,Calpha +0.1):}`
`K_(a) = (Calpha(Calpha+0.01))/(C(1-alpha)) = (Calpha^(2)+0.1alpha)/((1-alpha))`
Neglecting `alpha ` in comparison to unity and `Calpha^(2)` i.e ` 0.1 alpha^(2)`
`K_(a) = 0.1 alpha `
oe `alpha = (K_(a))/(0.1) = (1.75 xx10^(-5))/(0.1) = 1.75 xx 10^(-4)`
` [H^(+)] Total = 0.1 +Calpha ,Calpha ` is negligiable as compared to `0.1 `
` :. [ H^(+)] _(Total) = 0.1 `
` :. pH = 1 `
(ii) `6g NaOH = 6/40 = 0.15 ` mol
`0.1 ` mole of NaOH will be consumed by `0.1 ` mole of HCL . Thus , `0.05` mole of NaOH will react with acetic acid (No. of mol of `CH_(3)COOH = 1 xx 0.2 = 0.1 ` ) according to the equation
`{:(CH_(3)COOH,+,NaOH,to,CH_(3)COONa,+,H_(2)O),(0.1" mol",,0.05 " mol",,0,,0),(0.05 " mol",,0 " mol" ,,0.05 " mol",,0.05 " mol"):}`
Thus , solution of acetic acid and sodium acetate will become acidic buffer . So pH of the buffer will be
`pH = pK_(a) + log. (["salt"])/(["acid"]) = - log (1.75 xx 10^(-5)) + log 1 = 4.75`