Calculate the `pH` of `0.05 M KHC_(8)H_(4)O_(4)`
`H_(2)C_(8)H_(4)O_(4) + H_(2)O hArr H_(3)O^(o+) + HC_(8)H_(4)O_(4)^(o+) pK_(a_(1)) = 2.94`
`HC_(8)H_(4)O_(4)^(Theta) + H_(2)O hArr H_(3)O^(o+) + C_(8)H_(4)O_(4)^(2-) pK_(a_(2)) = 5.44`

Explanation

The conjugate base (here `HC_(8)H_(4)O_(4)^(Theta))` of polybasic acid (here `H_(2)C_(8)H_(4)O_(4))` is amphiorprotic and can act as either an acid or a basic because it can either donate its remaining acidic hydrohen atom or accept an acidic hydorgen atom and revert to the original acid.
Acid:
`HC_(8)H_(4)O_(4)^(Theta) (aq) + H_(2)O(l) hArr H_(3)O^(o+) + C_(8)H_(2)O_(4)^(2-), pK_(a_(2)) = 5.44`
`K_(a_(2)) = ([C_(8)H_(4)O_(4)^(-2)][H_(3)O^(o+)])/([HC_(8)H_(4)O_(4)^(Theta)]) ...(i)`
Hydrolysis:
`HC_(8)H_(4)O_(4)^(Theta) (aq) + H_(2)O(l) hArr H_(2)C_(8)H_(4)O_(4) + oH^(Theta), pK_(h) 11.06`
`K_(h) = ([H_(2)C_(8)H_(4)O_(4)][overset(Theta)OH])/([HC_(8)H_(4)O_(4)^(Theta)]) ...(ii)`
Also for `H_(2)C_(8)H_(4)O_(4)`
`H_(2)C_(8)H_(4)O_(4) hArr H^(o+) + HC_(8)H_(4)O_(4)^(Theta)`
`K_(a_(1)) = ([H^(o+)][HC_(8)H_(4)O_(4)^(Theta)])/([H_(2)C_(8)H_(4)O_(4)]) ...(iii)`
By equaitons (ii) and (iii), we get
`K_(h) xx K_(a_(1)) = [H^(o+)] [overset(Theta)OH] = K_(w) ....(iv)`
Assuming `HC_(8)H_(4)O_(4)^(Theta)` having low degree of dissociation as well as low degree of hydrolysis, acidic nature of `HC_(8)H_(4)O_(4)^(Theta)` and hydrolysis of `HC_(8)H_(4)O_(4)^(Theta)` producing `C_(8)H_(4)O_(4)^(2-)` and `H_(2)C_(8)H_(4)O_(4)` of almost equal concentration `[H_(2)C_(8)H_(4)O_(4)] = [C_(8)H_(4)O_(4)^(2-)] ...(v)`
From equations (i), (ii), and (v), `(K_(h))/(K_(a_(2)) = ([overset(Theta)OH])/([H^(o+)]) ...(iv)`
Also `K_(w) = [H^(o+) [overset(Theta)OH] ..(vii)`
By equations (vi) and (vii), `K_(h) xx K_(a_(1)) = ([H^(o+)]^(2) xx K_(h))/(K_(a_(2)))`
or `[H^(o+)]^(2) = K_(a_(1)) xx K_(a_(2))` or `[H^(o+)] = sqrt(K_(a_(1)xxK_(a_(2))))`
`pH = (1)/(2) [PK_(a_(1)) + pK_(a_(2))] = (1)/(2) [2.94 + 5.44] = 4.19`
Note:
a. The `pH` is independent of salt concentration in case of acidic salt solution.
b. All polyprotic acids (except `H_(2)SO_(4))` are weak acids and the relation `pH = (1)/(2) [pK_(a_(1))+pK_(a_(2))]` holds good.
c. The relation `pH = (1)/(2) [pK_(a_(1)) + pK_(a_(2))]` is derived by using reasonable assumptions and doing some intricate algebra. The assumptions include that if `pK_(a_(2))` is high `HA^(Theta)` is weal acid, the solution will have high `pH`. If `pK_(a_(1))` is large `HA^(Theta)` is likely to act as 'strong' weak base. Thus. if both `pK_(a_(1))` and `pK_(a_(2))` are large, `pH` will be high.