25 ml of a dilute aqueous solution of p-hydroxy benzoic acid is titrated with NaOH ( aq), the soloution has pH = 4.57 , when 8.12 ml of 0.0200 M NaOH had been added, and pH = 7.02 after 16.24 ml had been added ( the equivalent point). Use these data to determine `Ka_(1)` and `Ka_(2)` for p-hydroxy benoic acid.
`HOC_(6) H_(4) COOH + H_(2) O hArr H_(3) O^(+) + HOC_(6) H_(4) COO^(-)`
`HOC_(6) H_(4) COO^(-) + H_(2) O hArr H_(3) O^(+) + ""^(-)OC_(6)H_(5)COO^(-) , Ka_(2)`

Explanation

It is given that the equivalent point occurs when 16.24ml of NaOH had been added. Thus after 8.12 ml had been added, half neutralization occurs. At the half-neutralisation point, `[HOC_(6)H_(4)COO^(-)] = [ HOC_(6)H_(4)COOH^(-) ]`
then, pH `= pKa_(1) + log"" ( [ HOC_(6)H_(4)COO^(-)])/([HOC_(6)H_(4)COOH])=pKa_(1)`
`:. pKa_(1) = pH = 4.57 , log Ka_(1) = - 4.57`
`:. Ka_(1) = 2.7 xx 10^(-5)`
`HOC_(6)H_(4)COO^(-) ` is now amphoteric ion, a base as well as an acid
`HOC_(6)H_(4)COO^(-) + H_(2) O hArr O^(-) C_(6) H_(4) COO^(-) + H_(3)O^(-) `( acidic)
`HOC_(6)H_(4)COO^(-) + H_(2) O hArr HOC_(6)H_(4)COOH + OH^(-) `( basic )
For such cases, when equivalent point is obtained at pH = 7.02,
`pH = (( p Ka_(1)+ pKa_(2))/(2))`
`7.02 = ( 4.57 + pKa_(2))/( 2)`
`pKa_(2) = 9.47`
`:. Ka_(2) = 3.4 xx 10^(-10)`