Calculate pH of solution after calculating `A^(2-)` concentration in 0.1 `MH_(2)` A solution, if `K_(1) and K_(2)` for dissociation of `H_(2)A` is `4 xx 10^(-3) and 1 xx 10^(-5)`.
(Fill your answer upto 2 decimal places)

Answer

`1.75`

Explanation

`H_(2)A hArr H^(+) + HA^(-) " " K_(1) = 4 xx 10^(-3)`
`HA^(-) hArr H^(+) +A^(2-) " " K_(2) = 1 xx 10^(-5)`
For first step
`underset((0.1-x))(H_(2)A) hArr underset(x)(H^(+)) + underset(x)(HA^(-))`
`K_(1)=([H^(+)][HA^(-)])/([H_(2)A]) = 4 xx 10^(-3) = (x x)/((0.1-x))`
`:.x=0.018M`
`[H^(+)] =0.018M :.pH = 1.7447`.
`[HA^(-)] = 0.018M [H_(2)A] = 0.1 -0.018 = 0.082M`
For second step dissociation of `HA^(-)` occurs in presence of `H^(+)` ions and is suppressed
`underset((0.018-y))(HA^(-)) hArr underset((0.018+y))(H^(+)) +underset(y)(A^(2-))`
`K_(2)=([H^(+)] [A^(2-)])/([HA^(-)]) =10^(-5) = ((0.018 +y)(y))/((0.018-y))`
`(y lt lt lt 0.018)`
`:.10^(-5) =y = [A^(2-)]`
Concentration of anion formed after Ilnd dissociation of poly protic acid is almost equal to `K_(a_(II))` of acid.