Selenious acid `(H_(2)SeO_(3))`,a diprotoc acid has `Ka_(1)=10^(-6)` and `Ka_(2)=10^(-8)` respectively. Appoximate pH of 0.01M `NaHSeO_(3)` is given by :-

Options

A. `pH=7+(pKa_(1))/(2)+(logC)/(2)`
B. `pH=7-(pKa_(1))/(2)-(logC)/(2)`
C. `pH = (pKa_(1)+pKa_(2))/(2)`
D. `pH=7+(pKa_(1))/(2)-(pKa_(2))/(2)`

Answer

C

Explanation

`H2SeO_(3)overset(Ka_(1))hArroverset("Amphiprotic acid")overset(uarr)(HSeO_(3)^(-))overset(Ka_(2))hArrSeO_(3)^(2-)`
`Ka_(1)=([HSeO_(3)^(-)][H^(+)])/([H_(2)SeO_(3)]),Ka_(2)=([SeO_(3)^(2-)][H^(+)])/([H_(2)seO_(3)])`
`Ka_(1)xxKa_(2)=[H^(+)]^(2)xx([SeO_(3)^(2-)])` ....(1)
At isoelectric point `[SeO_(3)^(2-)]~~[H_(2)SeO_(3)]`
then from (1)
`Ka_(1)xxKa_(2)=[H^(+)]^(2)`
`rArr[H^(=)]=(Ka_(1)xxKa_(2))^(1//2)`
`becausepH-log(Ka_(1)xxKa_(2))^(1//2)`
`rArrpH=(1)/(2)(pKa_(1)+pKa_(2))`