What is the nature of 0.01M `NaHCO_(3)` solution? Calculate its pH `K_(1) and K_(2)" for "H_(2)CO_(3)" are "4.5 xx 10^(-7) and 4.7 xx 10^(-11)` respectively.

Explanation

`Na^(+)` ions do not hydrolyse. `HCO_(3)^(-)` may undergo hydrolysis as well as ionisation.
`HCO_(3)^(-)+H_(2)O Leftrightarrow OH^(-)+H_(2)CO_(3)`
`K_(h)=([OH^(-)] [H_(2)CO_(3)])/([HCO_(3)^(-)])=K_(w)/K_(1)=(1 xx 10^(-14))/(4.5 xx 10^(-7))=2.2 xx 10^(-8) .........(1)`
And, `HCO_(3)^(-) Leftrightarrow H^+ +CO_(3)^(2-) `
`K_(2)=([H^(+)] [CO_(3)^(2-)])/([HCO_(3)^(-)])=4.7 xx 10^(-11) ..........(2)`
As `K_(h) gt K_(2)`, i.e, `[OH^(-)] gt [H^+]`, the solution will be alkaline. In this solution, the electrical neutrality can be preserved by maintaining anionic charge as the catonic charge remains at 0.01M `(=[Na^(+)])` For every negative charge removed by converting `HCO_(3)^(-)" to "CO_(3)^(2-)`. Thus,
`[H_(2)CO_(3)]=[CO_(3)^(2-)]="x (say)"`
`[HCO_(3)^(-)]=0.01-2x approx 0.01M`.
Multiplying eqns. (1) and (2)
`K_(h) xx K_(2)=([H^+] [OH^(-)] [H_(2)CO_(3)] [CO_(3)^(2-)])/(HCO_(3)^(-))^2=(2.2 xx 10^(-8)) (4.7 xx 10^(-11))`
Substructing `[H^+] [OH^(-)] =K_(w)=1 xx 10^(-14)`
`x=1.02 xx 10^(-4)`
Now using the eqn. (2) again, we get
`K_(2)=[H^+] (1.02 xx 10^(-4))/(0.01)=4.7 xx 10^(-11)`
`of [H^+]=4.6 xx 10^(-9)`
`pH =-log (4.6 xx 10^(-9))=8.34`