`K_(1)` & `K_(2)` for oxalic acid are `6.5 xx 10^(-2)` and `6.1 xx 10^(-5)` respectively . What will be the `[OH^(-)]` in a `0.01 M` solution of sodium oxalate

Options

A. `9.6 xx 10^(-6)`
B. `1.4 xx 10^(-1)`
C. `1.2 xx 10^(-6)`
D. `1.3 xx 10^(-8)`

Answer

C

Explanation

The hydrolysis of `C_(2)O_(4^(2-))` is as follows
`C_(2)O_(4^(2-)) + H_(2)O hArr HC_(2)O_(4)^(-) + OH^(-)`
`[OH^(-)] = sqrt((K_(w) xx C)/(K_(2))) = sqrt((10^(-14) xx 10^(-2))/(6.1 xx 10^(-5)))`
`= 1.2 xx 10^(-6)`