`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. `9.6xx10^(-6)`
B. `1.4xx10^(-1)`
C. `1.3xx10^(-6)`
D. `1.3xx10^(-8)`

Answer

C

Explanation

The hydrolysis of `C_(2)O_(4)^(2-)` is as follows
`C_(2)O_(4)^(2-)+H_(2)OhArrHC_(2)O_(4)^(-)+H^(-)`
`K_(h)=(K_(w))/(K_(2))=(1.0xx10^(-14))/(6.0xx10^(-5))=(x^(2))/(0.01)`
`:.x=1.3xx10^(-6)`