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

K_(1) & K_(2) for oxalic acid are 6.5 xx 10^(-2) and 6.1 xx 10^(-5) re

1.	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
Answer» `9.6 xx 10^(-6)` `1.4 xx 10^(-1)` `1.2 xx 10^(-6)` `1.3 xx 10^(-8)` Solution :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)`