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`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 of sodium oxalateA. `9.6 xx 10^(-6)`B. `1.4 xx 10^(-1)`C. `1.2 xx 10^(-6)`D. `1.3 xx 10^(-8)` |
Answer» Correct Answer - C 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)` |
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