The `pH` of `0.01MNaOH_((aq))` solution will be

An aqueous solution of 0.01 M CH_(3) COOH has van't Hoff factor 1.01 . If pH = - "log"[H^(+)] , pH of 0.01 M CH_(3)COOH solution would be :

Calculate the pH of 0.1 MNH_(3) solution.

pH of 0.01 M HCOONH_4 aq solution changes

Find the pH of 0.4833% HCl (aq) solution

The pH of 0.01M HCN solution for which pK_(a) is 4 is

If the freezing point of 0.1 M HA(aq) solution is -0.2046^(@)C then pH of solution is ( If K_(f) water =1.86mol^(-1)kg^(-1))