A solution of `0.1 M NaZ` has `PH = 8.90`. The `K_a` of `HZ` is.

A solution of 0.1M NaZ has pH = 8.90 . The K_(a) of HZ is

A solution of 0.1M NaZ has pH = 8.90 . The K_(a) of HZ is

A 0.015 M solution of a weak acid has a pH of 3.52 . What is the value of the K_a for this acid?

Calculate pH solution: 0.1 M HCl

The freezing point of 0.20 M solution of week acid HA is 272.5 K . The molality of the solution is 0.263 "mol" kg^(-1) . Find the pH of the solution on adding 0.25 M sodium acetate solution. K_(f) of water = 1.86 K m^(-1)

The freezing point of 0.20 M solution of week acid HA is 272.5 K . The molality of the solution is 0.263 "mol" kg^(-1) . Find the pH of the solution on adding 0.25 M sodium acetate solution. K_(f) of water = 1.86 K m^(-1)

The freezing point of 0.20 M solution of week acid HA is 272.5 K . The molality of the solution is 0.263 "mol" kg^(-1) . Find the pH of the solution on adding 0.25 M sodium acetate solution. K_(f) of water = 1.86 K m^(-1)

Calculate pH of resultant solution of 0.1 M HA + 0.1 M HB [K_a (HA) =2 xx 10^(-5), K_a (HB)=4 xx 10^(-5)]

pH of a 0.01 M aq. Solution of potassium propanoate is 8 at 298 K. The Ka of propanoic acid at 298 K is

What should be the concentration of NaA. If its 50 mL solution is added to 50 mL of 0.2 M HA to have a pH = 4. K_a " for HA "= 1.0 xx 10^(-5) .