Calculate pH of 0.02 mol/L Ba(OH)2 solution. (Assume 100% dissociation).

To calculate the pH of a solution, we need to determine the concentration of hydrogen ions (H+) in the solution.

In this case, we have a 0.02 mol/L solution of Ba(OH)2. Since Ba(OH)2 is a strong base and dissociates completely, we can assume that it fully dissociates into Ba2+ and 2 OH- ions.

Since it is a 0.02 mol/L solution, the concentration of OH- ions will be 2 times the concentration of the Ba(OH)2. Therefore, the concentration of OH- ions is 0.02 mol/L x 2 = 0.04 mol/L.

To calculate the concentration of H+ ions, we need to use the self-ionization constant (Kw) of water, which is equal to 1 x 10^-14 at 25°C.

Since the concentration of OH- ions is 0.04 mol/L, we can use Kw to find the concentration of H+ ions:

Kw = [H+][OH-]

1 x 10^-14 = [H+][0.04]

[H+] = (1 x 10^-14) / 0.04

[H+] = 2.5 x 10^-13 mol/L

Now that we have the concentration of H+ ions, we can calculate the pH using the formula:

pH = -log [H+]

pH = -log (2.5 x 10^-13)

pH ≈ 12.60

Therefore, the pH of a 0.02 mol/L Ba(OH)2 solution is approximately 12.60.