Question

calculate the pH of a .72M KOCl solution at 25°C. the ka HOCl is 4x10^-8

Answer 1

KOCl is a salt of strong base KOH and weak acid HOCl.

Hence, it dissociates as follows

Now, the cation K+ does not contribute to pH as it comes from KOH which dissociates completely in water.

Now, OCl- will react with water as follows.

Now, the equilibrium constant of the above reaction can be expressed as

Where the concentrations are their equilibrium concentrations.

Now, given that Ka of HOCl is we can calculate the Kb of its conjugate base OCl^- as follows

Now, given that the initial concentration of KOCl = 0.72 M

Initial concentration of = 0.72 M

Hence, we can create the following ICE table for the reaction above.

Initial, M	0.72	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.72-x	x	x

Hence, we can write the equilibrium constant Kb as

Hence, the equilibrium OH- concentration of the solution is

We know that

Hence, the H+ concentration can be calculated as

pH is defined as the negative logarithm of the H+ concentration. Hence, pH of the solution can be calculated as

Hence, the pH of the solution is approximately 10.6.