Question

Consider a solution of 0.250 M KOCl. Ka of HOCl = 3.5 x 10^-8.

Identify the acid and base used to create the salt.

Write the balanced equation of KOCl in water.

Calculate the pH of the solution.

Answer 1

Solution:- KOCl(aq) + H₂O(l) <------> KOH(aq) + HOCl(aq)

From the hydrolysis of the salt, the acid and base used to form this salt are HOCl and KOH. From the ka value, HOCl is a weak acid and KOH is a strong base, so the solution would be basic in nature.

let's make the ice table to calculate the pH.

  KOCl(aq) + H₂O(l) <------> KOH(aq) + HOCl(aq)

I 0.250 0 0

C -X +X +X

E (0.250 - X) X X

Kb = [KOH][HOCl][KOCl]

Kb = Kw/ka = 1.0 x 10^-14/3.5 x 10^-8 = 2.9 x 10^-7

2.9 x 10^-7 = (X)²/(0.250 - X)

since the value of Kb is very low so we could use approxymation and the value of X on the bottom could be neglected and so 0.250 - X could be taken as 0.250

2.9 x 10^-7 = X²/0.250

X² = 2.9 x 10^-7 x 0.250

X² = 7.25 x 10^-8

taking square root to both sides...

X = 2.69 x 10^-4

KOH is a strong base and at equilibrium it's concentration is X = 2.69 x 10^-4

being strong base, it dissociates completely and so, [OH^-] = 2.69 x 10^-4

pOH = - log(2.69 x 10^-4) = 3.57

pH = 14 - pOH

pH = 14 - 3.57

pH = 10.43