Question

Determine the pH of each of the following solutions.

Part A) 0.25 M KCHO2

Part B) 0.15 M CH3NH3I

Part C) 0.22 M KI

Answer 1

1. KCHO2 is potassium formate, a salt. The K⁺ ion is not hydrolyzed but the formate ion is in which it acts as a base.

CHO₂^- + H₂O HCOOH + OH^-

Ka = 10-pKa = 10-3.744 =6.256= antilog (6.256)= 1.8 10^-4

Kb = Kw/Ka = 10^-14/1.8 10^-4 = 5.55 10^-11

let x = [OH-]

then x = [HCOOH]

and 0.25 - x = [HCOO-]

Kb = (Kw/Ka) = (OH^-)(HCHO2)/(CHO2^-)

5.55 10^-11 = x² / 0.25 - x

x² - 5.55 10^-11x - 1.380^-11 = 0 solving for x we get,

x= [OH^-] = 3.7 10^-6 M , pOH = -log[OH^-] = -log[3.710^-6] = 5.43

pH = 14-pOH = 14-5.43= 8.57

2. CH3NH3I is a salt. the cation gets hydrolysed as follows:

CH3NH3⁺ H⁺ + CH3NH2

let x = [H+], then x = [CH3NH2]
and (0.24 - x) = [CH3NH3+]

CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH-(aq)

Kb = 10-pKb = 10-3.36 = antilog (6.64) = 4.37 10^-4

Kb = Kw/Ka ==> Ka = Kw/Kb = [H+][OH-][CH3NH2]/[CH3NH3+][OH-] =

[H+][CH3NH2]/[CH3NH3+] = 10^-14/4.37 10^-4 = 2.29 10^-11

CH3NH3+(aq) --> CH3NH2(aq) + H+(aq)

let = x= [H+] = [CH2NH2]
and 0.24 - x = [CH3NH3+]

x²/(0.24 - x) = 2.29 10^-11

x² + 2.29 10^-11 - 5.49 10^-12 = 0 , solving for x we get,

x = [H+] = 2.34 10^-6 M ==> pH = -log[H+] = -log(2.34 10^-6) = 5.63

3. KI. neither the K ion nor the I ion is ionized; therefore, the solution is just like table salt in water and the pH is determined by the ionization of water. pH = 7.0