Question

Determine the pH of each of the following solutions: 0.25M of CH3NH3I and 0.25M KI

Answer 1

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

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 = 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

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

I am giving one example in case of formic acid dissociation, please follow the same procedure for calculating the pH for KI

HCOOH(aq) --> HCOO-(aq) + H+(aq)

Ka = 10^-pKa = 10-3.744 = 1.8 * 10^-4

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

HCOO-(aq) + H2O(l) --> HCOOH(aq) + OH-(aq)

let x = [OH-]

then x = [HCOOH]

and 0.25 - x = [HCOO-]

x^2/(0.25-x) = 5.55 * 10^-11

x^2 - 5.55 * 10-11x - 1.38 * 10^-11 = 0

x = [OH-] = 3.71 * 10^-6 M ==> pOH = -log[OH-] = -log(3.71 * 10^-6) = 5.43

pH = 14 - pOH = 14 - 5.43 = 8.57

Thank you and good luck .