Question

Determine the pH of each of the following solutions. Part A 0.17 M KCHO 2 Part B 0.24 M CH 3 NH 3 I Part C 0.16 M KI

Answer 1

1) 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 0.17 - x = [HCOO^-]

x²/(0.17-x) = 5.55 * 10^-11

x² - 5.55 * 10^-11x - 9.435 * 10^-12 = 0

x = [OH^-] = 3.07 * 10^-6 M ==> pOH = -log[OH^-] = -log(3.07 * 10^-6) = 5.513

pH = 14 - pOH = 14 - 5.513 = 8.487

2) 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