Question

The pH of a 0.050 ammonium chloride solution was experimentally determined to be 5.45.
a. write the equation for the chemical reaction that determines the pH?
I wrote the actual equation as NH4^+(aq) + H2O(l) ->/<- NH3 (aq) +H3O^+ the Ka =3.55 X 10^-6
pH = -log [3.55 X 10^-6]

PART B

Calculate the equilibrium constant

Answer 1

A solution with a pH = 5.45 would in fact have a hydronium ion concentration of 3.55 x 10ˉ⁶M, but for a 0.050M NH₄Cl(aq) solution the [H⁺] is closer to 5.27 x 10ˉ¹⁰M, pH = 5.28 and Ka(NH₄⁺) ~ 5.56 x 10 ˉ¹⁰.
NH₄Cl(aq) => NH₄⁺ + Clˉ(aq)
Clˉ + H₂O => no rxn
…………….NH₄⁺ + H₂O => NH₄OH + H⁺
C(eq)… 0.05M…….. --- ………x……...x
Ka(NH₄⁺) = Kw/Kb(NH₄OH) = 1 x 10ˉ¹⁴/1.8 x 10ˉ⁵ = 5.56 x 10ˉ¹⁰
= [NH₄OH][ H⁺]/[ NH₄⁺] = x²/0.050
=> x = [H⁺] = SqrRt[(0.050)(5.56 x 10ˉ¹⁰)] = 5.27 x 10ˉ⁶M => pH = 5.28 not 5.45.