CALCULATING pH QUESTION

Question

Calculate the pH for a   solution with a = x  .(aq) +  (l) <-->  (aq) +  (aq)

Answer 1

NH₃(aq) + H₂O(l) ==> NH₄⁺(aq) + OH^-(aq) ... balanced equation for hydrolysis of NH3

Kb = [NH₄⁺][OH^-]/[NH₃]

1.76x10^-5 = (x)(x) / 4.87x10^-6 - x and assuming x is small relative to 4.87x10^-6, we can ignore it

x² = 8.57x10^-11

x = [OH^-] = 9.26x10^-6 M which is NOT small relative to 4.87x10^-6, so we must go back and use quadratic

x² = 8.57x10^-11 - 1.76x10^-5x

x² + 1.76x10^-5x - 8.57x10^-11 = 0

x = 3.97x10^-6 = [OH-]

pOH = -log 3.97x10^-6 = 5.40

pH = 14 - pOH

pH = 8.60