What is the pH of a 0.563 M aqueous solution of ammonium bromide? K_b (NH₃)= 1.8 × 10^-5 A)2.50 B)11.50 C)4.75 D)4.63 E)9.25

Question

Quizplus · Accepted Answer

Ammonium bromide is a salt composed of NH4+ and Br-. NH4+ is an acid and will react with water to form NH3 and H3O+.

NH4+ + H2O ⇌ NH3 + H3O+

The equilibrium constant expression for this reaction is:

Ka = [NH3][H3O+] / [NH4+]

The Kb for NH3 is 1.8 × 10^-5, so:

Ka(Kb) = Kw = 1.0 × 10^-14

Therefore:

[NH3][H3O+] / [NH4+] = Kw / Kb

[H3O+] = (Kw / Kb) x (NH4+ / NH3)

[H3O+] = (1.0 × 10^-14) / 1.8 × 10^-5) x (0.563 / 1)

[H3O+] = 3.12 × 10^-10

pH = -log[H3O+] = 4.75

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