determine the molar solubility of copper (I) azide (CuN3) in a solution with a ph of 3.793. Ignore activities. The Ksp for CuN3 is 4.9x10^-9. The Ka for HN3 is 2.2x10^-5

Question

Use systematic treatment of equilibrium

Answer 1

To determine the molar solubility of copper(I) azide (CuN₃) in a solution with a pH of 3.793, I think we need to consider the dissociation of CuN₃(s) ==> Cu⁺(aq) + N₃^-(aq) and also the effect of the equilibrium for hydrazoic acid (the conjugate acid of N₃^-), HN₃(aq) ==> H⁺(aq) + N₃^-(aq). My best guess as to how to approach this is as follows:

Determine [H⁺] at pH 3.793...

pH = -log [H⁺]

[H⁺] = 1x10^-3.793 = 1.61x10^-4 M

Looking at the solubility of CuN₃ ...

CuN₃(s) ==> Cu⁺(aq) + N₃^-(aq)

Ksp = [Cu⁺][N₃^-]

4.9x10^-9 = [Cu⁺][N₃^-] and we need to find [N₃^-]

And since [N₃^-] is in equilibrium with HN₃, we can find [N₃^-]...

HN₃ <==> H⁺ + N₃^-

At equilibrium and pH 3.793, find [HN₃]

Ka = 2.2x10^-5 = [H+][N₃^-] / [HN₃] = (1.61x10^-4)² / [HN₃]

[HN₃] = 1.16x10^-3 M

At equilibrium and pH 3.793 find [N₃^-]

Ka = 2.2x10^-5 = (1.6x10^-4)(x) / 1.16x10^-3 - x

1.6x10^-4x = 2.6x10^-8 - 2.2x10^-5x

x = [N₃^-] = 1.43x10^-4 M

Going back to the Ksp expression and plugging in values, we have ...

Ksp = 4.9x10^-9 = [Cu⁺][1.43x10^-4]

[Cu⁺] = 3.4x10^-5 M = molar solubility of CuN₃