Question

Use systematic treatment of equilibrium to determine the molar solubility of copper(I) azide(CuN3) in solution with a pH of 11.88. The Ksp for CuN3 is 4.9 × 10–9. The Ka for HN3 is 2.2 × 10–5. Please show work, thanks

Answer 1

pH = -log[H+] = 11.88

[H+] = 1.32 x 10^-12 M

CuN3 <==> Cu+ + N3-   Ksp

N3- + H+ <==> HN3   1/Ka

adding the two equations,

CuN3 + H+ <==> Cu+ + HN3

Keq = [Cu+][HN3]/[H+]

Keq = Ksp/Ka

       = 4.9 x 10^-9/2.2 x 10^-5

       = 2.23 x 10^-4

let x be the change due to solubility of salt in solution

then,

2.23 x 10^-4 = x^2/(1.32 x 10^-12 - x)

2.94 x 10^-16 - 2.23 x 10^-4x = x^2

x^2 + 2.23 x 10^-4x - 2.94 x 10^-16 = 0

x = 1.315 x 10^-12 M

So,

molar solubility of copper(I) azide in solution at pH 11.88 = 1.315 x 10^-12 M