Question

Ignoring activities, determine the molar solubility of copper(I) azide (CuN3) in a solution with a pH of 12.00. The Ksp for CuN3 is 4.9 × 10–9. The Ka for HN3 is 2.2 × 10–5.

Answer 1

Answer

pH = 13.11 [H+] = 7.76 x 10^-14 
HN3 <===> H+ + N3- 
Ka = [H+][N3-]/ [HN3] = 2.2 x 10^-5 
Ksp = [Cu+][N3-] = 4.9 x 10^-9 

CuN3(s) ====⇒ Cu+ + N3- 
H+ + N3- ====⇒ HN3 
Add the two equations: 
----------------------------------- 
CuN3(s) + H+ ==⇒ Cu+ + HN3 Keq = [Cu+][HN3] / [H+] 
Rearrange Ka to have [HN3] / [H+] [N3-] 
Ka = [H+][N3-] / [HN3] 
[HN3] / [H+][N3-] = 1/Ka 
Keq = Ksp x 1/Ka = [Cu+][N3-] x [HN3] /[H+][N3-] = [Cu+][HN3] / [H+] 
Keq = 4.9 x 10^-9 / 2.2 x 10^-5 = 2.23 x 10^-4 
Let Y = molar solubility of CuN3 
Y = [Cu+] = [HN3] 
Keq = 2.23 x 10^-4 = Y(Y) / [H+] 
2.23 x 10^-4 = (Y)(Y) / 7.6 x 10^-14 
Y^2 = 1.69 x 10^-17 
Y = 4.1 x 10^-9 = Molar solubility of CuN3