Ignoring activities, determine the molar solubility of copper (I) azide (CuN 3 )
ID: 904266 • Letter: I
Question
Ignoring activities, determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 3.696. Ksp (CuN3) = 4.9x10-9 ; Ka (HN3)= 2.2x10-5.
List of steps my teacher suggests we follow:
1. Write the pertinent reactions.
2. Write the charge balance equation.
3. Write the mass balance equations.
4. Write the equilibrium constant expressions for each reaction.
5. Count the equations and unknowns.
6. Solve for all unknowns.
How far I've gotten:
1. Pertinent reactions:
CuN3 (s) <-> Cu+ (aq) + N3- (aq) Ksp= 4.9x10-9
N3- (aq) + H2O (l) <-> HN3 (aq) + OH- (aq) Kb = (Kw/Ka) = (1.0x10-14/2.2x10-5) = 4.55 x 10-10
H2O (l) <-> H+ (aq) + OH- (aq) Kw = 1.0x10-14
2. Write the charge balance equation:
Because the pH is fixed with no info given regarding the buffer used to fix the pH, there is no charge balance equation for this problem.
3. Write the mass balance equation(s):
[Total Cu+] = [Total N3-]
[Cu+] = [N3-] +[HN3]
4. Write the equilibrium constant expressions for each reaction:
Ksp = 4.9x10-9 = [Cu+] [N3-]
Kb = 4.55x10-10 = ([HN3] [OH-] / [N3-])
Kw = 1.0x10-14 = [H+] [OH-]
Now use the charge balance, mass balance, and equilibrium constant expressions to solve for [Cu+], which is the molar solubility of CuN3.
I tried to finish it 5 times and haven't been able to make any progress.
Explanation / Answer
CuN3 ===> Cu+ + N3-
HN3 ==> H+ + N3-
As H+ increases, H+ combines with N3- which shifts the solubility equilibrium to the right to make CuN3 more soluble.
If we let S = solubility of CuN3, then S = Cu+ ; S= N3- + HN3
Ka = 2.2 * 10-5 = (H+)( N3- )/( HN3).
pH = 3.696 , (H+) = 2*10-4. Substitute that into Ka and solve for (HN3) in terms of N3- .Knowing HN3 (in terms of N3-), use the above solubility equation of
S = N3- + HN3
and solve for N3-
Then (Cu+)( N3-)= Ksp
and solve for S.
Ans: 7.34*10-5