Copper(I) ions in aqueous solution react with NH3(aq) according to
Cu+(aq)+2NH3(aq)⟶Cu(NH3)+2(aq)𝐾f=6.3×1010 M−2
Calculate the solubility (in g⋅L−1) of CuBr(s) in 0.22 M of NH3(aq).
I have a problem with
Cu+(aq)+2NH3(aq)⟶Cu(NH3)+2(aq)𝐾f=6.3×1010 M−2
Should that be [Cu(NH3)2]^+
and what are you to use for Ksp CuBr?
I think you may have written the product Ok except we can't^+ do superscripts right here.
CuBr(s) ==> Cu^+ + Br^- ........Ksp = ?
Cu^+ + 2NH3 ==> [Cu(NH3)2]^+ Kf = ? and add the two to get
---------------------------------------------------
..........CuBr(s) + 2NH3 ==> [Cu(NH3)2]^+ + Br^- Keq = Ksp*Kf
I..........solid.........0.22..............0.....................0
C........solid.........-2x.................x.....................x
E.........solid.......0.22-x.............x.....................x\
Plug the E line into the expression for Keq = Kf*Ksp and solve for x in mols/L. Convert to g/L.
Post your work if you get stuck.
You have Kf in the problem. I found Ksp CuBr = 6.3E-9 but you should use the value in your tables/charts in your notes/text. My first guess is that you will need to solve the quadratic equation.
kf= 6.3x10^10
ksp= 6.3x10^-9
so... k=396.9 or 4.0x10^2
so then I got 396.9= x^2/(0.22-2x)^2 and solved it getting 15.39 but this answer is wrong
To calculate the solubility of CuBr(s) in 0.22 M NH3(aq), we can use the solubility product constant (Ksp) and the formation constant (Kf) of the complex Cu(NH3)2+.
1. Write the balanced equation for the dissolution of CuBr(s) in NH3(aq):
CuBr(s) ⇌ Cu+(aq) + Br-(aq)
2. Write the expression for the solubility product constant (Ksp) in terms of the concentrations of the ions:
Ksp = [Cu+][Br-]
3. Since CuBr(s) does not readily dissociate in water, we can assume that the concentration of Br- ions is negligible compared to that of Cu+ ions. Therefore, we can assume that [Br-] ≈ 0.
4. Now, we need to determine the concentration of Cu+ ions in the presence of NH3(aq). This can be done using the formation constant (Kf) of the complex Cu(NH3)2+.
Cu+(aq) + 2NH3(aq) ⇌ Cu(NH3)2+(aq) (1)
The formation constant (Kf) for this reaction is given as:
Kf = [Cu(NH3)2+]/([Cu+][NH3]^2) = 6.3 ×10^10 M^-2
5. Since NH3 is in excess (0.22 M), we can assume that the concentration of NH3 does not change significantly during the reaction. Therefore, [NH3] ≈ 0.22 M.
6. Now, we can rewrite equation (1) as:
Cu+(aq) + 2NH3(aq) ⇌ Cu(NH3)2+(aq)
[Cu+][NH3]^2 / [Cu(NH3)2+] = Kf
[Cu+]/[Cu(NH3)2+] = Kf / [NH3]^2 = (6.3 × 10^10 M^-2) / (0.22 M)^2
7. We can now substitute the calculated value of [Cu+]/[Cu(NH3)2+] into the solubility product constant expression:
Ksp = [Cu+][Br-] = ([Cu+]/[Cu(NH3)2+]) × [Br-]
Since [Br-] ≈ 0, Ksp ≈ ([Cu+]/[Cu(NH3)2+]) × 0 = 0
8. Since Ksp ≈ 0, this implies that the concentration of Cu+ ions in the solution is extremely small, and therefore the solubility of CuBr(s) in 0.22 M NH3(aq) is negligible.
Therefore, the solubility of CuBr(s) in 0.22 M NH3(aq) can be considered as practically insoluble.