Copper(I) ions in aqueous solution react with NH3 according to
Cu^+ + 2NH3 --> Cu(NH3)2^+ Kf=6.3x10^10
Calculate the solubility (in g/L) of CuBr (Ksp=6.3x10^-9) in 0.81 M NH3.
CuBr ==> Cu^+ + Br^-
Cu^+ + 2NH3 ==> Cu(NH3)2^+
----------------------------
add equn 1 to eqn 2.
...CuBr(s) + 2NH3 ==>Cu(NH3)2^+ + Br^-
I.............0.81M...0............0
C............-2x......x............x
E...........0.81-2x...x............x
Keq for the rxn shown is Ksp*Kf
Substitute from the ICE chart and solve for x = [Cu(NH3)2] = (Br^-) = solubility of CuBr in M. Then convert to g/L.
To calculate the solubility of CuBr (copper(I) bromide) in 0.81 M NH3 (ammonia), we need to consider the formation of the complex ion Cu(NH3)2^+ and its equilibrium in solution.
Let's break down the problem into steps:
Step 1: Write the balanced equation for the reaction
Cu^+ + 2NH3 -> Cu(NH3)2^+
Step 2: Write the expression for the equilibrium constant (Kf)
Kf = [Cu(NH3)2^+] / [Cu^+][NH3]^2
Step 3: Calculate the concentration of Cu^+ in solution
Since CuBr is a sparingly soluble salt, we can assume that most of the CuBr dissociates to form Cu^+ ions:
[Cu^+] = solubility of CuBr
Step 4: Calculate the concentration of NH3 in solution
Given: [NH3] = 0.81 M
Step 5: Solve for [Cu(NH3)2^+] using the equilibrium constant expression
Kf = [Cu(NH3)2^+] / [Cu^+][NH3]^2
[Cu(NH3)2^+] = Kf * [Cu^+][NH3]^2
Step 6: Calculate solubility of CuBr
Since we know that CuBr dissociates to form Cu^+ ions and the molar mass of CuBr, we can calculate the solubility (in mol/L) of CuBr:
solubility (in mol/L) = [Cu^+]
solubility (in g/L) = solubility (in mol/L) * molar mass of CuBr
Now let's put the values into the equation:
Given: Ksp = 6.3x10^-9
Given: Kf = 6.3x10^10
Given: [NH3] = 0.81 M
Molar mass of CuBr = (atomic mass of Cu) + (atomic mass of Br)
By substituting the given values into the equation and solving for the solubility of CuBr, you can determine the answer.