Copper(I) ions in aqueous solution react with NH3(aq) according to:
Cu+(aq) + 2NH3(aq) > Cu(NH3)2+ (aq)
Kf=6.3x10^10 M^-2
Calculate the solubility (in g·L–1) of CuBr(s) in 0.65 M NH3(aq).
To calculate the solubility of CuBr(s) in 0.65 M NH3(aq), we need to use the concept of solubility product constant (Ksp).
First, we need to write the balanced equation for the dissociation of CuBr(s) in water:
CuBr(s) ⇌ Cu+(aq) + Br-(aq)
Now, let's define the solubility product constant expression (Ksp) for CuBr:
Ksp = [Cu+][Br-]
Since we are given the concentration of NH3(aq) as 0.65 M, we need to calculate the concentration of Cu+(aq) and Br-(aq) using the given information:
Cu+(aq) + 2NH3(aq) ⇌ Cu(NH3)2+(aq)
We can see from the balanced equation that the concentration of Cu+ is two times the concentration of Cu(NH3)2+. Therefore, if we find the concentration of Cu(NH3)2+, we can determine the concentration of Cu+.
To find the concentration of Cu(NH3)2+, we need to use the equilibrium constant (Kf) given in the question:
Kf = [Cu(NH3)2+]/[Cu+][NH3]^2
Since we know the concentrations of Cu+ and NH3, we can rearrange the equation to solve for [Cu(NH3)2+]:
[Cu(NH3)2+] = Kf * [Cu+][NH3]^2
Substituting the given values:
[Cu(NH3)2+] = (6.3x10^10 M^-2) * [Cu+][NH3]^2
Now that we have the concentration of Cu(NH3)2+, we can calculate the concentration of Cu+:
[Cu+] = 2 * [Cu(NH3)2+]
Once we have the concentration of Cu+, we can plug it into the solubility product constant expression and solve for the concentration of Br-:
Ksp = [Cu+][Br-]
Now, we can solve for [Br-] and convert it to the solubility of CuBr(s) in g·L–1.
Please provide the necessary values for [Cu(NH3)2+] and [NH3] to proceed with the calculations.