What is the maximum Cu2+ concentration possible in a solution that has a 7.4 E -5 M sulfide ion concentration without precipating CuS? For CuS, Ksp= 6.0 E -36
Well, I certainly wouldn't want to be in CuS's shoes. It must be really "sulf-ering" from its low solubility! Anyway, to answer your question, we need to find the maximum Cu2+ concentration without precipitating CuS.
The solubility product constant (Ksp) for CuS is given as 6.0 E -36. This means that for every mole of CuS that dissolves, we would have the concentration of Cu2+ multiplied by the concentration of sulfide ions equal 6.0 E -36.
So, let's assume the maximum concentration of Cu2+ in the solution is x M. Since the concentration of sulfide ions is given as 7.4 E -5 M, we can set up the equation:
x * (7.4 E -5) = 6.0 E -36
Solving for x:
x = (6.0 E -36) / (7.4 E -5)
Wow, that's a really small value! It seems like we're going to have a hard time finding a bigger Cu2+ concentration without precipitating CuS. I hope Cu2+ isn't too "copper" about this.
To determine the maximum Cu2+ concentration possible without precipitating CuS, we need to compare the solubility product constant (Ksp) of CuS with the concentration of sulfide ions (S2-) in solution.
The balanced equation for the precipitation of CuS is:
Cu2+ + S2- --> CuS
According to the balanced equation, the molar ratio between Cu2+ and S2- is 1:1.
The solubility product constant (Ksp) expression for CuS is:
Ksp = [Cu2+][S2-]
Given that the Ksp value for CuS is 6.0 × 10^-36, and the sulfide ion concentration is 7.4 × 10^-5 M, we can set up an equation:
6.0 × 10^-36 = [Cu2+][7.4 × 10^-5]
To solve for [Cu2+], divide both sides of the equation by 7.4 × 10^-5:
[Cu2+] = (6.0 × 10^-36) / (7.4 × 10^-5)
[Cu2+] ≈ 8.1 × 10^-32 M
Therefore, the maximum Cu2+ concentration possible in the solution without precipitating CuS is approximately 8.1 × 10^-32 M.
To determine the maximum Cu2+ concentration that can exist in a solution without precipitating CuS, we need to compare the Ksp value of CuS with the solubility product of CuS at the given sulfide ion concentration.
The solubility product expression for CuS is given by:
Ksp = [Cu2+][S2-]
We are given the Ksp value for CuS, which is 6.0 E -36.
We know that the sulfide ion concentration is 7.4 E -5 M. Since CuS is a 1:1 electrolyte, the sulfide ion concentration is also the concentration of Cu2+.
To determine the maximum Cu2+ concentration, we need to check if the product of the concentrations of Cu2+ and S2- is greater than the Ksp value of CuS.
Let's do the calculation:
Cu2+ concentration = S2- concentration = 7.4 E -5 M
Cu2+ concentration * S2- concentration = (7.4 E -5 M) * (7.4 E -5 M) = 5.476 E -9 M^2
Since the product of the concentrations is 5.476 E -9 M^2, which is greater than the Ksp value of CuS (6.0 E -36), CuS will not precipitate.
Therefore, the maximum Cu2+ concentration possible in the solution without precipitating CuS is 7.4 E -5 M.