Final answer:
The concentration of Cu in a saturated solution of copper(I) bromide, given the Ksp of 6.3 x 10^-9, is 7.9 x 10^-5 M, which is option d) None of the above.
Step-by-step explanation:
The student has asked about the concentration of Cu in a saturated solution of copper(I) bromide. Given the Ksp (solubility product constant) for copper(I) bromide (CuBr) as 6.3 x 10^-9, we can calculate the molar solubility of CuBr in a solution. The dissolution of CuBr in water can be represented by the equation:
CuBr (s) → Cu+ (aq) + Br- (aq)
From the equation, it's evident that each mole of CuBr that dissolves produces one mole of Cu+ and one mole of Br-. Thus, if s is the solubility of CuBr in mol/L, then [Cu+] = s and [Br-] = s. The Ksp expression is:
Ksp = [Cu+] * [Br-]
Substituting s for each concentration in the equation, we get:
Ksp = s^2
Solving for s gives:
s = √(Ksp)
Therefore:
s = √(6.3 x 10^-9) = 7.9 x 10^-5 M.
This means that the concentration of Cu in a saturated solution of copper(I) bromide is 7.9 x 10^-5 M, which corresponds to the molar solubility. So, the correct answer is option d) None of the above.