The solubility product constant (Ksp) of copper(I) hydroxide (CuOH) is 2.00 × 10-15. What is its molar solubility (S0) in mol/L?  3.89 × 10-8 mol/L 4.47 × 10-8 mol/L 5.01 × 10-8 mol/L 5.23 × 10-8 mol/L

The solubility product constant (Ksp) is given by the product of the concentrations of the ions in a saturated solution of an ionic compound. For copper(I) hydroxide (CuOH), the dissolution can be represented as:

CuOH(s) ⇌ Cu+(aq) + OH-(aq)

The Ksp expression for this reaction is:

Ksp = [Cu+][OH-]

Since the molar solubility (S0) is the concentration of CuOH that can dissolve in solution, we can assume that for each mole of CuOH that dissolves, one mole of Cu+ and one mole of OH- are produced. Therefore, [Cu+] = [OH-] = S0.

Substituting these into the Ksp expression gives:

Ksp = S0^2

Solving for S0 gives:

S0 = sqrt(Ksp)

Substituting the given value of Ksp = 2.00 × 10^-15 gives:

S0 = sqrt(2.00 × 10^-15) = 1.41 × 10^-8 mol/L

None of the provided options match this result. There may be a mistake in the question or the provided options.