At 22 °C an excess amount of a generic metal hydroxide M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.49. What is the Ksp of the salt at 22 °C?

pH + pOH = pKw = 14.

Solve for pOH. Then
pOH = -log(OH^-)
(OH^-) = about 3E-4 but that's only approximate.
.........M(OH)2 ==> M^2+ + 2OH^-
equl......solid....x.......2x
The problem tell you 2x = (OH^-) = 3E-4
Ksp = (Mg^2+)(OH^-)^2
Ksp = (x)(3E-4)^2
Remember to recalculate that 3E-4 number.

To determine the Ksp of the salt M(OH)2 at 22 °C, we need to first understand the relationship between pH and the concentration of hydroxide ions (OH-) in solution.

The pH scale measures the concentration of hydrogen ions (H+) in a solution. In a neutral solution, the pH is 7, meaning the concentration of H+ ions is equal to the concentration of OH- ions. As the pH increases above 7, the concentration of OH- ions increases, indicating a basic solution.

In this case, the pH is given as 10.49, indicating a basic solution. The concentration of OH- ions in the solution can be calculated using the equation:

pOH = 14 - pH

In this case, pOH = 14 - 10.49 = 3.51

To convert pOH to the concentration of OH- ions, we use the equation:

OH- concentration = 10^(-pOH)

OH- concentration = 10^(-3.51)

Now, we can use the concentration of OH- ions to determine the concentration of the generic metal cations, denoted as [M2+]. Since the hydroxide and cation concentrations are equal according to the balanced chemical equation:

[M(OH)2] ↔ M2+ + 2OH-

We can say:

[M2+] = [OH-] = 10^(-3.51)

Now, we need to calculate the solubility product constant, Ksp, which is the product of the ion concentrations raised to the power of their respective stoichiometric coefficients.

Ksp = [M2+][OH-]^2

Substituting the values we obtained:

Ksp = (10^(-3.51))(10^(-3.51))^2

Finally, we can simplify this equation to determine the Ksp value.