At 22C(Celsius) an excess amount of a generic metal hydroxide M(OH)2 is mixed with pure water. The resulting equilibrium solution has a pH of 10.14. What is the Ksp of the salt at 22C? Help Please

..........M(OH)2 ==> M^2+ + 2OH^-

E.........solid......x.......2x
pH = 10.14 pOH = 14-10.14 = 3.86
pOH = -log(OH^-)
Solve for OH^- = about 1.4E-4 but that's approximate. That means (M^2+) = 1/2 that.
Ksp = (Mg^2+)(OH^-)^2
Substitute and solve for Ksp.

Mr. DrBob222,

would you mind explaining that a little more? I do not completely understand what is happening in the problem. Also, how do you know that Mg^2+ is half of OH^-?

To find the Ksp (solubility product constant) of the salt M(OH)2 at 22°C, we need to use the pH value of the equilibrium solution. The pH of a solution is related to the concentration of hydroxide ions (OH-) in the solution.

First, we need to find the concentration of hydroxide ions in the solution. Since the pH of the solution is given as 10.14, we can use the formula:

pOH = 14 - pH

pOH = 14 - 10.14
pOH ≈ 3.86

Next, we convert pOH back to [OH-] concentration using the formula:

[OH-] = 10^(-pOH)

[OH-] = 10^(-3.86)
[OH-] ≈ 1.21 x 10^(-4) M

The concentration of hydroxide ions in the equilibrium solution represents the solubility of M(OH)2 because each M(OH)2 molecule dissociates into 2 OH- ions.

Therefore, [M(OH)2] = 2 [OH-]
[M(OH)2] ≈ 2 * 1.21 x 10^(-4)
[M(OH)2] ≈ 2.42 x 10^(-4) M

Now, using the concentration of M(OH)2, we can write the Ksp expression for the salt's solubility equilibrium:

Ksp = [M] * [OH-]^2

Since the concentration of M(OH)2 is double the concentration of hydroxide ions, we can substitute the values into the expression:

Ksp = (2.42 x 10^(-4)) * (1.21 x 10^(-4))^2
Ksp ≈ 3.31 x 10^(-11) M^3

Therefore, the Ksp of the salt M(OH)2 at 22°C is approximately 3.31 x 10^(-11) M^3.