The solubility of Mg(OH)2 is measured and found to be 9.25×10-3 g/L. Use this information to calculate a Ksp value for magnesium hydroxide
Convert 9.25E-3 g/L to mols/L. I will call that x.
....Mg(OH)2 ==> Mg^2+ + 2OH^-
I...solid........0.......0
C...solid........x.......2x
E...solid........x.......2x
Substitute the E line into Ksp expression and solve for Ksp.
To calculate the Ksp value for magnesium hydroxide (Mg(OH)2), we need to use the solubility information provided.
The solubility of Mg(OH)2 is given as 9.25×10^(-3) g/L. This means that in each liter of solution, 9.25×10^(-3) grams of Mg(OH)2 dissolve.
The dissociation equation for Mg(OH)2 is:
Mg(OH)2 ⇌ Mg^2+ + 2OH^-
From this equation, we can see that one mole of Mg(OH)2 produces one mole of Mg^2+ ions and two moles of OH^- ions.
To find the concentration of Mg^2+ ions in solution, we assume that all the Mg(OH)2 dissociates completely. Thus, the concentration of Mg^2+ ions is equal to the solubility of Mg(OH)2, which is given as 9.25×10^(-3) g/L.
To find the concentration of OH^- ions in solution, we need to multiply the solubility of Mg(OH)2 by 2, as two OH^- ions are formed per mole of Mg(OH)2. Therefore, the concentration of OH^- ions is 2 × 9.25×10^(-3) g/L.
Now, we can use these concentrations to calculate the Ksp value:
Ksp = [Mg^2+][OH^-]^2 =
([Mg^2+]) × ([OH^-])^2 =
(9.25×10^(-3)) × (2 × 9.25×10^(-3))^2 =
(9.25×10^(-3)) × (4 × (9.25×10^(-3))^2) =
(9.25×10^(-3)) × (4 × 85.56×10^(-6)) =
(9.25×10^(-3)) × (342.24×10^(-6)) =
3.168×10^(-6)
Therefore, the Ksp value for magnesium hydroxide (Mg(OH)2) is 3.168×10^(-6).