What is the gram solubility of Mg(OH)2 in a 0.55mol/L solution of KOH if the Ksp of Mg(OH)2 is 9.0x10^-12?
This is a 1:2 ionization in pure water => Solubility of any 1:2 or 2:1 salt ionization = Cube Root(Ksp/4).
Solubility (pure water) of Mg(OH)2 = Cube Root[(9x10^-12)/4] = 1.3x10^4 M.
1:1 ionization => S = Sqr Rt(Ksp)
1:2 or 2:1 => S = Cube Rt(Ksp/4)
1:3 or 3:1 => S = 4th Rt(Ksp/27)
1:4 or 4:1 => S = 5th Rt(Ksp/256)
2:3 or 3:2 => S = 5th Rt(Ksp/108)
To find the gram solubility of Mg(OH)2 in a solution of KOH, we need to determine the concentration of Mg2+ ions in the solution. This can be done by using the principle of ionic equilibrium and the Ksp (solubility product constant) of Mg(OH)2.
The balanced equation for the dissociation of Mg(OH)2 in water is:
Mg(OH)2 ⇌ Mg2+ + 2OH-
From this equation, we can see that for every 1 mole of Mg(OH)2, you will get 1 mole of Mg2+ ions. Therefore, the concentration of Mg2+ ions will be equal to the concentration of Mg(OH)2.
Using the given Ksp value of Mg(OH)2, we can set up the equilibrium expression:
Ksp = [Mg2+][OH-]^2
Since the concentration of OH- ions in the solution is the same as the concentration of KOH (0.55 mol/L - assuming it fully dissociates), we can rewrite the equation as:
Ksp = [Mg2+][0.55]^2
Solving for [Mg2+], we get:
[Mg2+] = Ksp / [0.55]^2
[Mg2+] = (9.0x10^-12) / (0.55)^2
[Mg2+] ≈ 2.04x10^-11 mol/L
Now, to find the gram solubility of Mg(OH)2, we need to convert the concentration to grams per liter. The molar mass of Mg(OH)2 is 58.32 g/mol.
Gram solubility = [Mg2+] × molar mass of Mg(OH)2
Gram solubility = (2.04x10^-11 mol/L) × (58.32 g/mol)
Gram solubility ≈ 1.19x10^-9 g/L
Therefore, the gram solubility of Mg(OH)2 in a 0.55 mol/L solution of KOH is approximately 1.19x10^-9 g/L.