What is the molar solubility of Mg(OH)2 in a solution containing 0,10 M NH3 ?
Ksp Mg(OH)2 = 1.2x10^-11
Kp NH3 = 1.8x10^-5
To find the molar solubility of Mg(OH)2 in a solution containing 0.10 M NH3, we need to consider the reaction between Mg(OH)2 and NH3. Based on the given information, we can write the following equation:
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)
Since NH3 is a weak base, it can react with water and generate OH- ions:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
We can use the solubility product constant (Ksp) for Mg(OH)2 and the base dissociation constant (Kb) for NH3 to find the concentration of OH- ions, which will allow us to determine the molar solubility of Mg(OH)2.
First, let's write the expression for the Ksp of Mg(OH)2:
Ksp = [Mg2+] [OH-]^2
Since Mg(OH)2 dissolves to create Mg2+ and 2 OH- ions, the concentration of Mg2+ in the solution is equal to the molar solubility of Mg(OH)2, which we'll denote as "x". Therefore, [Mg2+] = x.
Using the stoichiometry of the reaction, the concentration of OH- ions is twice the concentration of Mg(OH)2. Therefore, [OH-] = 2x.
Now, let's write the expression for the Kb of NH3:
Kb = [NH4+] [OH-] / [NH3]
Since we know the value of Kb and the concentration of NH3 is given as 0.10 M, we can solve for [OH-].
The Kb expression becomes:
1.8x10^-5 = [NH4+] (2x) / 0.10
Simplifying:
1.8x10^-5 = (2x) [NH4+] / 0.10
Now we have two equations:
Ksp = x (2x)^2 = 4x^3 = 1.2x10^-11
1.8x10^-5 = (2x) [NH4+] / 0.10
We can solve these equations simultaneously to find the value of x, which represents the molar solubility of Mg(OH)2 in the given NH3 solution.
......Mg(OH)2 ==> Mg^2+ + 2OH^-
I......solid.......0.......0
C......solid.......x.......2x
E......solid.......x.......2x
Then NH3 + HOH ==> NH4^+ + OH^-
I....0.1...........0........0
C....-x............x........x
E....0.1-x.........x........x
First, determine the OH^- from the NH3 equation.
Then Ksp = (Mg^2+)(OH^-)
You know Ksp.
(Mg^2+) = x = solubility
(OH^-) = 2x from the Mg(OH)2 dissolution and OH^- from the NH3 for a total of 2x+OH^- from NH3.
Solve for x = Mg^2+ = solubility