The molar solubility of a generic substance, Mg(OH)2 ksp = (1.2x10^-11) in 0.10 M NH3 Kb = 1.8 x 10^-5

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What is solubility ****

To find the molar solubility of Mg(OH)2 in the presence of NH3, we need to determine the concentrations of Mg2+ and OH- ions in the solution.

First, let's write the balanced equation for the dissociation of Mg(OH)2:
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)

The solubility product constant (Ksp) expression for the equilibrium is:
Ksp = [Mg2+][OH-]^2

Given that the Ksp value is 1.2x10^-11, we can set up the following equation:
(1.2x10^-11) = [Mg2+][OH-]^2

Since the concentration of OH- ions is twice the concentration of Mg2+ ions, we can represent it as 2x:
(1.2x10^-11) = [Mg2+](2x)^2
(1.2x10^-11) = 4x^3

Solving for x, we get x ≈ 3.08x10^-4.

This value represents the concentration of OH- ions (and Mg2+ ions) in pure water. However, in the presence of NH3, the NH3 will react with OH- to form NH4+ and water according to the equation: OH- + NH3 ⇌ NH4OH

To consider this reaction, we need to use the Kb value of NH3, which is 1.8x10^-5. Kb is the base dissociation constant and represents the equilibrium constant for the reaction: NH4OH ⇌ NH4+ + OH-

Using the Kb expression, we have:
Kb = [NH4+][OH-]/[NH4OH]

Given that we want to find the concentration of OH-, which we'll denote as [OH-]eq (equilibrium concentration), we can rearrange the equation to isolate [OH-]:
[OH-]eq = (Kb × [NH4OH])/[NH4+]

Now, in this case, we have an excess of NH4OH due to the presence of NH3. NH4OH is weak and will only partially dissociate, so we can roughly assume that the equilibrium concentration of NH4OH will be equal to the initial concentration since it is already small.

Using the initial concentration of NH4OH as [NH4OH] = 0.10 M, and assuming [NH4+] is negligible, we can calculate the equilibrium concentration of OH-:
[OH-]eq ≈ (1.8x10^-5 × 0.10)/(0.10) = 1.8x10^-5

This means that the concentration of OH- in the presence of NH3 is approximately 1.8x10^-5 M.

Therefore, the molar solubility of Mg(OH)2 in the presence of 0.10 M NH3 is approximately 3.08x10^-4 M, and the equilibrium concentration of OH- is approximately 1.8x10^-5 M.