Calculate the molar solubility in NaOH Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.190M NaOH? (Ksp =5.61*10^-11)

Question

Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.

Answer 1

The K_sp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

 

The equation for the solubilization reaction is:

 

Mg(OH)₂ (s) <==> Mg²⁺ (aq) + 2 OH^– (aq)

 

And the associated equilibrium expression is:

 

K_sp = [Mg²⁺][OH^–]²    <== note that Mg(OH)₂(s) in not included because it is a solid. Solids have no molarity.

 

We know that:

• a small amount (let's call it "x") of Mg(OH)₂(s) will dissolve in solution
• the solution already contains a large amount of of OH^– from the presence of 0.190M NaOH

 

Using the information above, we can set up an ICE table for the equilibrium:

 

        Mg(OH)₂ (s) <==> Mg²⁺ (aq) + 2 OH^– (aq)

I      (irrelevant)^a                 0                 0.190M

C          -x                         +x                    +2x                

E     (irrelevant)^a                 x             0.190 + 2x  ~0.190^b

 

a) The "concentration" of a solid is meaningless, which is why it is not part of the K_sp expression. However, "x" will give you the molarity of Mg(OH)₂ (aq) present after the equilibration process. This value is equal to [Mg²⁺] (1:1 ratio)

b) the value of "OH^– + 2x" is essentially equal to the value of OH^– because "x" is going to be a very small number. This is predictable for two reasons: (1) K_sp is a very small value, meaning the left side of the equilibrium is favored, and (2) the presence of a large concentration of [OH^–] at the start will push the equilibrium to the left, thus "disfavoring" the dissolution reaction.

 

We can now plug the values from the ICE table into the K_sp expression:

 

       K_sp        = [Mg²⁺][OH^–]²

5.61x10^–11   = (x)(0.190)²

 

All that is left is to solve the equation for x

 

x = K_sp/[OH–]² = 5.61x10^–11 / (0.190)²