# Dissolution Versus Weak Electrolyte Formation

## Dissolution versus Weak Electrolyte Formation

We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Châtelier’s principle. For example, one way to control the concentration of manganese(II) ion, Mn2+, in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:

$$\text{Mn}{(\text{OH})}_{2}(s)⇌{\text{Mn}}^{\text{2+}}(aq)+2{\text{OH}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}[{\text{Mn}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^{2}$$

This could be important to a laundromat because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn2+ ion while increasing the amount of solid Mn(OH)2 in the equilibrium mixture, as predicted by Le Châtelier’s principle.

## Example

### Solubility Equilibrium of a Slightly Soluble Solid

What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH when each of the following are added to a mixture of solid Mg(OH)2 in water at equilibrium?

(a) MgCl2

(b) KOH

(c) an acid

(d) NaNO3

(e) Mg(OH)2

### Solution

The equilibrium among solid Mg(OH)2 and a solution of Mg2+ and OH is:

$$\text{Mg}{(\text{OH})}_{2}(s)⇌{\text{Mg}}^{\text{2+}}(aq)+2{\text{OH}}^{\text{−}}(aq)$$

(a) The reaction shifts to the left to relieve the stress produced by the additional Mg2+ ion, in accordance with Le Châtelier’s principle. In quantitative terms, the added Mg2+ causes the reaction quotient to be larger than the solubility product (Q > Ksp), and Mg(OH)2 forms until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is less and [Mg2+] is greater than in the solution of Mg(OH)2 in pure water. More solid Mg(OH)2 is present.

(b) The reaction shifts to the left to relieve the stress of the additional OH ion. Mg(OH)2 forms until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is greater and [Mg2+] is less than in the solution of Mg(OH)2 in pure water. More solid Mg(OH)2 is present.

(c) The concentration of OH is reduced as the OH reacts with the acid. The reaction shifts to the right to relieve the stress of less OH ion. In quantitative terms, the decrease in the OH concentration causes the reaction quotient to be smaller than the solubility product (Q < Ksp), and additional Mg(OH)2 dissolves until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is less and [Mg2+] is greater than in the solution of Mg(OH)2 in pure water. More Mg(OH)2 is dissolved.

(d) NaNO3 contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg2+ and OH. (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.)

(e) The addition of solid Mg(OH)2 has no effect on the solubility of Mg(OH)2 or on the concentration of Mg2+ and OH. The concentration of Mg(OH)2 does not appear in the equation for the reaction quotient:

$$Q=[{\text{Mg}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^{2}$$

Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.

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