Question

Magnesium hydroxide is used in several antacid formulations. When it is added to water it dissociates into magnesium and hydroxide ions. Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) The equilibrium constant at 25°C is 8.9 ´ 10-12. One hundred grams of magnesium hydroxide is added to 1.00 L of water and equilibrium is established. What happens to the solution if another 10 grams of Mg(OH)2 are now added to the mixture?

Answer 1

Adding another 10 grams of Mg(OH)₂ to a saturated solution will not change the concentration of ions in solution, as the solubility limit has already been reached and excess solid will remain undissolved.

Step-by-step explanation:

When 100 grams of magnesium hydroxide (Mg(OH)₂) is added to 1.00 L of water, it dissociates into magnesium (Mg²⁺) and hydroxide ions (OH⁻) until equilibrium is achieved based on its solubility product constant (Ksp) at 25°C, which is 8.9 × 10⁻¹². Given that the solubility of magnesium hydroxide is 0.0084 g per liter, initially, most of the 100 grams will remain undissolved as saturation is reached quickly. If an additional 10 grams of Mg(OH)₂ is added to the already saturated solution, no increase in the concentration of the ions in solution will occur since the system has already reached its solubility limit. The additional magnesium hydroxide will remain as a solid unless the equilibrium is disturbed by other reactions, such as with the addition of acid.

Answer 2

If you add 10 more grams of Mg(OH)₂, the solution will shift the equilibrium to the ions production to consume the excessive Mg(OH)₂.

Step-by-step explanation:

Mg(OH)₂(s) → Mg²⁺(aq) + 2OH⁻(aq)

K = 8.9 x 10⁻¹²

K = [Mg²⁺][OH⁻]²

The concentration of Mg(OH)₂ does not enter the equation because it is in the solid state.

[Mg²⁺] = x

[OH⁻]² = 2x

K = x.(2x)²

K = 4x³

4x³ = 8.9 x 10⁻¹²

x = 2.07 x 10⁻⁴ mol/L

At equilibrium:

[Mg²⁺] = x = 2.07 x 10⁻⁴ mol/L

[OH⁻]² = 2x = 4.14 x 10⁻⁴ mol/L