Question

What is the solubility in moles/liter for magnesium hydroxide at 25 oC given a Ksp value of 1.1 x 10-11. Write using scientific notation and use 1 or 2 decimal places (even though this is strictly incorrect!)

Answer 1

S = 0.00014 moles /L = 1.4 * 10^-4 moles/L

Step-by-step explanation:

Step 1: Data given

Temperature = 25.0 °C

Ksp = 1.1 * 10^-11

Step 2: The balanced equation

Mg(OH)2(s) ⇆ Mg^2+(aq) + 2OH-(aq)

Step 3: Define Ksp

[Mg(OH)2 = 1.11 * 10^-11 = S

[Mg^2+] = S

[OH-] = 2S

Ksp = [Mg^2+]*[OH-]²

Ksp = S * (2S)²

1.1 * 10^-11 = 4S³

S³ = 2.75 * 10^-12

S = 0.00014 moles /L

Answer 2

1.40 × 10⁻⁴ M

Step-by-step explanation:

Let's consider the solution of magnesium hydroxide.

Mg(OH)₂(s) = Mg²⁺(aq) + 2 OH⁻(aq)

We can relate the solubility (S) of the hydroxide with the solubility product (Ksp) using an ICE chart.

Mg(OH)₂(s) = Mg²⁺(aq) + 2 OH⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product is:

Ksp = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4 S³

S = ∛(Ksp/4) = ∛(1.1 × 10⁻¹¹/4)

S = 1.40 × 10⁻⁴ M