Question

The molar solubility of magnesium carbonate is 1.8 × 10–4 mol/l. What is ksp for this compound?

Answer 1

Answer:

3.2 × 10⁻⁸

Explanation:

Let's consider the solution of magnesium carbonate.

MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

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

         MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

I                             0                0

C                          +S              +S

E                            S                S

The Ksp is:

Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸

Answer 2

Answer:

Ksp = 3.2 * 10^-8

Explanation:

Step 1: Data given

magnesium carbonate = MgCO3

The molar solubility of magnesium carbonate = 1.8 * 10^-4 mol/L

Step 2: The balanced equation

MgCO3(s) ⇔ Mg^2+ + CO3^2-

For 1 mol MgCO3 we'll have 1 mol Mg^2+ and 1 mol CO3^2-

Step 3: Calculate Ksp for this compound

Ksp = [Mg^2+][Co3^2-]

For 1 mol MgCO3 we'll have 1 mol Mg^2+ and 1 mol CO3^2-

Ksp = X * X = X²

X = 1.8 * 10^-4

Ksp = (1.8 *10^-4 )* (1.8 * 10^-4)

Ksp = 3.2 * 10^-8