Question

The solubility of limestone, CaCO3, at 25˚C is 0.00067 g/100 mL. Write the chemical equation for the solubility equilibrium of this sparingly soluble salt in water. Then compute the molar solubility and the solubility product constant Ksp for CaCO3 at 25˚C.

Answer 1

Answer:

4.5 × 10⁻⁹

Explanation:

Step 1: Write the reaction for the solution of CaCO₃

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

Step 2: Convert the solubility of CaCO₃ from g/L to mol/L

We will use the following conversion factors:

• The molar mass of CaCO₃ is 100.09 g/mol.
• 1 L= 1000 mL.
• There are 0.00067 g of CaCO₃ per 100 mL of solution.

$\frac{0.00067 gCaCO_3}{100mLSol} \times \frac{1molCaCO_3}{100.09gCaCO_3} \times \frac{1000mLSol}{1LSol} = 6.7 \times 10^{-5} M$

Step 3: Calculate the solubility product constant (Ksp)

To relate Ksp and the molar solubility (S), we need to make an ICE chart.

        CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

I                               0                0

C                            +S               +S

E                              S                 S

The solubility product constant is:

Ksp = [Ca²⁺].[CO₃²⁻] = S² = (6.7 × 10⁻⁵)² = 4.5 × 10⁻⁹