Question

What is the molar solubility of calcium carbonate ( CaCO3 ) in water? The solubility-product constant for CaCO3 is 4.5 × 10-9 at 25 °C.
A) 8.35
B) 9.5 × 10-5
C) 9.0 × 10-9
D) 2.3 × 10-9
E) 6.7 × 10-5

Answer 1

Answer:

Option E)

• $6.7\times 10^{-5}M$

Explanation:

1. Write the solubility equation:

• CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻

2. Write the concentrations below the equation:

• CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻

             A - s               s            s

3. Write the Ksp equation:

• Ksp = [Ca²⁺] . [CO₃²⁻]       ↔ the solid substances do not appear

• Ksp = s × s

• 4.5 × 10⁻⁹ = s²

4. Solve the equation:

       $s=\sqrt{4.5\times 10^{-9}M^2}=6.7\times 10^{-5}M$