Question

The concentration of calcium carbonate in a saturated aqueous solution at 25°C is 4.24 x 10⁻⁵ M. What is the Ksp of this salt? the answer should have 3 sig figs. _____x 10______

Answer 1

The molar solubility of calcium carbonate is 2.06 x 10^-3 M, and the Ksp of calcium carbonate is 1.74 x 10^-8 M.

Step-by-step explanation:

The Ksp is the solubility product constant, which represents the equilibrium between a solid compound and its ions in a saturated solution. To calculate the Ksp of calcium carbonate, we need to set up an ICE (initial, change, equilibrium) table.

• CaCO3(s) → Ca2+(aq) + CO32-(aq)
• Initial concentration: 0 M
• Change in concentration: +x M (assuming the molar solubility of CaCO3 is x)
• Equilibrium concentration: x M for both Ca2+ and CO32-

Using the given concentration of 4.24 x 10-5 M, we can plug in the values to the Ksp expression:

Ksp = [Ca2+][CO32-]

Ksp = (x)(x) = x^2

Substituting the given concentration, we have:

4.24 x 10-5 = x^2

To solve for x, we take the square root of both sides:

x = √(4.24 x 10-5) = 2.06 x 10-3

Since we're asked to round to 3 significant figures, the molar solubility of CaCO3 is 2.06 x 10-3 M. Therefore, the Ksp of calcium carbonate is 4.24 x 10-5 M x 4.24 x 10-5 M = 1.74 x 10-8.