Final answer:
The solubility of calcite in grams per milliliter can be determined using the solubility product constant (Ksp). The Ksp of calcite is 9.9 x 10⁻⁹. By setting up the dissociation equation and solving for the molar solubility, we can calculate the solubility of calcite in grams per milliliter.
Step-by-step explanation:
The solubility of calcite, CaCO₃, in grams per milliliter of solution can be determined by using the solubility product constant (Ksp) value. In this case, the Ksp of calcite is given as 9.9 x 10⁻⁹. Ksp is the equilibrium constant for the dissolution of a solid salt into its constituent ions in a saturated solution. To find the solubility of calcite in grams per milliliter, we need to determine the molar solubility and then convert it to grams per milliliter using the molar mass of calcite.
First, let's set up the dissociation equation:
CaCO₃(s) → Ca²⁺(aq) + CO₃²⁻(aq)
Let's represent the molar solubility of CaCO₃ as s mol/L. Since the dissociation ratio is 1:1, the concentrations of Ca²⁺ and CO₃²⁻ ions at equilibrium will also be s mol/L.
From the dissociation equation, the expression for the solubility product constant is:
Ksp = [Ca²⁺][CO₃²⁻] = s * s = s²
Substituting the given Ksp value:
9.9 x 10⁻⁹ = s²
Taking the square root of both sides:
s ∼ 3.1 x 10⁻⁴ mol/L
To convert the molar solubility to grams per milliliter, we need to know the molar mass of calcite. Calcium has a molar mass of approximately 40 g/mol, carbon has a molar mass of approximately 12 g/mol, and oxygen has a molar mass of approximately 16 g/mol. So the molar mass of CaCO₃ is approximately 100 g/mol.
The solubility of calcite in grams per milliliter is calculated as follows:
3.1 x 10⁻⁴ mol/L × 100 g/mol = 3.1 x 10⁻² g/mL
Therefore, the solubility of calcite in grams per milliliter is approximately 3.1 x 10⁻² g/mL.