BaSO4 (s) ⇌ Ba2+ (aq) + SO4^2- (aq)
The Correct Answer and Explanation is:
To calculate the solubility of barium sulfate (BaSO4) in pure water, we will use its solubility product constant (Ksp), which is given as 1.1 × 10^-10 at 25°C.
The dissolution of BaSO4 in water can be represented by the following equilibrium equation:
BaSO4 (s) ⇌ Ba2+ (aq) + SO4^2- (aq)
Let the solubility of BaSO4 be S moles per liter. At equilibrium, the concentrations of Ba2+ and SO4^2- will both be S, since one mole of BaSO4 dissociates to give one mole of Ba2+ and one mole of SO4^2-.
The expression for the solubility product (Ksp) is:
Ksp = [Ba2+][SO4^2-]
Since both ion concentrations are equal to S, the equation becomes:
Ksp = S²
Substituting the given Ksp value:
1.1 × 10^-10 = S²
To find S, take the square root of both sides:
S = √(1.1 × 10^-10)
S ≈ 1.05 × 10^-5 M
Therefore, the solubility of BaSO4 in pure water is approximately 1.05 × 10^-5 moles per liter.
Now, to convert this into grams per liter, we need the molar mass of BaSO4. The molar mass of BaSO4 is:
Ba = 137.33 g/mol
S = 32.07 g/mol
O = 16.00 g/mol (×4 for SO4)
Molar mass of BaSO4 = 137.33 + 32.07 + 64.00 = 233.40 g/mol
The solubility in grams per liter is:
Solubility (g/L) = S × molar mass
Solubility ≈ (1.05 × 10^-5 mol/L) × (233.40 g/mol)
Solubility ≈ 2.45 × 10^-3 g/L
Therefore, the solubility of BaSO4 in pure water is approximately 2.45 × 10^-3 grams per liter.
