A 0.012-mol sample of Na₂SO₄ is added to 400 mL of each of two solutions. One solution contains 1.5 × 10^-3 M BaCl₂; the other contains 1.5 × 10^-3 M CaCl₂. K_sp for BaSO₄ = 1.5 × 10^-9 and K_sp for CaSO₄ = 6.1 × 10^-5. Which of the following statements is true? A) Both BaSO₄ and CaSO₄ would precipitate. B) Not enough information is given to determine whether precipitation would occur. C) CaSO₄ would precipitate but BaSO₄ would not. D) BaSO₄ would precipitate but CaSO₄ would not. E) Neither BaSO₄ nor CaSO₄ would precipitate.

Question

Quizplus · Accepted Answer

First, we need to write the equation of dissolution of Na2SO4:

Na2SO4(s) → 2Na+(aq) + SO42-(aq)

Next we need to write the dissociation equations for BaCl2 and CaCl2:

BaCl2(s) → Ba2+(aq) + 2Cl-(aq)
CaCl2(s) → Ca2+(aq) + 2Cl-(aq)

We can then write the solubility product expressions:

Ksp for BaSO4 = [Ba2+][SO42-]
Ksp for CaSO4 = [Ca2+][SO42-]

Using the given concentrations and the dissociation equations, we can determine the initial concentrations of Ba2+ and Ca2+ in each solution.

For BaCl2 solution:

[Ba2+] = 1.5 × 10-3 M
[SO42-] = 0.006 M (0.012 mol Na2SO4 in 400 mL solution)

Substituting these values into the Ksp expression for BaSO4 gives:

Ksp = [Ba2+][SO42-] = (1.5 × 10-3)(0.006) = 9 × 10-9

For CaCl2 solution:

[Ca2+] = 1.5 × 10-3 M
[SO42-] = 0.006 M (0.012 mol Na2SO4 in 400 mL solution)

Substituting these values into the Ksp expression for CaSO4 gives:

Ksp = [Ca2+][SO42-] = (1.5 × 10-3)(0.006) = 9 × 10-9

Comparing the calculated Ksp values to the Ksp values given in the question for BaSO4 and CaSO4, we can see that Ksp for BaSO4 is smaller than the given value (1.5 × 10-9) while Ksp for CaSO4 is larger than the given value (6.1 × 10-5). This means that BaSO4 will not precipitate from either solution while CaSO4 will precipitate from the CaCl2 solution.

Therefore, the correct answer is D) BaSO4 would precipitate but CaSO4 would not.

A 0012-Mol Sample of Na2SO4 Is Added to 400 ML of of Each

A 0012-Mol Sample of Na₂SO₄ Is Added to 400 ML of of Each