If 500 mL of a solution containing 1.00 g of barium nitrate (199.33 g/mol) is combined with
500 mL of a solution containing 1.00 g of sodium sulfate (142.03 g/mol) , will there be a precipitate? Barium sulfate has a K_sp value of 9.1 *10^-¹¹.

Question

Quizplus · Accepted Answer

The chemical equation for the reaction between barium nitrate and sodium sulfate is:

Ba(NO3)2 + Na2SO4 → BaSO4 + 2 NaNO3

When the two solutions are combined, the barium ions (Ba2+) and sulfate ions (SO42-) can potentially combine to form insoluble barium sulfate (BaSO4) precipitate.

To determine whether a precipitate will form, we need to calculate the ion product (Qsp) for barium sulfate and compare it to its solubility product constant (Ksp).

Qsp = [Ba2+][SO42-]

The initial concentration of Ba2+ and SO42- in the combined solution is:

[Ba2+] = 1.00 g Ba(NO3)2 / (199.33 g/mol * 0.500 L) = 0.0100 M
[SO42-] = 1.00 g Na2SO4 / (142.03 g/mol * 0.500 L) = 0.0141 M

Therefore, Qsp = (0.0100 M)(0.0141 M) = 1.41 × 10^-4

Since Qsp < Ksp, the solution is unsaturated and no precipitate will form. However, if Qsp > Ksp, the solution is supersaturated and a precipitate will form.

The Ksp value for barium sulfate is given as 9.1 × 10^-10. Since Qsp < Ksp, the solution is unsaturated and no precipitate will form. Thus, the best choice is A.

If 500 ML of a Solution Containing 1