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 $\times$ 10^-11.

Question

Quizplus · Accepted Answer

To determine if a precipitate will form, we need to calculate the ion concentrations (in mol/L) of Ba2+ and SO42- in the combined solution and compare them to the Ksp value for BaSO4.

First, we need to calculate the number of moles of each compound present in their respective solutions:

moles Ba(NO3)2 = 1.00 g / 199.33 g/mol = 0.00501 mol 
moles Na2SO4 = 1.00 g / 142.03 g/mol = 0.00705 mol

When the two solutions are mixed together, the total volume becomes 1000 mL = 1 L. Thus, the concentrations of Ba2+ and SO42- are:

[Ba2+] = 0.00501 mol / 1 L = 0.00501 M 
[SO42-] = 0.00705 mol / 1 L = 0.00705 M

Now we can calculate the ion product of BaSO4:

IP = [Ba2+][SO42-] = (0.00501 M)(0.00705 M) = 3.52845 x 10^-5

Since the IP is smaller than the Ksp value of 9.1 x 10^-10, we can conclude that a precipitate of BaSO4 will form. Therefore, the correct answer is A) Barium sulfate will precipitate.

If 500 ML of a Solution Containing 1 ×\times× 10-11

If 500 ML of a Solution Containing 1 $\times$ 10^-11