What is the maximum mass (in grams) of each of the following soluble salts that can be added to 120 mL of 0.050 M BaCl2 without causing a precipitate to form? (Assume that the addition of solid causes no change in volume.)

Question

(a) (NH4)2SO4

× 10
g
(Enter your answer in scientific notation.)

(b) Pb(NO3)2

g

(c) NaF

g

Answer 1

Here is how you do the first one. I'll be glad to check your work for the others. You can add enough of each salt so that the Ksp of the ppt is not exceeded.
(NH4)2SO4 + BaCl2 ==> BaSO4(s) + 2NH4Cl
The ppt will be BaSO4. Look up the Ksp for BaSO4. It is approx 1.1E-10 but you should use the value shown in your class/text book.
The Ksp expression is (Ba^2+)(SO4^2-) = 1.1E-10
The problem tells you that the BaCl2 is 0.05M. Plug that into the Ksp expression for (Ba^2+) and solve for (SO4)^2-. That will tell you the sulfate value that will just meet the Ksp for BaSO4. Anothr moleculer over that and the BaSO4 will ppt.
(0.05)(SO4^2-) = 1.1E-10
(SO4^2-) = 1.1E-10/0.05 = 2.2E-9 M = 2.2E-9 mols/L solution.
In 120 mL of solution there will be 2.2E-9 mols x (120/1000) = 2.64E-10 mols/120 mL. Convert that to grams (NH4)2SO4.
2.64E-10 mols x molar mass (NH4)2SO4 = grams (NH4)2SO4 in 120 mL =?
Good luck.