How many grams of MgSO4 can be added to the solution BaSO4 begins to precipitate?

Question

Starting with 14.0 ml of 3.70 × 10−2 M BaI2, how many grams of MgSO4 can be added to the solution before BaSO4 begins to precipitate ?

Answer 1

Impossible to say.

To get is MgSO4 soluble you have to form the heptahydrate (epson salts essentialy). By this time you have added quite a bit of MgSO4 to the solution So much of the MgSO4 will still be sitting as a solid at the bottom of you flask (or in a lump), and you will NEVER see the SMALL conc of BaSO4 ppt.

Answer 2

1) Reaction is:

BaI2 + MgSO4 -----> BaSO4+ MgI2

2) We should know the Kps of the BaSO4

Kps (BaSO4) = 1.08 x 10^-10

We know also that:

Kps = [Ba++] [SO4=]

we can compute concentration of Ba++ at equilibrium:

[Ba++] =V1.08x10^-10 = 1.039x10-5 mol/L

3) We know that we start with:
(0.014mL)(0.037 M) = 5.18 x 10-4 moles of BaI2

Hence, we can conclude that reacted:

5.18 x 10-4 - 1.039x10-5 moles = 5.076 x 10-4 moles of BaI2

that should be the same amount of MgSO4 that reacted before to form the precipitate:

mass of MgSO4 = (5.076x10-4 moles) (120.36 g/mol) = 0.0611 grams

4) Finally, we conclude that we can add 0.0611 grams of MgSO4 to BaI2 before it is formed the BaSO4 pp.

Hope it helps!

Good luck!