Starting with 14.2 ml of 3.90 × 10−2 M BaCl2?

Question

how many grams of ZnSO4 can be added to the solution before BaSO4 begins to precipitate ?

Answer 1

First of all we need to know the Ksp of BaSO4
According to
http://www.ktf-split.hr/periodni/en/abc/kpt.html
It is Ksp=1.08*10^-10

We know the concentration of Ba(+2) at equilibrium (BaCl2 dissociates 100% with a stoichiometry 1 BaCl2 : 1 Ba(+2), thus [Ba(+2)]=0.0390 M) So as a first step we need to find the concentration of [SO4(-2)] in order to have a saturated BaSO4 solution for the given Ba(+2) concentration

Ksp= [Ba(+2)][SO4(-2)] =>
[SO4(-2)]= Ksp/[Ba(+2)] =>
[SO4(-2)]= (1.08*10^-10)/0.039 =2.77*10^-9 mole/L

We have 14.2*10^-3 L, thus in total 2.77*(10^-9)*14.2*10^-3 =3.93*10^-11 mole SO4(-2)

1 mole ZnSO4 contains 1 mole SO4(-2) and it is a strong electrolyte.
Thus in order to have 3.93*10^-11 mole SO4(-2) in solution you need 3.93*10^-11 mole ZnSO4.

But mass= mole*MW= 3.93*(10^-11)*161.39 = 6.34*10^-9 g
This is the maximum amount you can add before precipitation of BaSO4 begins.

Answer 2

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