An aqueous solution of Sr(NO3)2 is added slowly?

Question

to 1.0 litre of a well-stirred solution containing 0.020 mole F- and 0.10 mole SO42- at 25°C. You may as¬sume that the added Sr(NO3)2 solution does not ma¬terially affect the total volume of the system.)

1. Which salt precipitates first?

2. What is the concentration of strontium ion, Sr2+, in the solution when the first

precipitate be¬gins to form?

(d) As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At that stage, what percent of the anion of the first precipitate remains in solution?

Answer 1

1) Find molar solubilities of each salt in this solution using Ksp (see source below)

Ksp (SrF2) = 4.35x10^-9 = [Sr++] [F-]²

Let molar solubility = x, then

4.35x10^-9 = (x) (2x + 0.02)²

Since 0.02>>x, we can ignore x in the right parentheses:

4.35x10^-9 = (x) (0.02)²

Solving, x = 1.09x10^-5 M.

Ksp (SrSO4) = 3.44x10^-7 = [Sr++] [SO4--]

Let molar solubility = x, then

3.44x10^-7 = (x) (x + 0.1)

Since 0.1>>x, we can ignore x in the right parentheses:

3.44x10^-7 = (x) (0.1)

Solving, x = 3.44x10^-6 M.

Since the molar solubility of SrSO4 is smaller in this solution, it will precipitate first.

2. From calculations above, [Sr++] when SrSO4 starts to precipitate is 3.44x10^-6 M.

Again, from the calculations above, [Sr++] when SrF2 starts to precipitate is 1.09x10^-5 M. [SO4--] when this occurs is:

Ksp (SrSO4) = 3.44x10^-7 = [Sr++] [SO4--];

3.44x10^-7 = (1.09x10^-5 M) [SO4--]; solving,

[SO4--] = 0.0316 M. Since [SO4--] was 0.10 M initially, then

100% * 0.0316 / 0.1 = 31.6 % of [SO4--] is still in solution when SrF2 starts to precipitate.