[sinthreck]

A 100.-mL sample of solution contains 10.0 mmol of Ca2+ ion. How many mmol of solid Na2SO4 must be added in order to cause precipitation of 99.9% of the calcium as CaSO4? The Ksp of CaSO4 is 6.1X10^–5. Assume the volume remains constant.
   a)   61.0
   b)   71.0
   ANS:   b)   71.0

I keep getting the answer as (a) 61, but the proper answer is 71!!! Help me please....

My calculations:

[Ca2+]= 0.01 mol/ 0.1 L= 0.1 M
(100%-99.9%)/100 x 0.1 M= 0.0001

6.1 x 10^-5 = 0.0001 [SO4^2-]

[SO4^2-] = 0.61 M

moles: Na2SO4 = 0.61M x 0.1L= 0.061 mol
mmoles: Na2SO4 x 1000= 61 mmol

[Borek]

You have correctly calculated concentration of sulfate LEFT IN SOLUTION.

[sinthreck]

ok, I think I have it, unless it is a fluke.

Using an ICE table

           Ca²⁺            SO₄^2-
I           0.1                x
C         -.0999            -.0999
E         .0001             .61

solving for x = .61 + .0999 = 0.71

moles: Na2SO4 = 0.71M x 0.1L= 0.071 mol
mmoles: Na2SO4 x 1000= 71 mmol


Is there a more efficient way of calculating this?

[Borek]

Seems to me like the result is close by accident.

How much precipitate and what is its composition?

[sinthreck]

99.9% of the Ca²⁺ ions precipitate, so:

0.999 * 10mmol = 9.99 mmol

So you will have 9.99mmol of CaSO₄ as a solid.

not sure where you're going with this?

[Borek]

Where does SO₄ in CaSO₄ came from?

Note: could be your approach was in some twisted way correct, but there really is no need for ICE table, it is much simpler.

[sinthreck]

Good, I like simple calculations

When you add the solid Na₂SO₄ it dissolves and you are left with Na⁺ and SO₄^2- ions in the solution.

These SO₄^2- ions then attach to the Ca²⁺ ions to form CaSO₄ solid.

[sinthreck]

ha!

I just realised how simple it is!

Whilst we have 61mmol of SO₄ in solution, we have already caused the precipitation of 9.99mmol of Ca²⁺

Hence, the answer = 61mmol + 9.99mmol = 71mmol

[Borek]

Right