30.0 mL of 0.10 M Ca(NO3)2 and 15.0 mL of 0.20 M Na3PO4 solutions are mixed. After the reaction is complete, which of these ions has the lowest concentration in the final solution?

Question

A) Na+ B) NO3- C) Ca+2 D) PO4-3

Answer is C.. How do you solve it though?!

Answer 1

If you imagine the process occuring in 2 discrete steps, mixing and precipitation, you get the following.

The mixing step gives you:
30.0 mL of 0.10 M Ca(NO3)2 = 3 mmol Ca(NO3)2 = 3 mmol Ca(2+) + 6 mmol NO3(-).
15.0 mL of 0.20 M Na3PO4 = 3 mmol Na3PO4 = 9 mmol Na(+) + 3 mmol PO4(3-)

Then the precipitation step is
3 Ca(2+) + 2 PO4(3-) ---> Ca3(PO4)2 which forms a solid.
Note that you have a 1:1 ratio in solution but a 3:2 ratio in the reaction. Ca(2+) is the limiting reagent. So it will be used up in the precipitation reaction faster than PO4(3-).

This means that Ca(2+) will have the lowest concentration in the end.