Question

Please help! 200.mL of 0.200 M potassium phosphate is mixed with 300.mL of 0.250 M calcium chloride. Determine the concentration of all ions remaining in solution.

Answer 1

@mathstudent55

Answer 2

Total (final) volume -- 500 mL

The reaction:

2K3PO4 + 3CaCl2 --> 6KCl + Ca3(PO4)2
The Ca3(PO4)2 precipitates out
K+ and Cl- remain in solution

The final concentration of Cl- will be computed from 300 mL * 0.250 M *2 = 500 mL * X M or
0.30 M. (It is higher because there are two Cl- ions in every mole of CaCl2)

Similarly for the K+ -- 200 mL * 0.200 M * 3 = 500 mL * X M or
0.24 M

You originally had 200 mL * 0.200 M PO4 -- or 0.040 moles of phosphate
Similarly you had 300 mL * 0.250 M Ca -- or 0.075 moles of calcium

For every three phosphates, you need two calciums -- so all the PO4 precipitates out and leaves 0.048 moles of calcium. This is the number of moles in 500 mL, so the [Ca+2] is 0.096 M

Answer 3

Thank you so much. The very last bit was a bit off but I was able to get the right answer.

Answer 4

Construct a balanced chemical equation

2 K3PO4 + 3 CaCl2 → Ca3(PO4)2(s) + 6 KCl

Change ml to dm^3 by dividing by 1000
(200 mL/1000) x (0.200 M K3PO4) = 0.040 mmol K3PO4
(300 mL/1000) x (0.250 M CaCl2) = 0.075 mmol CaCl2

0.040 moles of K3PO4 would react completely with 0.040 x (3/2) = 0.060 moles of CaCl2, but there is more CaCl2 present than that, so CaCl2 is in excess and K3PO4 is the limiting reactant.
add volumes: 200 mL + 300 mL = 500 mL
K3PO4-->3K+ + PO4-3
(0.040 mol K3PO4) x (3 mol K{+} / 1 mol K3PO4) / (500/1000 mL) = 0.240 mol/L K{+}
CaCl2-->Ca+2 + 2Cl-
(0.075 mol CaCl2) x (2 mol Cl{-} / 1 mol CaCl2) / (500 mL/1000) = 0.300 mol/L Cl{-}
(0.040 mol K3PO4)*(1 mol PO4[-3]/ 1mol K3PO4/0.5 L)=0.15 mol/L PO4[-3]

Intial Moles-Final moles=moles left unreacted
((0.075 mol CaCl2 initially) - (0.060 mol CaCl2 reacted)) x (1 mol Ca{2+} / 1 mol CaCl2) / (0.5 L) = 0.03 mol/L Ca{+2}

Answer 5

Have you go a mark scheme or an answer book to check my work :)

Answer 6

Yes, all of your answers are correct. Thank you.

Answer 7

your most welcome :)