what mass of silver chloride can be prepared by the reaction of 100 ml 0,20 M of silver nitrate with 100,0 ml of 0.15 M calcium chloride calculate the concentration of each ion reamaining in solution after the precipitation is complete

Question

DrBob222 · Answer

This is a limiting reagent (LR) problem wrapped up in a Ksp problem wrapped up in a common ion problem. First calculate how much AgCl we can get and identify the LR.

2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2

millimols AgNO3 = mL x M = 100 x 0.2 = 20
mmols CaCl2 = mL x M = 100 x 0.15 = 15

mmols AgCl formed IF we used the AgNO3 and all of the CaCl2 we needed.
20 x (2 mols AgCl/2 mols AgNO3) = 20

mmols AgCl formed IF we used the CaCl2 and all of the AgNO3 we needed.
15 x (2 mols AgCl/1 mol CaCl2) = 30.
You see we obtained different numbers; in LR problems the smaller value is ALWAYS the correct one and the reagent providing that number is the LR. So AgNO3 is the LR and we form 20 mmols AgCl. You can convert that to g by g = mols x molar mass = ?

How much remains in solution. We didn't touch the NO3^- and we have 2*20 = 40 mmoles in 200 mL so (NO3^-) = mmols/mL = ?
We didn't touch the Ca^2+. That concn is
mmols/mL = 30/200 = ?

...2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
I...20......15..........0........0
C..-20.....-10.........+20.....+10
E...0.......5..........20(pptd)...10

So we have 5 mmols CaCl2 not used which is 10 mmols Cl^- not used and that is the common ion for the solubility of AgCl.
..........AgCl ==> Ag^+ + Cl^-
I........solid.....0.......0
C........solid.....x.......x
E........solic.....x.......x
Ksp AgCl = (Ag^+)(Cl^-) = Look up Ksp.
Substitute (Ag^+) = x
and (Cl^-) = (x + 10/200) = (x+0.05)
Solve for x to find (Ag^+). The (Cl^-) will be 0.05 + whatever Ag^+ is but that's a small number and probably can be ignored and just call Cl = 0.05
Note: The equation Ksp = (Ag^+)(Cl^-) = Ksp = (x)(x+0.05) is a quadratic if you solve it exactly BUT usually x is small and you can assume x + 0.05 = 0.05. That keeps the quadratic out of the way. I don't know how advanced this course is but the author of the problem MAY NOT have intended to have you go through this last part on the Ksp. Post your work if you get stuck.