These really are difficult problems because they are Ksp, common ion effect, and limiting reagent rolled into one. I didn't use the Ksp data except to estimate the Ag^+ and the small amount of additional Cl^-
mols AgNO3 = M x L = about 0.04.
mols CaCl2 = M x L = about 0.022 but you need to do these and all other calculations to do them more accurately since I've estimated here and there.
2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO)3)2
Which is the limiting reagent?
mols AgNO3 x (2 mol AgCl/2 mols AgNO3) = 0.04 x 2/2 = about 0.04 mols AgCl.
mols CaCl2 x (2 mols AgCl/1 mol CaCl2) = about 0.044
Both 0.04 and 0.044 can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value. Therefore, AgNO3 is the limiting reagent and there will be some CaCl2 left un-reacted.
How much CaCl2 is left unreacted?
0.04 mols AgNO3 x (1 mol CaCl2/2 mol AgNO3) = 0.04 x 1/2 = about 0.02 mols USED which leaves 0.022-0.02 = 0.002.
Let's make a table but I will use millimoles instead of mols..
...2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
I...40.......22.........0.......0
C..-40......-20........40.......20
E...0.........2........40.......20
volume is 160 mL + 110 = 270 mL.
Note the NO3^- didn't enter into the reaction; therefore, it was 40 mmols to start and finish. Final M = 40mmol/270 mL = ?
Ca didn't enter into the reaction either; therefore, it started at 22 and ended at 22/270 = ?
Cl is a Ksp problem. You must recognize that you have pptd AgCl (40 mmols of it) BUT you have an excess of 4 mmols Cl left unused from the CaCl2 (that's 2 mmols CaCl2 x 2 Cl/mol CaCl2 = 4 mmol Cl).
Cl = 4mmol/270 mL = ?
There will be a small additional amount of Cl from the dissolved AgCl but that amounts to only about 2.4E-8M.
What mass of silver chloride can be prepared by the reaction of 160.0 mL of 0.25 M silver nitrate with 110.0 mL of 0.20 M calcium chloride?
Calculate the concentrations of each ion remaining in solution after precipitation is complete. (Enter a 0 if none of the ion remains.)
Cl ‾
NO3‾
Ca2+
1 answer