this is the rest of the calculations
Q= 1.4x10^-4
ksp= 1.77x10^-10
Q>ksp so precipitation will form
Tap water that contains Cl− at a concentration
of 50 ppm is used to prepare a 0.100 mol/L
AgNO3 solution. Does a precipitate form? How much?
I don't know how to find out how much precipitate will form but a precipitate will form based on my calculations
50 ppm 0.0014 mol/l
0.1 Mol/l = AgNO3
Ag+ = 0.1 Mol/l
Cl-=0.0014 mol/l
Q= [ag+][Cl-]
2 answers
You're ok as far as you went.
Who ever made up this problem forgot to tell us how much solution is being prepared. It makes sense, doesn't it, that we can get more AgCl if we made up 5L than if we made up 1L. So I guess we assume we make 1 L solution.
........Ag^+ + Cl^- ==> AgCl
I......0.1mol.0.00141.....0
C..-0.00141.-0.00141....0.00141
E...............0........0.00141
So we will have 0.00141 mols AgCl ppt and that x molar mass AgCl gives you the grams AgCl. I assume it's obvious that Cl^- is the limiting reagent.
Who ever made up this problem forgot to tell us how much solution is being prepared. It makes sense, doesn't it, that we can get more AgCl if we made up 5L than if we made up 1L. So I guess we assume we make 1 L solution.
........Ag^+ + Cl^- ==> AgCl
I......0.1mol.0.00141.....0
C..-0.00141.-0.00141....0.00141
E...............0........0.00141
So we will have 0.00141 mols AgCl ppt and that x molar mass AgCl gives you the grams AgCl. I assume it's obvious that Cl^- is the limiting reagent.
2 answers