posted by Nick on .
A certain municipal water sample contains 46.1 mg SO42-/L and 30.6 mg Cl -/L. How many drops (1 drop = 0.05 mL) of 0.0010 M AgNO3 must be added to 1.00 L of this water to just produce a precipitate?
What will this precipitate be?
how do i do this problem?
agcl ksp = 1.8 e-10
ag2so4 ksp = 1.5 e10-5
Ksp AgCl = (Ag^+)(Cl^-) = 1.8 x 10^-10
(Cl^-) = 30.6 mg from the problem.
moles Cl = 30.6 x 10^-3 g/35.45 = ??moles and since that is in 1 L of the water sample, that is the molarity.
Plug Cl^- (in moles/L) into Ksp and solve for (Ag^+) required for AgCl to ppt.
Then M Ag^+ x 1000 mL = moles Ag^+.
Since moles = M x L. You know moles Ag^+ needed and you know M of the AgNO3 solution, calculate L and convert that to drops. Post your work if you get stuck.
You work the Ag2SO4 the same way.
I did the moles of Cl found/1 L = .00863.
1.8e-10 / .00863 = 2.1e-8 * 1 L = 2.1e-8 moles
.0010 / 2.1e-8 = 47619 L*1000 mL / .5
but the answer comes out wrong=/
I think (Cl^-) = 0.0306/35.45 = 8.63 x 10^-4 moles which is a factor of 10 different from your value. That will make (Ag^+) needed to ppt AgCl be (1.8 x 10^-10/8.63 x 10^-4 = 2.085 x 10^-7 M instead of your value. (I know that's too many significant figures but I carry too many and round at the end). From here on out, I don't follow your work at all.
Since this is 2.085 x 10^-7 moles/L and we have 1 L of solution, we have 2.085 x 10^-7 moles in the 1 L. Then L = (moles/M) = 2.085 x 10^-7/0.001) = 2.085 x 10^-4 L or 2.085 x 10^-1 mL = 0.2085 mL. Check my work
Finally. 0.2085 mL x (1 drop/0.05 mL) = 4.17 drops which I would round to 5 drops. (4.17 rounds to 4 drops to 1 significant figure which is all we are allowed based on the 0.05 mL/drop but 4 drops will be too small to cause a ppt and we must take 5 drops). So 5 drops it is. Check my arithmetic.