posted by brad .
A 30.00mL sample of a clear saturated solution of PbI2 requires 14.7mL of a certain AgNO3 for its titration.
I^-(saturated PbI2)+Ag^+(from AgNO3)--> AgI(s)
What is the molarity of this AgNO3?
I'm not really sure where to start.
Let's work backwards.
M = moles/L
Do we know L AgNO3? Yes, 14.7 mL or 0.0147 L. Do we know mole? No. Where can we get that? Look at the PbI2 (that's all that's left. right?)
PbI2(sat'd) ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2 = Ksp. You know Ksp, You can calculate solubility PbI2 and from that the (Pb^2+) as well as the (I^-). How many mols I^- do you have? That's mols x L (30 mL sample) = ?
The titration step is Ag^+ + I^- ==> AfI
mols I^- = moles Ag^+ from the balanced equation which gives you the moles you need back at the top. Solve for M Ag^+
how can i calculate the solubility with the information given ? in class we use concentrations to find solubility
Ksp PbI2 = about 1E-10 but you need to look it up.
Let x = solubility
...........PbI2 ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2
1E-10 = (x)(2x)^2
Solve for x = solubility
So i found the solubility to be x=0.001296, is this the concentration of I^-? can i multiply this by 30mL to get my moles?