This is a limiting reagent problem and a Ksp problem rolled into one.
Pb(NO3)2 + 2NaI ==> PbI2 + 2NaNO3
mols Pb(NO3)2 = M x L = /
mols NaI = M x L = ?
Convert mols Pb(NO3)2 to mols PbI2.
Convert mols NaI to mols mols PbI2.
It is likely that the two values for mols PbI2 will not be the same. The correct value to use in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. That process gives you the mols PbI2 formed and that NaI is the limiting reagent.
Next you want to find how much Pb(NO3)2 (the non-limiting reagent) remains unreacted. That is done the same way; i.e., you use the coefficients in the balanced equation to convert mols NaI to mols Pb(NO3)2, then subtract this amount from the initial amount.
The excess Pb^2+ will act as a common ion for PbI2. Use that to substitute for Pb^+2 in the Ksp expression and solve for I^-. The Pb^+2 and I^- gives you the Ksp part, the others are spectator ions Don't forget that everything is diluted by the TOTAL volume; ie., 20 mL + 15 mL = 25 mL.
20.00 ml of 0.100F Pb(No3)2 (FW 331.2) are mixed with 15.00 ml of 0.125 F Nal (FW 149.9). Lead (II) and iodide form an insoluble salt pbl2 (FW 461.0) Calculate the concentration of each of the remaining ions after mixing. [Pb+2], [i-],[na+],[n03-]. ksp pbi2 = 7.1 *10^-9
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