Question

I'm sorry here is another I do not understand.
How do you find out exactly how much of a solution you would need to completely precipitate another compound in the reaction?
We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. If we have 500. mL of 0.632 M lead(II) nitrate solution, how much 1.50 M potassium iodide must we add to exactly precipitate all the lead(II) ions?

2 KI(aq) + Pb(NO3)2(aq) ¨ PbI2(s) + 2 KNO3(aq)

Answer 1

We worked the problem last night but they wanted to know grams PbI2 pptd. In this problem, the only difference is that they want to know how much 1.50M KI must be added.

1. mols Pb(NO3)2 = M x L = ?
2. Convert mols Pb(NO3)2 to mols KI using the coefficients in the balanced equation.
3. The M KI = mols KI/L KI. You know M and mols, solve for L. Convert to mL if you wish.