[Alex B]

When 38 mL of 0.1250 M H₂SO₄ (sulfuric acid) is added to 100 mL of solution of PbI₂ (Lead(II) Iodide) a precipitate of PbSO₄ forms. The PbSO₄ (Lead(II) Sulfate) is then filtered from the solution, dried, and weighed. If the recovered PbSO₄ is found to have a mass of 0.0471 g, what was the concentration of iodide ions in the original solution?

So I started with an equation: H₂SO₄ + PbI₂ = PbSO₄ + 2HI

I treated this problem like a limiting reagent. I assumed that the amount recovered was 100% yield (Perfect world. I know haha)

If Sulfuric acid was the limiting reagent, then the amount of moles of Lead(II) Sulfate would be 1.44 grams; which is 0.00475 moles of Lead(II) Sulfate. So that means sulfuric acid is in excess and the Lead(II) Iodide was the limiting reagent. To figure out the Molarity of the PbI₂ I worked backwards from moles of PbSO₄ recovered: 0.1L "X" = 1.553X10^-4 moles. This gave me 1.553X10^-3 M PbI₂. Then, I changed this from M of PbI[/sup]2[/sub] to M of Iodine: (1.553X10^-3 M PbI₂) (2mol I/ 1mol PbI₂) = 3.11X10^-3 M I. To get moles of I, I multiplied by 0.1 L PbI₂ solution, and I got 3.11X10^-4 moles I. finally, I divided that by Liters of beginning solution which was 100mL + 38mL: (3.11X10^-4 moles I)/(.138L) = 0.00225 M of Iodine. 

Are my steps correct? I know this was a kind of a long process, but if you could please give me feedback that would be much appreciated!

[ooosh]

Your steps are correct, but,they are not the correct answer of your question"what was the concentration of iodide ions in the original solution?"
Since you know the  Lead(II) Iodide is the limiting reagent,and you have the quantity of PbI2 1.553X10-3 M,so the quantity of I^- is 2X1.553X10-3 M ,finally the concentration of iodide ions in the original solution=n(I^-)/L(original solution).