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apbuiii

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## Homework Statement

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?

## Homework Equations

So I started with an equation: H2SO4 + PbI2 = PbSO4 + 2HI; Molarity(M)= Moles of solute/moles of solution

## The Attempt at a Solution

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!