Asked by Hel P.
If 32.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.545 g precipitate, what is the molarity of lead(II) ion in the original solution?
Answers
Answered by
bobpursley
The precipate is Lead(II)iodide, hopefully, and you have .545grams lead iodide.
The easy way is to figure the number of moles that you have.
moles= massPbI2/molmassPbI2
Next, because the moles of Pb+2 ion is the same as the moles of PbI2 which was formed, then
Molarityoriginal= molesPb+2/.032
The easy way is to figure the number of moles that you have.
moles= massPbI2/molmassPbI2
Next, because the moles of Pb+2 ion is the same as the moles of PbI2 which was formed, then
Molarityoriginal= molesPb+2/.032
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