Post a New Question

Chemistry

posted by .

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?

  • Chemistry -

    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

  • Chemistry -

    55

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question