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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

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