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Posted by on Thursday, September 15, 2011 at 4:23am.

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 - , Thursday, September 15, 2011 at 8:40am

    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 - , Sunday, November 9, 2014 at 9:19pm


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