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Finding the iodide ion concentration

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data
    What is the iodide ion concentration in a solution if the addition of an excess of 0.100 M Pb(NO3)2 to 37.2 mL of the solution produces 539.7 mg of PbI2?

    2. Relevant equations
    possibly c(1)*V(1)=c(2)*V(2) is pertinent to the problem

    3. The attempt at a solution

    I don't know if this is relevant to finding the concentration of Iodine, but I went ahead and converted total number of 539 mg PbI2 => the number of mmols of I^1-

    n= 539 mg PbI2 *(1 mmol PbI2/461 mg PbI2) *(2 mmol I^1- /1 mmol PbI2)= 2.34 mmol I^1-

    .100 M Pb(NO3)2 = .100 mmol Pb(NO3)2/mL

    .110 mmol Pb(NO3)2/mL*(37.2 mL)=3.72 mmol Pb(NO3)2

    Perhaps I didn't need to find the total mmol of Pb(NO3)2 and maybe just needed to divide the number of millimoles of ion by the total volume of the solution.

    Maybe the concentration of [I^1-] = 2.34 mmol I^1- /( 37.2 mL of solution)

    [I^1-] = .063 M
  2. jcsd
  3. Sep 7, 2007 #2


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    Staff: Mentor

    This is simple stoichiometry, start with reaction equation.
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