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29 January, 01:59

If 73.5 mL of 0.200 M KI (aq) was required to precipitate all of the lead (II) ion from an aqueous solution of lead (II) nitrate, how many moles of Pb2 + were originally in the solution?

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  1. 29 January, 02:23
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    There were 0.00735 moles Pb^2 + in the solution

    Explanation:

    Step 1: Data given

    Volume of the KI solution = 73.5 mL = 0.0735 L

    Molarity of the KI solution = 0.200 M

    Step 2: The balanced equation

    2KI + Pb2 + → PbI2 + 2K+

    Step 3: Calculate moles KI

    moles = Molarity * volume

    moles KI = 0.200M * 0.0735L = 0.0147 moles KI

    Ste p 4: Calculate moles Pb^2+

    For 2 moles KI we need 1 mol Pb^2 + to produce 1 mol PbI2 and 2 moles K+

    For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

    There were 0.00735 moles Pb^2 + in the solution
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