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3 October, 07:20

What is the iodide ion concentration in a solution if the addition of an excess of 0.100 m pb (no3) 2 to 42.9 ml of the solution produces 861.8 mg of pbi2?

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  1. 3 October, 11:08
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    Moles Pbl2 = 0.8628 g : 461.01 g/mol = 0.001871

    moles I = 2 x 0.001871 = 0.003742

    [I-] = 0.003742 / 0.0429

    = 0.0872 M
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