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10 August, 20:19

770. mL of 0.00639 M NaI (aq) is combined with 285. mL of 0.00146 M Pb (NO3) 2 (aq). Determine if a precipitate will form given that the Ksp of Pbl2 is 1.40x10-8. Precipitation will occur because Qsp Ksp Precipitation will occur because Qsp = Ksp Precipitation will occur because Qsp > Ksp Precipitation will not occur because Qsp < Ksp

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  1. 10 August, 20:45
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    Precipitation will not occur because Qsp < Ksp

    Explanation:

    Step 1: Data given

    Volume of 0.00639 M NaI = 770 mL

    Volume of 0.00146 M Pb (NO3) 2 = 285 mL

    Ksp of PbI2 = 1.40 * 10^-8

    Step 2: The balanced equation

    Pb (NO3) 2 + 2NaI → PbI2 + 2NaNO3

    Step 3: Calculate new concentrations

    [NaI] = 0.00639 * (770/1055)

    [NaI] = 0.004664 M

    [Pb (NO3) 2] = 0.00146 * (285/1055)

    [Pb (NO3) 2] = 0.0003944 M

    Step 4: Calculate Q

    Q (pbI2) = [Pb2+]*[I-]² = [Pb (NO3) 2] * [NaI]²

    Q = 0.0003944 * 0.004664²

    Q = 0,0000000085793421824 = 8.58 * 10^-9

    Q < Ksp

    This means a precipitate will not be formed

    Precipitation will not occur because Qsp < Ksp
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