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22 May, 17:13

Ml of 0.00237 m nai (aq) is combined with 625. ml of 0.00785 m pb (no3) 2 (aq). determine if a precipitate will form given that the ksp of pbl2 is 1.40x10-8.

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  1. 22 May, 17:25
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    The volume of I - is missing in your question by assuming it = 1L

    moles I - = molarity * volume

    = 0.00237 * 1 L

    = 0.00237 mol

    [I-] = moles / total volume

    = 0.00237 / 1.625L

    = 0.00146 M

    moles Pb2 + = molarity * volume

    = 0.00785 * 0.625 L

    = 0.0049 mol

    [Pb2+] = 0.0049 / 1.625L

    = 0.003 M

    when PbI2 (s) ↔ Pb2 + (aq) + 2I - (aq)

    when Q = [Pb2+][I-]^2 and we neglect [PbI2] as it is solid

    ∴ Q = 0.003 * (0.00146) ^2

    = 6.4 x 10^-9

    by comparing the value of Q with Ksp value we will found that:

    Q < Ksp which mean that more solid will dissolve, and this is an unsaturated solution which has ion concentrations < equilibrium concentrations, so the reaction will go forward until achieving equilibrium.
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