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Calculate whether or not a precipitate will form if 1.40 mL of 0.62 M NH3 are added to 4.4 L of 1.0 10-3 M FeSO4.

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    Each solution will be diluted; therefore, calculate that first.
    (NH3) = 0.62*(1.4 mL/4,400) = about 2E-4 but you recalculate it.
    (FeSO4) = 1.0E-3 x (0.0014/4.4) = essentially 1.oE-3

    What is the (OH^-) in the NH3 solution?
    .......NH3 + H2O ==> NH4^+ + OH^-
    I.....2E-4............0.......0
    C......-x.............x........x
    E.....2.0E-4-x........x........x

    Kb = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
    Substitute the E line into the Kb expression and solve for x = OH^-. Then
    Qsp = (Fe^2+)(OH^-)^2 and compare with Ksp for Fe(OH)2.
    Ksp > Qsp, no ppt.
    Ksp = Qsp, equilibrium between both states.
    Ksp < Qsp, gives ppt.

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