Posted by Annie on Tuesday, March 19, 2013 at 2:33pm.
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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