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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 16.47

What [CrO42–] must be present in 0.00105 M AgNO3(aq) to just cause Ag2CrO4(s) to precipitate?




Answer:

The desired reaction is

2 AgNO3(aq) + CrO42–(aq) Ag2CrO4(s) + 2 NO3(aq)

Since silver chromate has some solubility, we need to use the solubility equilibrium associated with solver chromate to solve the problem and treat the silver nitrate as a common ion.
AgNO3(aq)
Ag+(aq)
+
NO3(aq)
Ag2CrO4(s)
2 Ag+(aq)
+
CrO42–(aq)
Ksp = [Ag+]e2[CrO42–]e = 1.1×10–12
Initial
0.00105
0
Change
+ 0
+ x
Equilibrium
0.00105
x

(Note: the silver ion only arises from the silver nitrate; none comes from silver chromate because no silver chromate was introduced by the experiment.)

1.1×10–12 = [0.00105]2[x]

x = 1.0×10–6 M = the concentration of chromate required to just start precipitation.