Cliffs AP Chemistry, 3rd Edition

Step 4: Substitute the equilibrium concentrations of the ions into the equilibrium expression.

Ksp = [Pb2+][I–]^2 = (x)(2x)^2 = 1.08 × 10 –7

4 x^3 = 1.08 × 10 –7

x^3 = 27 × 10 –9

x = 3.0 × 10 –3

10. Will a precipitate form when one mixes 75.0 mL of 0.050 M K 2 CrO 4 solution with 75.0

mL of 0.10 M Sr(NO 3 ) 2? Kspfor SrCrO 4 = 3.6 × 10 –5

A. Yes, a precipitate will form, Q > Ksp

B. Yes, a precipitate will form, Q < Ksp

C. Yes, a precipitate will form, Q = Ksp

D. No, a precipitate will not form, Q > Ksp

E. No, a precipitate will not form, Q < Ksp

Answer:

Recognize that this problem is one involving the ion product, Q. We calculate Q in the same
manner as Ksp, except that we use initial concentrations of the species instead of equilibrium
concentrations. We then compare the value of Q to that of Ksp:

If Q < Ksp— no precipitate.

If Q = Ksp— no precipitate.

If Q > Ksp— a precipitate forms.

At this point you can rule out choices B, C, and D because they do not make sense. If you have
forgotten how to do the problem mathematically, you should guess now; you have a 50%
chance of getting the answer right.

Step 1: Realize that this problem involves a possible double displacement, the possible precipi-
tate being either KNO 3 or SrCrO 4. Rule out the KNO 3 since most nitrates are soluble and that
you were provided with the Kspfor SrCrO 4. To answer these questions, you must know your
solubility rules!

Step 2:Write the net ionic equation.

Sr+-^2 ()aq+CrO 42 ()aq"SrCrO 4 ()s

Step 3:Write the equilibrium expression.

Ksp=[][ ]Sr^2 + CrO 42 -

Equilibrium