South Pasadena • AP Chemistry
Name
3 ▪ Chemical Equilibrium
Period
3.2
PROBLEMS
–
1. Predict, on the basis of the solubility rules, which
of the following salts are soluble and which are
insoluble. For those that are “insoluble” look up
their solubility products.
−16
(a) AgI
Insoluble. Ksp = 1.5 × 10
(b) Na3PO4
Soluble
(c) BaSO4
Insoluble. Ksp = 1.5 × 10−9
Date
SOLUBILITY EQUILIBRIUM
6. The Ksp of strontium fluoride, SrF2, is 7.9 × 10−10.
What is the [Sr2+] and [F−] in a saturated solution
of SrF2? What is the molar solubility of SrF2?
Ksp = [Sr2+][F−]2 = 7.9 × 10−10
(s)(2s)2 = 7.9 × 10−10
s = 5.8 × 10−4
(d) (NH4)2SO4 Soluble
−7
(e) NiCO3
Insoluble. Ksp = 1.4 × 10
(f) Cu(OH)2
Insoluble. Ksp = 1.6 × 10−19
7. From which of the following mixtures of silver
nitrate and sodium sulfite would silver sulfite
precipitate? The Ksp for silver sulfite = 1.5 × 10−14.
(a) 50 mL of 1.0 × 10−4 M Ag+ and 50 mL of
1.0 × 10−4 M SO32−.
2. Write solubility product expressions for the
following salts:
[Ag+] =
(1.0 × 10−4 M)(50 mL)
= 5.0 ×10−5 M
(100 mL)
Ksp = [Pb2+]eq[SO42−]eq
(a) PbSO4
(b) Ca3(PO4)2
Ksp = [Ca2+]3eq[PO43−]2eq
(c) CuS
Ksp = [Cu2+]eq[S2−]eq
(d) CaF2
Ksp = [Ca2+]eq[F−]2eq
[SO32−] =
(1.0 × 10−4 M)(50 mL)
= 5.0 ×10−5 M
(100 mL)
Q = [Ag+]2[SO32−] = (5.0 × 10−5)2(5.0 × 10−5)
= 1.3 × 10−13 > 1.5 × 10−14
Precipitate Forms
3. If the molar concentration of lead bromide, PbBr2,
in an aqueous solution is 4.8 × 10−6 M, what is
[Pb2+] and [Br−]?
(b) 25 mL of 1.0 × 10−3 M Ag+ and 25 mL of
1.0 × 10−5 M SO32−.
[Ag+] =
PbBr2 (s)  Pb2+ (aq) + 2 Br− (aq)
−6
2+
[Pb ] = 4.8 × 10 M
[Br−] = 2(4.8 × 10−6 M) = 9.6 × 10−6 M
4. If the molar solubility of silver iodide is
1.5 × 10−16 M, what is the solubility product for
AgI?
Ksp = [Ag+][I−] = (s)(s)
= (1.5 × 10−16)2 = 2.3 × 10−32
2+
2−
Ksp = [Cd ][S ] = (s)(s) = 1.0 × 10
s = 1.0 × 10−14 M
[SO32−] =
−28
(1.0 × 10−5 M)(25 mL)
= 5.0 ×10−6 M
(50 mL)
Q = [Ag+]2[SO32−] = (5.0 × 10−4)2(5.0 × 10−6)
= 1.3 × 10−12 > 1.5 × 10−14
Precipitate Forms
(c) 50 mL of 1.0 × 10−5 M Ag+ and 100 mL of
1.0 × 10−3 M SO32−.
[Ag+] =
5. What is the molar solubility of cadmium sulfide,
CdS, if its Ksp = 1.0 × 10−28?
(1.0 × 10−3 M)(25 mL)
= 5.0 ×10−4 M
(50 mL)
(1.0 × 10−5 M)(50 mL)
= 3.3 ×10−6 M
(150 mL)
[SO32−] =
(1.0 × 10−3 M)(100 mL)
= 6.7 ×10−4 M
(150 mL)
Q = [Ag+]2[SO32−] = (3.3 × 10−6)2(6.7 × 10−4)
= 7.4 × 10−15 < 1.5 × 10−14
No Precipitation
AP Chemistry 2006 #1
Answer the following questions that relate to solubility of salts of lead and barium.
(a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution at 25˚C.
The concentration of Pb2+(aq) in the saturated solution is found to be 1.3 × 10–3 M. The chemical equation for
the dissolution of PbI2(s) in water is shown below.
PbI2(s)  Pb2+(aq) + 2 I–(aq)
(i) Write the equilibrium-constant expression for the equation.
Ksp = [Pb2+][I−]2
(ii) Calculate the molar concentration of I–(aq) in the solution.
[I−] = 2 × [Pb2+] = 2(1.3 × 10−3 M) = 2.6 × 10−3 M
(iii) Calculate the value of the equilibrium constant, Ksp.
Ksp = (1.3 × 10−3)(2.6 × 10−3)2 = 8.8 ×10−9
(b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25˚C. What
are the molar concentrations of Pb2+(aq) and I–(aq) in the solution? Justify your answer.
[Pb2+] = 1.3 ×10−3 M, [I−] = 2.6 × 10−3 M
Saturated solutions have the same concentrations of dissolved ions, regardless of the volume.
(c) Solid NaI is added to a saturated solution of PbI2 at 25˚C. Assuming that the volume of the solution does not
change, does the molar concentration of Pb2+(aq) in the solution increase, decrease, or remain the same?
Justify your answer.
If NaI is added to the solution, the [I−] increases, so the reaction shifts to the left and [Pb2+] decreases.
(d) The value of Ksp for the salt BaCrO4 is 1.2 × 10–10. When a 500. mL sample of 8.2 × 10–6 M Ba(NO3)2 is
added to 500. mL of 8.2 × 10–6 M Na2CrO4, no precipitate is observed.
(i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+(aq) and CrO42–(aq) in the
1.00 L of solution.
[Ba2+] =
(8.2 × 10−6 M)(500 mL)
= 4.1 × 10−6 M
1000 mL
[CrO42−] =
(8.2 × 10−6 M)(500 mL)
= 4.1 × 10−6 M
1000 mL
(ii) Use the molar concentrations of Ba2+(aq) ions and CrO42−(aq) ions as determined above to show why a
precipitate does not form. You must include a calculation as part of your answer.
Q = [Ba2+][CrO42−] = (4.1 × 10−6 M)(4.1 × 10−6 M) = 1.7 × 10−11
Because Q < Ksp, no precipitate forms.