PSI AP Chemistry: Equilibrium
Chapter Problems
Name _______________________
Part A Classwork:
1. A student desires to produce oxygen gas by heating a 40.0 gram sample of potassium chlorate in
a closed rigid container with volume of 10.0 L to a temperature of 800 K.
a. Write the balanced equation for the process and the expression for Kp.
b. What would happen to the pressure in the container as the reaction progresses? Justify
your answer.
c. When the reaction reaches equilibrium after 13 seconds, there is found to be 13.5
grams of potassium chlorate remaining in the container:
i. What would be the value of Kp for this reaction?
ii. Draw a graph plotting PO2 gas over time for this reaction clearly showing when
the reaction reaches equilibrium. Take care to label axis properly and to provide
a reasonable scale.
2. A 5.00 L container was filled with differing mole amounts of nitrogen and hydrogen gas at a
temperature of 18.1 C. The reaction was sparked and the following graph was created using
date from the experiment.
15
[ ] vs. time
[ ]
10
[NH3]
5
[N2]
[H2]
0
0
5
10
15
20
Time (sec)
a. Initially, what would have been the pressure of the nitrogen gas in the container?
b. Write the balanced equation for this reaction and calculate the value of the equilibrium
constant, Kc.
c. How would the value of the equilibrium constant be affected if:
i. The initial concentrations were all increased by a factor of two? Justify your
reasoning.
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ii. The initial concentrations of nitrogen and hydrogen were zero and there was a
10 M concentration of ammonia instead.
d. Explain why the slope is greater from 0-5 seconds for the change in hydrogen gas
compared to the change in nitrogen gas.
3. Silver phosphate dissociates in water by the reaction shown below:
Ag3PO4(s)  3Ag+(aq) + PO43-(aq)
Given that the concentration of phosphate ions was determined to be 6.88 x 10-3 M in a
saturated solution of silver phosphate:
a. What is the value of Ksp for this reaction?
b. Would silver phosphate be more or less soluble in water than copper(I)phosphate which
has a Ksp of 3.41 x 10-18? Justify your answer.
4. A sample of SO3 is placed in a rigid container at 25 C with a volume of 5.00 L. The initial pressure
of the container is 2.0 atm. Sulfur trioxide is known to decompose by the reaction below:
2SO3(g)  2SO2(g) + O2(g)
The reaction is initiated and the pressure of the container is measured and graphed below:
3
P vs time
Pressure (atm)
2.5
2
1.5
Pressure
1
0.5
0
0
5
10
15
20
25
30
35
Time (Sec)
a. What is the pressure of each gas at equilibrium?
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Equilibrium
b.
c.
d.
e.
What is the value of the equilibrium constant, Kp, for the reaction?
What was the % decomposition of the SO3?
Approximately when does this reaction reach equilibrium?
How would the value of the equilibrium constant you calculated in part “b” be affected if
the container developed a leak during the experiment? Justify your answer.
5. Agree or disagree with the following statements and justify your answer:
a. A reaction reaches equilibrium when the concentrations of reactants and products are
equal.
b. Reactions with large equilibrium constants will have a greater rate of production of
products at equilibrium than rate of production of reactants.
Part A Homework:
6. A student wants to fill a metal cartridge with carbon dioxide gas. To do this, they thermally
decompose a 20 gram sample of calcium carbonate in a reaction chamber and collect the
carbon dioxide gas via a tube into the cartridge at 500 C.
a. Write the balanced equation for the process and the expression for Kp.
b. Draw a graph plotting the pressure in the container, with points corresponding to 50%
decomposition of the calcium carbonate, 75% decomposition, and 90% decomposition.
c. When the reaction reaches equilibrium, there was found to be 11.9 grams of solid
material in the reaction chamber. The pressure in the 300.0 mL cartridge was 38.9 atm.
i. How would the experimenter know they reached equilibrium?
ii. What % of the calcium carbonate had decomposed?
iii. Why is the mass of the solid in the reaction chamber greater than the mass of
leftover calcium carbonate?
iv. What would be the value of Kp for this reaction?
v. Explain how the value of Kp for this reaction would be affected if the reaction
were carried out at room temperature. Justify your answer.
7. A 10.00 L container was filled with 5.00 grams each of chlorine and hydrogen gas at a
temperature of 25.1 C. The reaction was sparked and the following data was obtained as the
experiment was monitored. The student was negligent and missed two measurements!!
Time
0s
10 s
15 s
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Container Pressure
?
6.29 atm
?
AP Chemistry
Equilibrium
a. Initially, what would have been the pressure in the container?
b. Write the balanced equation for this reaction and write the expression for Kc.
c. What would be the expected pressure in the container at 15 s assuming constant
temperature and volume? Justify your reasoning.
d. Explain why determining when the reaction reaches equilibrium would be difficult for
this reaction.
e. When the equilibrium mixture was sent through a gas chromatograph, it was
determined that the molar concentration of chlorine gas was 0.0061 M.
i. What were the equilibrium concentrations of the other two gases?
ii. What is the value of Kc for this reaction @25.1 C?
iii. How would the value of Kc for this reaction be affected if the original gram
amounts of each reactant were doubled? Justify your answer.
8. Iron(II)hydroxode dissociates in water by the following reaction:
Fe(OH)2(s)  Fe2+(aq) + 2 OH-(aq)
Given that a saturated solution had a pH of 11.2:
a. What is the value of Ksp for this reaction?
b. Cd(OH)2 has a Ksp of 5.3 x 10-15. Is it more or less soluble in distilled water than
iron(II)hydroxide at a given temperature? Justify your answer.
9. It is well known that nitrous acid decomposes in soil to produce nitrogen oxide gases and water
and can be represented by the reaction below:
2HNO2(aq) → NO(g) + NO2(g) + H2O(l)
The reaction is initiated and the pressure of the container is measured and graphed below:
a. What is the pressure of each gas at equilibrium?
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b. What is the value of the equilibrium constant, Kp, for the reaction?
c. Approximately when does this reaction reach equilibrium?
d. How would the value of the equilibrium constant you calculated in part “b” be affected if
the container developed a leak during the experiment? Justify your answer.
10. Agree or disagree with the following statements and justify your answer:
a. If the forward process is occurring at a greater rate than the reverse process, the
reactant concentrations will decrease until they are the same as the product
concentrations.
b. Reactions with small equilibrium constants will have a greater rate of production of
reactants at equilibrium than rate of production of products.
Part B Classwork:
11. A student mixes 120.0 uL of 0.0020 M Ca(NO3)2 with 80.0 uL of 0.0040 M NaF
expecting to form a precipitate:
a. Write the proper net-ionic reaction for the reaction they expected to see.
b. Predict whether they would expect to form a precipitate or not. Justify this with
a calculation. (Ksp for CaF2(s)= 3.45 x 10-11)
c. Agree or disagree with the following statements and justify your answer:
i.
Evaporating water from the mixture would make the precipitate more
likely to form.
ii.
The ions remaining in solution in order of increasing concentration
would be Ca2+< F-< Na+< NO3d. Given that the Ksp for MgF2 is 3.50 x 10-9, what would be the Ksp for the reaction
below:
Mg2+(aq) + CaF2(s)  Ca2+(aq) + MgF2(s)
12. Potassium chlorate can be thermally decomposed into potassium chloride and
oxygen gas. The equilibrium constant (Kp) for the formation of three moles of
oxygen gas reaction is equal to 18.9 at 1200 K
a. What would be the equilibrium constant (Kp) for the:
i.
Formation of 1 mole of potassium chlorate from potassium chloride
and oxygen gas.
ii.
Formation of 1 mole of oxygen from the decomposition of potassium
chlorate
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b. If a student were to mix 10.0 grams of both solid potassium chloride and
potassium chlorate in a rigid 10.0 L container filled with 0.45 grams of oxygen
gasat a temperature of 1200 K:
i.
What is the initial pressure of oxygen gas in the container?
ii.
Would you expect the partial pressure of oxygen gas to decrease or
increase as the reaction proceeds? Justify your answer with a
calculation.
13. The Haber process for the production of ammonia is represented by the following
equation:
3/2H2(g) + 1/2N2(g) NH3(g)
The equilibrium constant for this reaction is higher at higher temperatures.
a. Is the reaction endo or exothermic? Justify your reasoning using Le-Chatelier’s
principle.
b. What would be the effect of increasing the partial pressure of nitrogen gas on the
equilibrium constant at a given temperature? Explain your reasoning.
c. A fellow student proposes increasing the yield of ammonia by decreasing the
volume of the container. Do you agree with their method? Explain why or why
not.
d. Assuming a student began with initial partial pressures of each gas that were
1.00 atm and that the Kp for this reaction is 0.056 at 273 K:
i.
How would the partial pressure of the ammonia compare at equilibrium
to what it was initially? Justify your answer with a calculation.
ii.
Would the total pressure increase or decrease inside the container,
assuming the reaction occurred at constant temperature and pressure?
Explain.
14. Ammonia is a base that can react with acetic acid by the reaction below:
NH3(aq) + HC2H3O2(aq)  NH4+(aq) + C2H3O2-(aq)
Given that:
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) Kb = 1.8 x 10-5
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C2H3O2-(aq) + H2O(l)  HC2H3O2(aq) + OH-(aq) Kb = 5.6 x 10-10
a. What would be the equilibrium constant for the overall reaction between
ammonia and acetic acid?
b. How would adding solid sodium acetate influence the concentration of
ammonium ion assuming no change in overall solution volume? Explain.
c. Given that the solution got warmer as the reaction occurred, what would be the
effect of increasing the temperature on the equilibrium constant? Explain.
Part B Homework:
15. A student mixes 4.5 mL of 0.0020 M NaOH with 4.5 mL of 0.0010 M Cd(NO3)2
expecting to form a precipitate:
a. Write the proper net-ionic reaction for the reaction they expected to see.
b. Predict whether they would expect to form a precipitate or not. Justify this with
a calculation. (Ksp for Cd(OH)2(s)= 5.3 x 10-15)
c. Agree or disagree with the following statements and justify your answer:
i.
Increasing the volume of solutions added will make the precipitate
more likely to form.
ii.
Increasing the concentration of solutions added will make the
precipitate more likely to form.
iii.
Draw a picture representing the relative amounts of ions left in
solution after the solutions have been mixed.
d. Given that the Ksp for CdC2O4 is 1.4 x 10-8, what would be the K for the reaction
below:
Cd(OH)2(s) + C2O42-(aq) --> 2OH-(aq) + CdC2O4(s)
16. Nitrogen dioxide gas will decompose into two gaseous products by the following
reaction:
2 NO2 → 2 NO + O2
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Kp = 4.6 @ 440 K
Equilibrium
a. What would be the equilibrium constant (Kp) for the:
i.
Formation of 1 mole of nitrogen dioxide from oxygen and nitrogen
monoxide gas?
ii.
Formation of 2 mole of oxygen from the decomposition of nitrogen
dioxide.
b. If a student were to mix 6.00 grams of each gas in a rigid 10.0 L container @440
K:
i.
What is the initial pressure of oxygen gas in the container?
ii.
Would you expect the partial pressure of oxygen gas to decrease or
increase as the reaction proceeds? Justify your answer with a calculation.
c. If the equilibrium partial pressures of each gas were found to 0.2 atm @600 K, is
the reaction endo or exothermic? Justify your reasoning using Le-Chatelier’s
principle.
d. What would be the effect of increasing the partial pressure of oxygen gas on the
equilibrium constant at a given temperature? Explain your reasoning.
e. A fellow student proposes increasing the yield of oxygen gas by increasing the
volume of the container. Do you agree with their method? Explain why or why not.
f. Assuming a student began with initial partial pressures of each gas that were 1.00
atm and that the Kp for this reaction is 0.056 at 500 K:
i.
How would the partial pressure of the oxygen gas compare at equilibrium
to what it was initially? Justify your answer with a calculation.
ii.
Would the total pressure increase or decrease inside the container,
assuming the reaction occurred at constant temperature and pressure?
Explain.
17. Baking soda (NaHCO3) will react with lactic acid to produce carbon dioxide gas by
the reaction below:
HCO3-(aq) + HC3H6O3(aq) → H2O(l) + CO2(g) + C3H6O3-(aq)
Given the following:
HCO3-(aq) + H2O(l) → CO2(g) + H2O(l) + OH-(aq) Kb = 2.4 x 10-8
HC3H6O3(aq) → H+(aq) + C3H6O3-(aq) Ka = 1.4 x 10-4
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a. What would be the equilibrium constant for the overall reaction between
bicarbonate and lactic acid?
b. How would adding additional solid sodium bicarbonate influence the
concentration of aqueous lactic acid assuming no change in overall solution
volume? Explain.
c. Given that the solution got cooler as the reaction occurred, what would be the
effect of increasing the temperature on the equilibrium constant? Explain.
Part C Classwork:
18. NaCl is known to be a soluble salt. To test this a middle school student adds salt to 100
mL of water at 20 C and to their surprise finds that after adding the NaCl consistently for
10.5 minutes, it stops dissolving and starts accumulating at the bottom of the beaker. The
student continued to add NaCl for another 2.5 minutes and quit.
a) Explain this observation in terms of equilibrium concepts:
b) A high school student performed the same experiment above except monitored
the conductivity of the solution as the NaCl was added. On a graph with clearly
labeled axis, show what the expected data would look like.
c) On the same graph drawn in “b”, show the expected data if the experiment was
done with an insoluble salt such as AgCl.
19. A graduate student is given a beaker containing a solution taken from a brownfield site
that is thought to be 0.34 M Cd2+ and 0.21 M Pb2+. In the literature, the student found the
following Ksp values:
Ksp
CdC2O4
1.4 x 10-8
PbC2O4
4.8 x 10-12
CdS
8.0 x 10-27
PbS
8.0 x 10-28
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The student has been asked by their advisor to devise a method to selectively precipitate
the Cd2+ and Pb2+ ions from solution.
a) Which solution should the student add to selectively precipitate the ions (0.1 M
Na2C2O4 or 0.1 M Na2S)? Justify your answer.
b) Write an experimental procedure to selectively precipitate these ions and
indicate what observations would be seen at each step and identify the materials
used in each step.
c) Based on the solution you selected to use in “a”:
i) What would be the minimum concentration of the anion needed to
precipitate the Pb2+ ion?
ii) What would be the concentration of Pb2+ ion when the Cd2+ starts to
precipitate?
20. The relative molar solubilities of Sr3(PO4)2 in various solutions are listed below:
Solution
Molar Solubility of Sr3(PO4)2
Distilled water
1.29 x 10-5 M
0.1 M Na3PO4
1.14 x 10-9 M
0.1 M HCl
> 1.29 x 10-5 M
a) What would be the Ksp for Sr3(PO4)2?
b) Explain why the molar solubility is:
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i) Higher in 0.1 M HCl
ii) Lower in 0.1 M Na3PO4
c) If the following changes were made to a saturated solution of Sr3(PO4)2 in distilled
water, explain what would happen to the following:
i) The Ksp if additional distilled water was added
ii) The concentration of dissolved ions if half of the water was evaporated
away
iii) The mole quantity of dissolved ions if additional distilled water was
added
iv) The molar solubility of Sr3(PO4)2 if an aqueous solution of Ca(NO3)2 was
added. (The Ksp of Ca3(PO4)2 is 1.0 x 10-26 at this temperature)
Part C Homework:
21. A student adds solid sodium carbonate into a beaker containing 500.0 mL of water and
measures the conductivity over time. The student notices the conductivity increase steadily
and then reach a constant value despite the addition of more solid sodium carbonate.
a) Explain why the conductivity stopped increasing in terms of equilibrium
concepts.
b) What other observation should the student have noticed when the conductivity
stopped increasing?
c) Draw a graph plotting conductivity vs time for the same experiment except using
solid glucose (C6H12O6) instead of sodium carbonate. Be careful to label your axis.
22. Seawater is roughly 0.054 M Mg2+ and 0.018 M Ca2+. A graduate student has been asked
by their advisor to devise a method to selectively precipitate the Mg2+ and Ca2+ ions from
solution. The student prepares a 0.4 M NaCl solution, a 0.4 M Na2CO3 solution, and a 0.4 M
NaOH solution. In the literature, they find some relevant Ksp values:
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Ksp @25 C
CaCO3
3.4 x 10-9
MgCO3
6.8 x 10-6
Mg(OH)2
5.6 x 10-12
Ca(OH)2
5.0 x 10-6
a) Why would you advise the student NOT use the NaCl solution?
b) Which solution (Na2CO3 or NaOH) would you advise the student to use and why?
c) Based on whichever solution you choose to use in “b”
i. What would be the concentration of the anion needed to start to precipitate
the Mg2+ ion? The Ca2+ ion?
ii. What would be the concentration of Mg2+ ions when the Ca2+ first starts to
precipitate?
iii. Based on your answer to “c ii”, was this an effective separation? Explain:
23. A student knows from their solubility rules that silver chloride aught to have very
limited solubility in water @25 C. To test this, they plan to add 10.000 grams of AgCl to a
beaker with 1.0 L of distilled water. Then they plan to filter the mixture and collect the solid
undissociated AgCl and expect it to be only slightly less than the original mass amount.
Their data can be found below:
Mass of filter paper: 2.3411 grams
Mass of filter paper and collected AgCl: 12.3385 grams
a) What is the molar solubility of AgCl?
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b) What would be the Ksp of AgCl at 25 C?
c) How would the calculated value of Ksp be different if the student used tap water
(contains Cl- ions) instead of distilled water? Explain:
d) If PbCl2 was used instead of AgCl, what would have been the expected mass of the
filter paper? (Ksp of PbCl2 is 1.70 x 10-5)
24. The relative molar solubilities of ZnCO3 in various solutions are listed below:
Solution
Molar Solubility of ZnCO3
Distilled water
0.1 M ZnCl2
1.21 x 10-5 M
?
a) What would be the Ksp for ZnCO3?
b) What would be the molar solubility of zinc carbonate in 0.1 M ZnCl2?
c) Explain the difference in molar solubilities between the two solutions:
c) If the following changes were made to a saturated solution of ZnCO3 in distilled
water, explain what would happen to the following:
i) The concentration of Zn2+ ions if water was evaporated.
ii) The Ksp if water was evaporated.
iii) The mole quantity of dissolved ions if additional distilled water was
added
iv) The molar solubility of ZnCO3 if an aqueous solution of MnCl2 was added.
(The Ksp of MnCO3 is 2.24 x 10-11 at this temperature)
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Answers
1.
a. 2KClO3(s)  2KCl(s) + 3O2(g)
Kp = (PO2)3
b. The pressure would increase due to the increased moles of gas being
produced.
c.
i. 0.326 moles oxygen gas produced = P = 2.14 atm = Kp = 9.80
ii. The graph should have an appropriate y axis scale with the pressure
of oxygen being 2.14 when 13 seconds have elapsed. The pressure of
oxygen gas should be constant after 13 seconds.
2.
a. There would have been 50 moles of N2 = P = 238 atm
b. 3H2(g) + N2(g)  2NH3(g)
At equilibrium, [N2] = 6 M, [H2] = 2 M, [NH3] = 8 M
Kc = 1.33
c. How would the value of the equilibrium constant be affected if:
i. There would be no effect, the equilibrium constant is dependent on
the favorability of the reaction which is NOT influenced by initial
concentrations.
ii. Again, there would be no effect for the same reason given above in
part “I”
d. hydrogen gas molecules are reacting for every one nitrogen molecule
3.
a. Ksp = 8.30 x 10-11
b. More soluble, the large K value indicates a more favorable reaction.
4.
a.
b.
c.
d.
e.
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PSO3 = 0.8 atm, PSO2 = 1.2 atm, PO2 = 0.6 atm
Kp = 1.35
1.2/2.0 x 100 = 60%
23 seconds
This would result in a smaller final pressure measurement, thereby
decreasing the value of “x” which results in a smaller measured Kp value.
AP Chemistry
Equilibrium
5.
a. Disagree, equilibrium is reached when the rates of forward and reverse rates
are equal.
b. Disagree, the rates must be the same at equilibrium.
6.
a. CaCO3(s) → CO2(g) + CaO(s)
b. 50 % decomposition - P = 21.1 atm
75% decomposition - P = 31.7 atm
90% decomposition - P = 38.1 atm
Graph should plot the pressure on y axis and % decomposition on x axis and
these three points must be clearly plotted.
c.
i. The pressure in the container would stop increasing
ii. 92%
iii. Much of the mass was converted into calcium oxide - a solid and
thereby still present in the reaction container.
iv. 38.9
v. Since the reaction is endothermic, a drop in temperature would shift
the reaction left and decrease the pressure of carbon dioxide thereby
lowering Kp.
7.
a. 6.29 atm
b. 2HCl(g) → H2(g) + Cl2(g) Kc = [H2][Cl2]/[HCl]2
c. 6.29 atm, since the overall moles do not change in the reaction, neither will
the pressure
d. The pressure would always be constant throughout making it difficult to
determine when the reaction reaches equilibrium. An individual gas or other
metric would have to monitored.
e.
i. 0.033 M
ii. 28.1
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iii. It would not be affected as changing the concentrations only affects
the equilibrium position - not the constant.
8.
a. 2.02 x 10-9
b. Less soluble due to the smaller Ksp
9.
a.
b.
c.
d.
Each gas would have a pressure of 0.8 atm
0.64
After about 11 seconds
The pressure of the gases would be diminished and so would the value of Kp
10.
a. Disagree, equilibrium is NOT defined as when the concentrations of reactants
and products are equal.
b. Disagree, at equilibrium the rates of the opposing processes are equal!
11.
a. Ca2+(aq) + 2F-(aq)  CaF2(s)
b. Q = .0012*(.0016)^2 = 3 x 10-9 since Q > K, reaction will shift to precipitate.
c.
iii.
Agree, this would increase the Molarities and thereby increase Q and
promote formation of precipitate.
iv.
Disagree, the F- would be the limiting reactant so it’s concentration
would be less than the calcium ion.
d. K = 9.86 x 10-4
12.
a.
i.
ii.
0.23
2.66
c.
i.
0.139 m
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ii.
Q = 0.139^3 = 0.00269 so Q < K so reaction will proceed towards
product and the partial pressure of oxygen gas will increase.
13.
a. Since the reaction is shifting towards products with increasing temperature, the
energy must be a reactant so the reaction is endothermic.
b. None, a new equilibrium position would be established, not a new equilibrium
constant.
c. Agree, Decreasing the volume will cause the reaction to reduce the number of
moles which can be accomplished by shifting toward product thereby increasing
the yield of ammonia.
d.
i.
Calculate Q = 1 Q > K so the reaction shifts left thereby
diminishing the partial pressure of ammonia.
ii.
Since the reaction shifted to the side with more moles, the overall
pressure would increase as there are more moles of reactants than
products.
14.
a. 3.21 x 104
b. The acetate ion concentration would increase and shift the reaction left thereby
decreasing the ammonium ion concentration.
c. The reaction must be exothermic since energy is being produced so increasing
the temperature will add energy and shift the reaction left and diminish the size
of the equilibrium constant.
15.
a. 2OH-(aq) + Cd2+(aq) → Cd(OH)2(s)
b. Calculate Q = 5 x 10-10 for dissociation reaction, react will shift left to make
precipitate!
c.
i.
No, this will not change the concentrations and therefore will not
change Q
ii.
No, a precipitate will form at existing concentrations. Although an
increase in concentrations will form MORE precipitate.
iii.
The picture must have very few OH- or Cd2+ ions as they are both
equally precipitated. If they are drawn, there must be 2x hydroxide for each
Cd2+. There would be twice as many nitrate ions in solution than hydroxide.
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d. 3.78 x 10-7
16.
a.
i.
ii.
0.47
21.2
b.
i.
0.68 atm
ii.
Calculate Q = 1.59 which is smaller than K, so the reaction will shift
right and the partial pressure of oxygen gas will increase.
c. Exothermic, as the Kp is lower at higher temperatures so the reaction must be
shifting left at these higher temperatures meaning the energy must be a product
d. None, this will only shift the reaction to a new equilibrium position.
e. I agree, at a higher volume, the reaction will prefer the side with more moles,
which is the products and the oxygen gas.
f.
i.
Q = 1 which is greater than K, so the reaction will shift left and diminish
the partial pressure of oxygen gas.
ii.
The total pressure would decrease as the moles decreases as the reaction
shifts left.
17.
a. 3.36 x 10-12
b. The concentration would go down as it would react with the additional sodium
bicarbonate.
c. The reaction must be endothermic so increasing the temperature would shift the
reaction right, making more product and a higher K value.
18.
a) The solution has become saturated - an eventuality even for soluble salts. When
this happens the undissociated solid has reached equilibrium with the dissolved
ions and the rate of precipitation and dissociation are equal.
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b) The graph should plot conductivity on the y axis and time on the x axis. The graph
must show a linear increase in conductivity over time up to 10.5 minutes after
which there would be no more increase.
c) The graph would show an initial conductivity far less than in the NaCl and no
linear increase as the solution will be immediately at equilibrium.
19.
a) The student should use the Na2C2O4 solution as the differences in Ksp are 4 orders
of magnitude apart meaning the lead ion should be almost entirely precipitated
before the cadmium ion starts to precipitate making for an effective separation.
b) Slowly add microliter quantities of Na2C2O4 to the beaker, filter the PbC2O4
precipitate out, then slowly add microliter quantities of NaC2O4 to the beaker and
filter out the CdC2O4. Materials - micropipette, filter paper, and funnel.
c)
i) 2.28 x 10-11 M
ii) 1.17 x 10-4 M
20.
a) 4 x 10-28
b)
i) The H+ reacts with the phosphate ion, shifting the equilibrium to the right,
causing more of the strontium phosphate to dissolve.
ii) The additional phosphate ions shift the equilibrium left, decreasing the
amount of dissociated ions.
c)
i) Nothing, the concentration of ions would remain the same.
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ii) Nothing, they would remain constant. As water is evaporated away, the
concentrations would tend to increase but to restore equilibrium, the reaction shifts
towards the solid state, thereby keeping the concentration of ions constant.
iii) This would increase, although the concentration will stay the same - there
will now be more ions dissolved in the larger volume of water.
iv) Nothing, the Ksp of calcium phosphate is larger than that of strontium
phosphate thereby preventing the calcium ions from precipitating with the
phosphate so status quo is maintained.
21.
a) The solution became saturated and reached equilibrium so the concentration of
ions stopped increasing.
b) Solid sodium carbonate would be present in the beaker.
c) The conductivity would not change throughout the experiment as glucose does
not dissociate into ions.
22.
a) No precipitate will form with either ion
b) The NaOH as the difference in Ksp’s are greater for the hydroxide salts thereby
ensuring a greater precipitation of the first ion before precipitation of the 2nd.
c)
i. 1.02 x 10-5, 1.17 x 10-2
ii. 4.1 x 10-8
iii. Yes, most of the Mg2+ had been precipitated out before precipitating the
Ca2+
23.
a) 1.83 x 10-5 M
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b) 3.35 x 10-10
c) This would have lowered the molar solubility and given a lower Ksp value.
d) 7.838 g
24.
a) 1.46 x 10-10
b) 1.46 x 10-9
c) Since, Zn2+ act as a common ion in the 0.1 M ZnCl2 solution, the equilibrium is
shifted left resulting in a fewer dissociated ions.
d)
i) Nothing, the equilibrium would simply shift left to dissipate the higher
concentration created by the evaporation of water.
ii) Nothing, again, the reaction will shift to a new equilibrium position but the
Ksp would be unaltered.
iii) These would increase as the absolute number of dissociated ions would
increase
iv) Increase, since MnCO3 has a lower Ksp, the Mn2+ ions will preferentially
precipitate with the carbonate ions thereby shifting the equilibrium to the
right.
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