BRING YOUR REG CARD OR PICTURE ID TO THE FINAL EXAM
FINAL EXAM:
Friday, Dec. 13: 8 – 11 AM,
Last Names: A – B
Last Names: C – Z
North Hall 1105
Chem. 1179
It is very important for everyone to work and study all the examples and problems assigned in the text. Work the problems in
previous reviews, midterms, and quizzes. You are allowed to use one 8.5 x 11 inch page of notes during the exam. Formulas,
gas constants, table of Ka values and other possibly useful information will be given on the exam.
OFFICE HOURS:
Monday, Dec. 9, 10–11:15
Thursday, Dec. 12, 11 – 12 and 3 – 4 PM PSBN 3670B
REVIEW PROBLEMS
1. Which of the following is not a base?
a) NH3
b) C2H6NH
c) C2H5COOH
d) C2H6NNH2
2. Which of the following salts would produce an aqueous solution with the highest pH?
a) NH4Cl
b) NaCl
c) NaNO2
d) NaNO3
3. What is the conjugate acid of ammonia, NH3?
4. What is the conjugate base of acetic acid?
5. The H+ concentration in a certain solution is 6.6 x 10-4 M. What is the OH– concentration? Is the solution acidic, basic or neutral?
6. Which of the following pairs of 1.0 M solutions would result in a buffer if they were mixed?
a) NH3 and H2NNH2
b) HCl and HOCl
c) HNO3 and HNO2
d) CH3COOH and NaCH3COO
7. Circle the correct answer for each of the following questions.
a) Which is the strongest acid?
HCN H2O
H2S
HF
b) Which is the strongest base?
CN–
PO43– H2O
C5H5N
c) Which salt, when dissolved in water, will produce the most basic solution?
KCl
KClO
KClO2
KClO3
8. Indicate whether the following salts will give acidic, basic or neutral solutions. For those that are not neutral, write balanced
chemical equations for the reactions causing the solution to be acidic or basic.
a) NH4Cl
b) NaCN
c) NaI
d) CaCO3
9. A buffer solution is made with carbonic acid (H2CO3) and the salt of its conjugate base (NaHCO3). Determine the pH of the
solution when [H2CO3] = [HCO3–].
10. A strong acid, HA, is titrated with NaOH. When 30 mL of 1.0 M NaOH has been added to 40 mL of 1.0 M HA what is the
pH of the solution? a) pH < 7
b) pH = pKa
c) pH > 7
d) pH = 7
11. A weak acid, HA, is titrated with NaOH. When 20 mL of 1.0 M NaOH has been added to 40 mL of 1.0 HA what is the pH
of the solution?
a) pH < 7
b) pH = pKa
c) pH > 7
d) pH = 7
12. 50 mL of 1.0 M HCl is titrated with 0.5 M NaOH. Calculate the pH after 60 mL of NaOH has been added.
13. Using the data in the Table of Acid Ionization Constants determine the equilibrium constant for the following reaction.
HSO4– (aq) + HCO3– (aq)
SO42– (aq) + H2CO3 (aq)
Identify the stronger acid and the stronger base.
14. For a 0.13 M solution of formic acid, HCOOH, the pH = 2.40. Calculate Ka for formic acid.
HCOOH
HCOO – (aq) + H+ (aq)
15. Consider the following reaction at equilibrium. CH3COOH
CH3COO – (aq) + H+ (aq)
If NaCH3COO is added to the solution will the following quantities increase, decrease or remain the same?
a) [H+]
b) [OH–]
c) [CH3COOH]
d) pH
e) pKa
16. If 50 mL of 0.2 M HCl is titrated with 0.1 M NaOH
a) Calculate the pH of the solution after 10 mL 0.1 M NaOH was added.
b) Calculate the pH of the solution at the equivalence point.
c) Calculate the pH of the solution after 110 mL 0.1 M NaOH was added.
17. a) A buffer solution is made by mixing 60 mL of 0.5 M HF and 50 mL of 0.6 M NaF. Calculate the pH of this solution.
b) For the buffer solution, made by mixing 60 mL of 0.5 M HF and 50 mL of 0.6 M NaF, calculate the pH after 10 mL of
0.1 M NaOH has been added.
c) Is the difference in pH reasonable for these two solutions?
18. In the titration of 100 mL of 0.2 M of CH3COOH with NaOH, calculate the pH after 40 mL of the 0.5 M NaOH is added. Is
the solution acidic, basic or neutral?
19. Ksp = 1 x 10–10 for BaSO4 (s). If 50 mL of 2.0 x 10–4 M BaCl2 is mixed with 50 mL of 2.0 x 10–4 M Na2SO4, will the
precipitate BaSO4 (s) be formed? You must show your work to determine if a precipitate forms.
20. a) Calculate the solubility of CaF2 (s) in water. Ksp = 4.0 x 10–11 for CaF2 (s).
b) Calculate the solubility of CaF2 (s) in a 1.0 M NaF solution.
c) Compare the solubility obtained in part a) and b). Are the results reasonable? Explain why or why not.
21. Which of the following compounds are more soluble in acidic solution than in pure water?
a) AgCN
b) PbI2
c) Al(OH)3
d) Ca3(PO4)2
SOLUTIONS TO PROBLEMS (NOTE: Even though I try hard not to make mistakes, it happens. Please email me if
you find an answer that is not correct. Thanks.)
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c) C2H5COOH is an acid
c) NaNO2 produces the most basic solution => highest pH. NO2– + H2O
HNO2 + OH–
+
NH4
CH3COO–
[H+] = 6.6 x 10–4 M [OH–] = Kw / [H+] = 1.5 x 10–11
[H+] > [OH–] => the solution is acidic
d) CH3COOH /NaCH3COO
a) HF
b) PO43–
c) KClO, (ClO– produces the most basic solution because it is the strongest base, largest Kb)
a) acidic
NH4+
H+ + NH3
–
b) basic
CN + H2O
HCN + OH–
c) neutral
d) basic
CO32– + H2O
HCO3–
+ OH–
–
–
a) pH = pKa + log([A ]/[HA]) when [A ] = [HA] then pH = pKa. From the table of acid ionization constants pKa = 6.37 for
H2CO3. Therefore pH = 6.37.
When a strong acid is titrated with a strong base, the reaction is
H+ + OH– —> H2O This is a stoichiometry problem.
+
–
–
Determine the moles of H and OH . In this case, OH is the limiting reagent and is completely consumed. Excess H+ =>
pH < 7. You can show this by calculating [H+] = 0.14 M and pH = 0.84.
Major species in solution: HA, Na+, OH– , H2O
Reaction: HA + OH– —> A– + H2O
Stoichiometry problem.
Determine moles of HA and OH–. After the reaction goes to completion you will see that [HA] = [A–] and thus, pH = pKa
Major species in solution: H+, Cl–, Na+, OH–, H2O
Reaction: H+ + OH – —> H 2O This is a stoichiometry problem. Determine moles of H+ and OH –. After the reaction goes
to completion, calculate concentration, [H+], using the total volume. [H+] = 20 mmol / 110 mL = 0.18 M and pH = 0.74.
K = Ka (HSO4–) / Ka (H2CO3)
K = 1.2 x 10–2 / 4.3 x 10–7 = 2.79 x 104
–
Of the two acids: HSO4 and H2CO3, HSO4– is stronger. Of the two bases: SO42– and HCO3–, HCO3– is stronger.
When you are given the pH, it allows you to calculate [H+] at equilibrium. [H+] = 10–pH = 3.98 x 10–3 M
Ka = x2 / (0.13 – x) = (3.98 x 10–3)2 / (0.13 – 3.98 x 10–3) = 1.3 x 10–4
a) [H+] decrease
b) [OH–] increases c) [CH3COOH] increases d) pH increases
e) pKa remains the same
+
–
Reaction: H + OH —> H 2O This is a stoichiometry problem. Determine moles of H + and OH – and the limiting reagent.
Reaction goes to completion. Determine [H+] and pH.
a) pH = 0.82 b) pH = 7.0
c) pH = 11.8
a) Remember when two solutions are mixed together the concentrations of HF and F– are calculated using the total volume.
[HF] = 0.5M (60 mL / 110 mL) = 0.27 M and [F–] = 0.27 M For a buffer solution, when [HF] = [F–], pH = pKa = 3.18
b) Major species in solution: HF, OH–, F–, H2O
Reaction: HF + OH– —> F– + H2O Stoichiometry problem.
After the reaction goes to completion, [HF] = 0.26 M and [F–] = 0.28 M. Calculate pH at equilibrium. pH = 3.21
c) pH increased from 3.18 to 3.21, a small change is a reasonable change for a buffer solution.
pH = 8.9 basic solution. (If you need help, this type of problem is like problem 2, on the back of the Ch. 8 Buffer hand-out).
BaSO4(s)
Ba2+ + SO42–
Ksp = 1.0 x 10–10
NOTE: You must write the reaction as shown with BaSO4(s) on the left. If you write the reverse reaction, the equilibrium
constant is the inverse of the solubility product constant, 1/Ksp.
Also remember that when you add two solutions together the concentrations are calculated using the total volume. The initial
concentrations are: [Ba2+] = [SO42–] = 1.0 x 10 –4 M. Q = [Ba 2+] [SO42–] = 1.0 x 10–8. Because Q > K, the reaction shifts to the
left and the precipitate, BaSO4 (s), is formed.
a) CaF2(s)
Ca2+ + 2 F–
Ksp = 4.0 x 10–11
Ksp = s (2s)2 = 4.0 x 10–11
Solubility = s = 2.2 x 10–4 M
2
–11
–11
b) Ksp = s (1.0 – 2s) = 4.0 x 10
Solubility = s = 4.0 x 10
c) The solubility of CaF2 is much less in 1.0 M NaF. This is due to the common ion effect.
AgCN, Al(OH)3 and Ca3(PO4)2 are more soluble in acidic solution than in pure water. The solubility of PbI2 does not
increase in acidic solution, because I– is an ineffective base.
Good Luck on the Final and Happy Holidays
BRING YOUR REG CARD OR PICTURE ID TO THE FINAL EXAM