Chem 112, Fall 05
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Exam 3A
Before you begin, make sure that your exam has all 10 pages.
There are 32 required problems (3 points each, unless noted otherwise) and two extra credit
problems (3 points each).
Stay focused on your exam.
YOU MUST:
Put your name and student ID on the bubble sheet correctly.
Put all your answers on the bubble sheet; nothing on this exam will be used for grading.
Sign the statement on the last page of the exam.
Turn in both the exam and bubble sheet when you are done. Good Luck!
Some values of Ksp
Carbonates
Hydroxides
BaCO3
8.1 x 10-9
NiCO3
6.6 x 10-9
Ni(OH)2 2.8 x 10-16
Fe(OH)2 7.9 x 10-15
Fe(OH)3 6.3 x 10-38
page 1
Sulfates
BaS04
1.1 x 10-10
Sulfides
CaS
8 x 10-6
NiS
3.0 x 10-21
Chem 112, Fall 05
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Exam 3A
1. According to the Brønsted-Lowry definition, a base
A.
B.
C.
D.
E.
increases the H3O+ concentration in an aqueous solution.
increases the OH- concentration in an aqueous solution.
is a proton acceptor. Ans.
is a proton donor.
is an electron-pair donor.
Brønsted acids are proton donors and bases are proton acceptors
2. Which of the following is never a Brønsted-Lowry acid in an aqueous solution?
A.
B.
C.
D.
E.
hydrogen chloride, HCl(g)
dihydrogen sulfide, H2S(g)
ammonium chloride, NH4Cl(s)
hydrogen fluoride, HF(g)
magnesium oxide, MgO(s)
Ans. Each of the others are recognized either as
strong or as weak Brønsted-Lowry acids
3. Which salt forms a 0.10 M aqueous solution with the highest pH?
A. NaNO3
B. NH4Cl
C. FeCl3
D. Ca(ClO4)2
E. NaF
Ans.
+
NaNO3 and Ca(ClO4)2 form neutral salts. NH4 is a weak acid and Cl is the conjugate base
of HCl NH4Cl is acidic. Hydrated Fe3+, Fe3+(H2O)6, is also acidic. F-, the conjugate base
of a weak acid, is a weak base. NaF forms a basic solution.
4. Consider the following series of acids:
CH3CO2H, ClCH2CO2H, Cl2CHCO2H, Cl3CCO2H
Which statement explains the trend the pKAs (left to right).
A.
B.
C.
D.
E.
pKAs increase due to inductive effects generated by the substitution of H by Cl.
pKAs decrease due to inductive effects generated by the substitution of H by Cl. Ans.
pKAs increase due to the larger number of resonance structures possible in the anion.
pKAs decrease due to the larger number of resonance structures possible in the anion.
Both B and D.
The greater electronegativity of Cl relative to H stabilizes the conjugate base anion by an
inductive effect; pKas decrease as more Cl is substituted for H. Resonance also stabilizes
the anion, but the resonance structures are the same in all members of the series, and thus
resonance does not contribute to the trend.
page 2
Chem 112, Fall 05
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Exam 3A
5. At 15 °C, the water ionization constant, Kw, is 4.5 × 10-15. What is the H3O+ concentration in
neutral water at this temperature?
A.
B.
C.
D.
E.
2.0 × 10-29 M
4.5 × 10-15 M
6.7 × 10-8 M
1.5 × 10-7 M
2.2 M
Kw = [H3O+][OH-]
For ‘neutral’ water, [H3O+] = [OH-]
Kw = [H3O+]2
[H3O+] = sqrt(Kw)
[H3O+] = sqrt(4.5 x 10-15) = 6.7 x 10-8
Ans.
6. If you mix equal molar quantities of HF (Ka = 7.2 × 10-4) and KCN (Kb = 2.5 × 10-5), the
resulting solution will be
A.
B.
C.
D.
E.
acidic because Ka of HF is greater than Kb of CN-. Ans.
basic because Ka of HF is greater than Ka of HCN.
basic because Ka of HF is greater than Kb of CN-.
basic because Kb of F- is less than Kb of CN-.
neutral because the weak acid neutralizes the weak base.
HF is a stronger acid than
CN- is a base. The solution
becomes acidic, since both
are present in equal
amounts. HF wins.
7. Determine the equilibrium constant for the reaction below. (Ka (HF) = 7.2 × 10-4,
Ka (NH4+) = 5.6 × 10-10)
HF(aq) + NH3(aq) ' NH4+(aq) + F-(aq)
A. 4.0 × 10-13
B. 1.3 × 10-8
C. 7.8 × 10-7
D. 1.3 × 106
Ans.
E. 2.5 × 1012
K is given by Ka(HF)/Ka(NH4+) = (7.2 × 10-4)/(5.6 × 10-10) = 1.3 x 106
8. Which of the following chemical equations corresponds to Ka3 for phosphoric acid?
A.
B.
C.
D.
E.
HPO42-(aq) + H2O(l) ' PO43-(aq) + H3O+(aq) Ans.
PO43-(aq) + H2O(l) ' HPO42-(aq) + OH-(aq)
H3PO4(aq) + H2O(l) ' H2PO4-(aq) + H3O+(aq)
H3PO4(aq) + 3 H2O(l) ' PO43-(aq) + 3 H3O+(aq)
H2PO4-(aq) + H2O(l) ' HPO42-(aq) + H3O+(aq)
page 3
3rd Ionization, Ka3
Kb3 = Kw/Ka3
1st Ionization, Ka1
Overall Rxn, Ka1Ka2Ka3
2nd Ionization, Ka2
Chem 112, Fall 05
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Exam 3A
9. Boric acid (H3BO3) has a pKa value of 9.14. What is the Kb of sodium borate (NaH2BO3) at
25 oC?
A. 7.2 × 10-10
B. 1.4 × 10-5
Ans.
pKb = pKw- pKa = 14 – 9.14 = 4.86
C. 1.1 ×10-4
D. -4.86
E. 4.86
Kb = 10-4.86 = 1.4 x 10-5
10. In the following acid-base reaction equilibrium
C9H7NH+ + C6H5COO- ' C9H7N + C6H5COOH
☺
☺
+
A. C9H7NH is the conjugate acid of C9H7N.
B. C6H5COO- is the conjugate acid of C6H5COOH.
C. C6H5COOH and C6H5COO- represent a conjugate acid-base pair
D. Both A and B
E. Both A and C
Ans.
11. Suppose that K < 1 for the equilibrium in problem 10.
☺
☺
+
A. C6H5COO- is the weaker base and C9H7NH is the stronger acid
+
B. C6H5COO- is the weaker base and C9H7NH is the weaker acid
C. C9H7N is the weaker base and C6H5COOH is the weaker acid
D. C9H7N is the stronger base and C6H5COOH is the stronger acid
E. Both B and D
Ans.
12. (4 pts) For the following three reactions, K > 1
HF (aq) + C6H5COO- (aq) '
HC9H7O4 (aq) + C6H5COO- (aq) '
HF (aq) + C9H7O4- (aq)
F- (aq) + C6H5COOH (aq)
C6H5COOH (aq) + C9H7O4- (aq)
' F- (aq) + HC9H7O4 (aq)
Among the three acids and three conjugate bases, which are the strongest acid and base?
A.
B.
C.
D.
E.
HC9H7O4 and C6H5COOK > 1, therefore strong acids & bases are on left.
HF and C6H5COO
Ans. The strongest acid has the weakest conj. base.
HF and C9H7O4The weakest acid has the strongest conj. base.
C6H5COOH and F
Identify strongest and weakest acids.
HF, always on the left, is the strongest acid.
C9H7CO4- and FC6H5OO-, always on the left, is the strongest base.
page 4
Chem 112, Fall 05
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Exam 3A
13. Two solutions of the same weak acid (pKA = 6) were made to have different concentrations
(10-4 M and 0.1 M). The 10-4 M solution was found to have a
A.
B.
C.
D.
smaller pH than the 0.1 M solution
As more acid is added, the [H3O+] will increase
larger pH than the 0.1 M solution (Ans.) and the pH will decrease. The pH of the 10-4 M
the same pH as the 0.1 M solution
solution will be larger (smaller [H3O+]).
Not enough information is provided to determine if a difference in pH exists.
14. Two buffer solutions were made with equal molar mixtures of a weak acid its conjugate
base, and differed only by the total concentration of acid plus base. The buffer solution with
the larger total acid plus base concentration had
A.
B.
C.
D.
a pH that was larger than the buffer with the smaller total concentration.
a pH that was smaller than the buffer with the smaller total concentration.
a pH that was the same as the buffer with the smaller total concentration. Ans.
Not enough information is provided to make this decision.
pH is determined by the [conj. base]/[acid] ratio. Doubling both concentrations will leave
the pH unchanged.
15. Each of the following mixtures can produce a buffer solution EXCEPT
A.
B.
C.
D.
E.
HClO4 and NaClO4
HF and NaF
NaHCO3 and Na2CO3
Na2HPO4 and Na3PO4
NH4Cl and NH3
Ans.
HClO4 is the only strong acid. All the
others are weak acids and will form buffers.
16. Which of the following mathematical expressions is the Henderson-Hasselbalch equation?
⎛ [ conjugate base] ⎞
A. pK a = pH + log ⎜
⎟⎟
⎜
[acid ]
⎝
⎠
⎛ ⎡OH - ⎤ ⎞
⎦ ⎟
B. pH = pK a + log ⎜ ⎣
⎜ ⎡ H 3O + ⎤ ⎟
⎦⎠
⎝⎣
⎛
⎞
[acid ]
C. pH = pK a + log ⎜⎜
⎟⎟
⎝ [ conjugate base] ⎠
⎛
⎞
[acid ]
D. pK a = pH - log ⎜⎜
⎟⎟
⎝ [ conjugate base] ⎠
⎛ [ conjugate base] ⎞
E. pH = pK a + log ⎜⎜
⎟⎟
[acid ]
⎝
⎠
Ans.: E. Definition of the HH equation.
page 5
Chem 112, Fall 05
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Exam 3A
17. What is the pH of a solution that results from adding 25 mL of 0.15 M HCl to 25 mL of
0.52 M NH3? (Kb of NH3 = 1.8 × 10-5)
A. 2.74
B. 4.35
C. 9.65
Ans.
D. 11.26
E. 11.41
Rxn of HCl with NH3 is complete to from 0.15/2 M NH4+ and 0.37/2 M NH3. Use HH eqn:
pH = pKa + log [NH3]/[ NH4+] = 14 + log(1.8 × 10-5) + log(0.37/0.15) = 9.65
18. Which of the following acid-base pairs is most suited to make a buffer with a pH = 7.68?
A.
B.
C.
D.
E.
HCO2H/NaHCO2
H2CO3/NaHCO3
HOCl/NaOCl
H3BO3/NaH2BO3
NH4Cl/NH3
pKA
pKB
3.74
6.38
7.46
9.14
9.25
10.26
7.62
6.54 Ans.
4.86
4.75
The HOCl/NaOCl acidconjugate base pair has
the pKa closest to 7.68.
19. 500 mL of a buffer solution formed with NH4Cl and NH3 has a pH = 9.64 and the
[NH3] = 0.5 M. What is the number of grams of NH4Cl required to make this solution?
(MWNH3 = 17.0 g/mol, MWNH4Cl = 53.5 g/mol) (Kb of NH3 = 1.8 × 10-5)
A.
B.
C.
D.
E.
10.90 g
1.73 g
3.46 g
5.57 g
26.7 g
Ans.
Find [NH4+], then moles of NH4+, then the mass of NH4Cl.
Ka = [H3O+][NH3]/[NH4+]
[NH4+] = [H3O+][NH3]/Ka
[NH4+] = [H3O+][NH3]Kb/Kw
[NH4+] = (10-9.64)(0.5)(1.8x10-5)/(10-14) = 0.206 M
0.206 mole/L x 0.5 L x 53.5 g/mole = 5.52 g
20. If the ratio of acid to base in a buffer increases by a factor of 10, the pH of the buffer
A. decreases by 1 Ans.
D. increases by 1
B. decreases by 10
E. remains unchanged
C. increases by 10
When the acid/base ratio increases by 10, the pH will decrease by 1.
log([base]/[acid]) = log(0.1) = -1
21. What is the pH of the buffer that results when 11 g of NaCH3CO2 is mixed with 85 mL of
1.0 M CH3CO2H and diluted with water to 1.0 L? (Ka of CH3CO2H = 1.8 × 10-5; MW of
CH3CO2Na = 82.04 g/mol)
A. 2.91
B. 3.86
C. 4.55
D. 4.74
E. 4.94
Ans.
Moles acetate = (11 g)/(82.04 g/mol) = 0.13. Moles Acetic acid = 0.085 L x 1 M = 0.085.
pH = pKa + log[(moles base)/(moles acid)] = -log(1.8 x 10-5) + log(0.13/0.085) = 4.95
page 6
Chem 112, Fall 05
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Exam 3A
22. Which one of the following conditions is always met at the equivalence point of the titration
of a monoprotic weak base with a strong acid?
A.
B.
C.
D.
E.
The pH of the solution is equal to 7.00.
The volume of acid added from the buret equals the volume of base titrated.
The molarity of the acid equals the initial molarity of the weak base.
The percent ionization of the acid equals the percent ionization of the base.
The moles of acid added from the buret equals the initial moles of weak base.
Ans.
At the equivalence point moles weak base = moles strong acid.
23. Which one of the following conditions is always true for a titration of a weak acid with a
strong base?
A.
B.
C.
D.
E.
A colored indicator with a pKa less than 7 should be used.
If a colored indicator is used, it must change color rapidly in the weak acid’s buffer region.
Equal volumes of weak acid and strong base are required to reach the equivalence point.
The equivalence point occurs at a pH equal to 7.
The equivalence point occurs at a pH greater than 7. Ans.
At the equivalence point, the solution consists of the conjugate base of the weak acid that
was present initially. The solution has a pH > 7.
24. A 25.0 mL sample of 0.0200 M NH3(aq) is titrated with 0.0100 M HCl(aq). What is the pH
at the equivalence point? (Kb of NH3 = 1.8 × 10-5)
A. 3.46
B. 5.48
C. 5.72
D. 8.25
E. 10.54
Ans.
The titration generates a weak acid solution (NH4+). First, calculate the [NH4+] at the
equivalence point, and then carry out an ICE calculation of the NH4+ solution.
Moles of NH3 = moles of NH4+ = moles of HCl added = 0.025 x 0.02 M = 5 x 10-4 moles,
Liters 0.01 M HCl added = 5 x 10-4 moles/(0.01 moles/L) = 0.05 L,
Final volume = 0.025 + 0.05 L = 0.075 L; [NH4+] = 5 x 10-4 moles/0.075 = 0.0067 M,
pH = 5.71
[H3O+] = sqrt(Ka x 0.0067) = sqrt(5.56 x 10-10 x 0.0067) = 1.92 x 10-6
25. Which of the following equations is the solubility product for magnesium iodate, Mg(IO3)2?
A. K sp = [Mg 2+ ][I- ]2 [O 2- ]6
B. K sp = [Mg 2+ ][I- ]2 [3O 2- ]2
C. K sp = [Mg 2+ ][IO3- ]
D. K sp = [Mg 2+ ]2 [IO3- ]
E. K sp = [Mg 2+ ][IO3- ]2
Ans.
Follows directly from the definition of Ksp
page 7
Chem 112, Fall 05
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Exam 3A
26. The solubility of SrSO4 in water is 0.107 g in 1.0 L at 25 °C. What is the value of Ksp for
SrSO4?
A. 3.4 × 10-7
Ans.
B. 5.8 × 10-4
C. 1.2 × 10-3
D. 1.1 × 10-2
E. 2.1 × 10-1
MW(SrSO4) = 87.62 + 32.07 + 4 x 16.00 = 183.69 g/mole
[Sr2+] = (0.107 g)/[(183.69 g/mole)*(1 L)] = 5.82 x 10-4 M = X
Ksp = [Sr2+][SO42-] = X2 = (5.82 x 10-4)2 = 3.4 x 10-7
27. What is the molar solubility of Fe(OH)3(s) in a solution that is buffered at pH 2.75 at 25 °C?
The Ksp of Fe(OH)3 is 6.3 × 10-38 at 25 °C.
A.
B.
C.
D.
E.
1.1 × 10-29 mol/L
1.1 × 10-26 mol/L
2.0 × 10-15 mol/L
2.2 × 10-10 mol/L
3.5 × 10-4 mol/L
Fe(OH)3
Fe3+(aq) + 3OH- (molar solubility equals [Fe3+])
3+
Ksp = [Fe ][OH-]3; pOH = 14 – pH = 11.25; [OH-] = 10-11.25
[Fe3+] = Ksp/[OH-]3 = 6.3 × 10-38/(10-11.25)3 = 6.3 × 10-38/(10-33.75)
[Fe3+] = 6.3 × 10-4.25= 3.5 × 10-4
Ans.: E
28. The following anions can be separated by precipitation as silver salts: Cl-, Br-, I-, CrO42-. If
Ag+ is added to a solution containing the four anions, each at a concentration of 0.10 M, in
what order will they precipitate?
Compound
AgCl
Ag2CrO4
AgBr
AgI
A.
B.
C.
D.
E.
Ksp
1.8 × 10-10
1.1 × 10-12
5.4 × 10-13
1.5 × 10-16
Ksp
[Ag+][Cl-]
[Ag+]2[CrO42-]
[Ag+][Br-]
[Ag+][I-]
AgCl → Ag2CrO4 → AgBr → AgI
AgI → AgBr → Ag2CrO4 → AgCl
Ag2CrO4 → AgCl → AgBr → AgI
Ag2CrO4 → AgI → AgBr → AgCl
AgI → AgBr → AgCl → Ag2CrO4
Ans.
page 8
[Ag+]
Ksp/[Cl-]
(Ksp/[CrO42-])0.5
Ksp/[Br-]
Ksp/[I-]
[Ag+] (M)
1.8 × 10-11
3.3 × 10-6
5.4 × 10-12
1.5 × 10-15
Chem 112, Fall 05
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Exam 3A
29. (4 pts) Barium sulfite (s) and barium fluoride (s) are in equilibrium with a solution
containing 8.70 x 10-3 M ammonium fluoride (completely soluble). Calculate the
concentration of sulfite ion present in this solution. Ksp (BaSO3) = 6.0 x 10-7, Ksp (BaF2) =
1.7 x 10-6.
A.
B.
C.
D.
E.
0.35 M
0.023 M
2.7 x 10-5 M
3.1 x 10-3 M
6.0 x 10-7 M
[F-] is used to calculate [Ba2+], which is then
used to compute [SO32-]:
[Ba2+] = (1.7 x 10-6)/(8.70 x 10-3)2 = 0.0225 M
[SO32-] = (6.0 x 10-7)/(0.0225) = 2.67 x 10-5 M
Ans.
30. A solution contains 0.10 M potassium sulfide and 0.10 M sodium carbonate. Solid nickel
acetate is added slowly to this mixture. What substance precipitates first?
A.
B.
C.
D.
E.
K2CO3
NiS
Ans.
CH3CO2K
NiCO3
No precipitate will form.
Soluble
Ksp = [Ni2+][S2-] = 3.0 x 10-21 (from exam front page)
Soluble
Ksp = [Ni2+][CO32-] = 6.6 x 10-9 (from exam front page)
The solubility product for NiS is much smaller, NiS will precipitate first, since S2- and
CO32- are present in equivalent concentrations.
31. The ionic solids AgNO3 and KI are dissolved in water and mixed together and in various
proportions to form solutions with the Ag+ and I- concentrations listed below:
1
2
3
[Ag+]
(M)
10-5
10-8
10-16
[I-]
(M)
10-10
10-8
2
[Ag+][I-]
Precipitate?
10-15
10-16
2 x 10-16
Yes
No
Yes
Which of these solutions will form a precipitate? (Ksp = 1.5 x 10-16)
A.
B.
C.
D.
E.
1
1 and 3 Ans.
1 and 2
all three
None of the solutions form precipitate.
page 9
Chem 112, Fall 05
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Exam 3A
32. (4pts) A solution contains 6.8 x 10-4 M KOH. Solid iron(III) nitrate is added slowly to this
mixture. What is the concentration of iron(III) ion when precipitation first begins ?
A.
B.
C.
D.
E.
6.8 x 10-4 M
1.2 x 10-11 M
1.4 x 10-31 M
2.0 x 10-28 M
1.7 x 10-8 M
Ans.
Take the Ksp for (Fe(OH)3 and solve for [Fe3+].
Ksp = 6.3 x 10-38 (from the exam front page)
[Fe3+] = Ksp/[OH-]3 = 6.3 x 10-38/(6.8 x 10-4)3
[Fe3+] = 2.0 x 10-28
EXTRA CREDIT
33. (3 pts) At 25 °C, only 1.04 x 10-3 g Cu(NO3)2 will dissolve per liter of a solution that is
buffered at pH 6.80. What is the value of Ksp for Cu(OH)2? The molar mass of Cu(NO3)2 is
187.6 g/mol.
A. 2.2 × 10-20
Ans.
B. 1.4 × 10-19
C. 4.1 × 10-18
D. 3.5 × 10-13
E. 8.8 × 10-13
pOH = 14 – 6.8 = 7.2
[OH-] = 10-7.2
moles Cu(NO3)2 dissolved = (1.04 x 10-3 g)/(187.6 g/mol) = 5.54 x 10-6 Cu2+ moles dissolved/L
Ksp = [Cu2+][OH-]2 = (5.54 x 10-6)(10-7.2)2 = 2.21 x 10-20
34. (3 pts) What is the concentration of PO43- in a solution created by mixing equal volumes of
0.1 M NaH2PO4 and 0.2 M Na2HPO4? (The acid dissociation constants for phosphoric acid
are Ka1 = 7.5 × 10-3, Ka2 = 6.2 × 10-8, and Ka3 = 3.6 × 10-13)
A. 6.2 × 10-8 M B. 0.2 M
C. 2.3 × 10-6
Ans.
D. 5.8 × 10-5
E. zero
[H2PO4-] = 0.1 M; [HPO42-] = 0.2 M
Find [H3O+] with Ka2, then use Ka3 to find [PO43-]
Ka2 = [H3O+][HPO42-]/[H2PO4-]
[H3O+] = (Ka2[H2PO4-])/[HPO42-] = (6.2 x 10-8)(0.05)/(0.1) = (3.1 x 10-8)
Ka3 = [H3O+][PO43-]/[HPO42-]
[PO43-] = Ka3[HPO42-]/[H3O+] = (3.6 x 10-13)(0.1)/(3.1 x 10-8) = 1.16 x 10-6 M
The closest ‘correct’ answer, C, was computed incorrectly, 0.2 M was used mistakenly for
[HPO42-] to compute [PO43-].
page 10