Chemistry 1412
EXAM # 2A Sample
1
CHEM 1412 Exam #2A Sample
(Chapters 15, 16, and 18)
Name:___________________
Score:
PART I - (3 points each) - Please write your correct answer next to each question
number. DO NOT CIRCLE
_____1. Calculate the pH of a solution if it's [OH] = 0.000700 M and indicate whether the
solution is acidic, basic, or neutral.
A.
3.15, acidic
B. 17.2, basic
C. 10.8, basic
D. 11, basic
_____2. Which one of the conjugate bases of the following Brønsted-Lowry acids is incorrect?
A. ClO- for HClO
B. HS- for H2S
C. H2SO4 for HSO4-
D. NH3 for NH4+
_____3. At normal body temperature, 37°C, Kw = 2.4 x 10-14.
Calculate [H+] if [OH-] = 1.3 x 10-9 M at this temperature.
A. 1.3 x 10-9 M
B. 1.0 x 10-7 M
C.7.7 x 10-6 M
D. 1.9 x 10-5 M
_____4. Calculate the pH of a 0.025 M solution of propanoic acid (Ka = 1.3 x 10-5).
A. 5.7 x 10-4
B. 0.025
C. 1.6
D. 3.25
_____5. Predict whether aqueous solutions of the following compounds are acidic, basic, or
neutral. Find the incorrect answer.
A. KNO3 (neutral)
B NaC2H3O2 (basic)
C. KClO (acidic)
D. NaCN (basic)
_____6. Which of the following acids consider as a strong acid?
A. HCN
B. HClO4
C. HF
D. HNO2
_____7. Does the pH increase, decrease, or remain the same on addition of each of the
following? (i) NaNO2 to a solution of HNO2, (ii) HCl to a solution of NaC2H3O2
A. (i) decrease, (ii) increase
C. (i) decrease, (ii) decrease
B. (i) increase, (ii) increase
D. (i) increase, (ii) decrease
_____ 8. Calculate the pH of a solution of 0.080 M H2SO4. (Ignore the second ionization of
H2SO4)
A. 1.10
B. 0.80
C. -1.10
D. 0.080
_____ 9. How many mL of 0.0350 M NaOH are required to titrate 65.0 mL of 0.0620M HBr to
the equivalence point?
A. 8.69 x 10-3 mL
B. 33.4 mL
C. 36.7 mL
D. 115 mL
2
____ 10. Calculate the pH of a buffer that is 0.20 M in formic acid and 0.15 M in sodium formate
(Ka = 1.8 x 10-4).
A. 2.4 x 10-4
B. 3.63
C. 0.82
D. 8.33
____11 Which of the following acid dissociation constants represents a stronger acid?
A. Ka = 1.8x10-5
B. Ka = 7.4x10-4
C. Ka = 1.45x10-6
D. Ka = 6.67x1 0-10
____12. If X is the solubility of a salt in moles per liter, which one of the following Ksp expressions
is incorrect?
A. Ag2S , Ksp = 4X3
C. BaSO4 , Ksp = 4 X2
B. AgCl , Ksp = X2
D. Cu(OH)2 , Ksp = 4X3
____13. How does the entropy of the system change when each of the following occurs (i) a solid is
melted, (ii) a liquid vaporizes, (iii) a gas liquefies?
A.
B.
C.
D.
(i) increases, (ii) increases, (iii) decreases
(i) increases, (ii) increases, (iii) increases
(i) decreases, (ii) increases, (iii) increases
(i) increases, (ii) decreases, (iii) increases
____ 14. For each of the following processes, predict whether entropy, S , is positive or negative:
(i) 2K(s) + Br2(l)  2KBr(s),
(ii) 2MnO2(s)  2MnO(s) + O2(g)
A. (i) positive, (ii) positive
C. (i) positive, (ii) negative
B. (i) negative, (ii) positive
D. (i) negative, (ii) negative
____ 15. Using the given values of S° (J/mol.K) , calculate the value of S°
for the reaction;
C2H4(g) + H2(g)  C2H6(g)
0
S = 219.4
131.0
229.5
A. -120.5 J/ K
B. 120.5 J/ K
C. 10.1 J/ K
D. -10.1 J/ K
____16. For a certain reaction,  H = 45 kJ and S = 125 J/K. Assume that H and S do not
vary with temperature. At what temperature will the reaction have G = 0?
A.
0.36 °C
B.
0.36 K
C. 360 °C
D. 360 K
____ 17. In which case does the entropy of the system decrease significantly?
A. H2O(l)  H2O(g)
C. SnO2(s) + 4C(s)  Sn(s) + 4CO2(g)
B. C(s) + O2(g)  CO2(g)
D. N2(g) + 3 H2(g)  2NH3(g)
3
____18. Calculate the equilibrium constant at 298 K for the reaction
CaCO3(s)  CaO(s) + CO2(g)
G0 = - 254.8 KJ
A. 4.61
B. 1.02x102
C. 4.61x1044
D. none of these
____19. At equilibrium which of the following is zero?
A. H
B. G
C. S
D. all of these
____20. Which of the following represents a spontaneous reaction ?
A. K < 1
B. G< 0
C. S >0
D. all of these
PART II - ( 8 points each) Please show all your work.
21. If the pH of a solution is 6.30, what are the molar concentrations of H+(aq), OH-(aq) , and
pOH in the solution?
22. Lactic acid, HC3H5O3, has one acidic hydrogen. A 0.10 M solution of lactic acid has a pH of
2.44. Calculate Ka.
23. Calculate the percent ionization of 0.10 M butanoic acid (Ka = 1.5 x 10-5) in a solution
containing 0.050 M sodium butanoate.
24. A 50.0-mL sample of 0.50-M acetic acid, CH3COOH, is titrated with a 0.150M NaOH
solution. Calculate the pH after 25.0 mL of the base have been added (Ka = 1.8 x 10-5).
25. Given calcium hydroxide, Ca(OH)2 . a) What is the molar solubility? ( Ksp = 5.5x10-6)
b) What is the pH of this solution?
4
BONUS QUESTION - (10 points)
Given 20.0 ml , 0.05 M acetic acid ( Ka = 1.8x10-5) titrated with 0.05 M NaOH. Calculate the pH
Of the following solutions.
a)
b)
c)
d)
e)
initial solution
mixture after adding 5.0 ml , 0.05 M NaOH solution
mixture at one-half equivalent point
mixture at equivalent point
mixture after adding 25 ml , 0.05 M NaOH
5
CHEM 1412 Exam # 2A (Chapters 15,16, and 18) ( Answers)
PART - I
1. C
2. C
3. D
4. D
5. C
6. B
7. D
8. A
9. D
10. B
11. B
12. C
13. A
14. B
15. A
16. D
17. D
18. C
19. D
20. B
PART - II
21. [H + ] = 10 - pH = 10 - 6.30 = 5.0 x 10 -7 M
1.0 x 10 -14
[OH
-]
=
= 2.0 x 10 -8 M;
5.0 x 10
pOH = 14 - 6.30 = 7.7
-7
22. [H + ] = 10 - pH = 10 - 2.44 = 3.63 x 10 -3 M
______
+
[H ] =  MaKa ;
23. pka = 4.82 ;
(3.63 x 10 -3) 2
ka =
= 1.32 x 10 -4
(0.10)
PH = 4.82 + log ( 0.050 / 0.10) = 4.82 - 0.30 = 4.52
[H + ] = 10 - pH = 10 - 4.52 = 3.02 x 10 -5 M
[H + ] x 100
% ionization =
( 3.02 x 10 -5 ) ( 100 )
=
Ma
= 3.02 x10 -2 %
or .0302 %
(0.10)
0R,
MaKa
[H
+]
(0.10) (1.5 x 10 -5)
=
=
= 3.0 x 10 - 5 M, % I = 0.0300%
Ms
(0.05)
24.
(0.50)(50.0)
[ H + ] = Ma =
= 0.33 M (MORE ACIDIC), pKa = 4.75
(50.0 + 25.0)
(0.150) (25.0)
[OH
-]
= Mb =
= 0.05 M
(50.0 + 25.0)
Mb
pH =4.75 +log
0.05
= 4.75 + log
Ma - Mb
= 4.75 - 0.748 = 4.00
0.33 - 0.05
6
25. (a) Ca(OH)2 (S)  Ca 2+(aq) +2OH- (aq) ; Ksp=[Ca 2+ ] [OH- ]2 = (X)(2X)2=4X3
X = 3√ 5.5 x 10 -6 /4
= 1.11 x 10 - 2 M
(b)[OH- ]= 2x = 2(1.11x10 -2) = 2.22 x10-2 M;
pOH =1.65; pH = 12.35
BONUS
Kw (1.0 x 10 -14 )
Ka = 1.8 x
10 -5;
Kb =
= 5.56 X 10 -10 ;
pKa = 4.75
10 -5
Ka (1.8 x
)
__________________
 ( 0.05 ) ( 1.8 x 10 -5 ) = 9.49 x 10 -4 M ;
______
+
(a) [H ] =  MaKa =
(b)
PH=
3.02
(0.05) (20.0)
Ma =
= 0.04 M (WEAK ACID )
(20.0 + 5.0)
(0.05) (5.0)
Mb =
= 0.01 M (STRONG BASE)
(20.0 + 5.0)
Mb
0.01
pH = 4.75 + log
= 4.75 + log
Ma - Mb
= 4.75 - 0.477 = 4.27
0.04 - 0.01
(c) At one-half equilibrium, pH = pKa = 3.75
(d)
(0.05) (20.0)
Mb =
= 0.025 M; at equilibrium strong base dominates.
(20.0 + 20.0)
______
________________
[OH ]= MbKb = (0.025) (5.56x10-10) =3.72x10-6 M; pOH=5.42; pH = 8.57
(e)
(0.05)(20.0)
Ma =
= 0.022 M (WEAK ACID)
(20.0 +25.0)
(0.05)(25.0)
Mb =
= 0.028 M (STRONG BASE)
(20.0 + 25.0)
Diff [OH-] = Mb – Ma = 0.028 – 0.022 = 0.006 M
pH=14 + log (0.006) = 14 – 2.22  pH = 11.78
7
Chemistry 1412
EXAM # 2B Sample
8
CHEM 1412 Exam # 2B Sample
(Chapters 16, 17, and 19)
Name: _____________________
Score:
PART I - (3 points each) - Please write your correct answer next to each question
number. DO NOT CIRCLE
_____ 1. What is the conjugate acid of HSO4-?
A. H2SO4–
B. HSO4
C. H2SO4
D. SO42-
_____ 2. The hydronium ion concentration of a glass of beer is 5.0 x 10-5 M. What is the pH of the
beer?
A. 5.70
B. 5.30
C. 3.30
D. 4.30
_____ 3. A sample of orange juice has a hydroxide concentration of 3.5 x 10-11 M. What is the pH?
A. 3.11
B. 3.31
C. 3.54
D. 10.46
_____ 4. A wine has a pH of 3.85. What is the [H3O+]?
A. 7.1 x 10-3
B. 1.4 x 10-3
C. 1.4 x 10-4
D. 7.1 x 10-4
_____ 5. The value of Ka in water at 25oC for hypochlorous acid is 2.95 x 10-8. Calculate the pH of
an aqueous solution with a total concentration of HClO equal to 0.15 M.
A. 9.80
B. 10.80
C. 3.76
D. 4.18
_____ 6. If Na2CO3 solid sodium carbonate is dissociated in pure water, the solution will be…
A. acidic
B. basic
C. neutral
D. cannot be determined
_____ 7. If Ka for HCN is 6.2x10-10, what is Kb for CN- ?
A. 6.2x10-24
B. 6.2x104
C. 1.6x10-5
D. 1.6x1023
_____ 8. What is the pH of a solution of 0.46 M acid and 0.36 M of its conjugate base if the
pKa = 5.51?
A. 4.90
B. 5.40
C. 5.20
D. 5.62
C. H2SO4
D. H2SeO3
_____ 9. Which of the following is the weakest acid?
A. H2SO3
B. H2SeO4
_____10. What is the percent ionization of a 0.010 M HNO2 solution? (Ka = 4.4 x 10-4)
A. 15%
B. 10%
C. 21%
D. 19%
9
_____11. A 20.00 mL sample of 0.100 M CH3CO2H is titrated with 0.100 M NaOH. What is the
pH of the solution at the points where 10.00 mL of NaOH have been added.
(Ka = 1.8 x 10-5)
A. 5.61
B. 4.74
C. 9.26
D. 7.00
_____12. A 20.00 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. What is the pH of
the solution at the points where 20.00 ml of NaOH have been added.
A. 10.70
B. 7.00
C. 10.30
D. 3.30
_____13. You have a saturated solution of Ag3PO4. Its molar solubility (X) follows that:
A. X = 1/3 Ksp
B. X = (Ksp)1/4
C. X = 3 (Ksp)1/4
D. X = (Ksp/27)1/4
_____14. What is the molar solubility of MgC2O4 (Ksp = 8.5 x 10-5) in a 0.020 M C2O42- solution?
A. 1.7 x 10-6
B. 1.7 x 10-4
C. 4.2 x 10-5
D. 4.2 x 10-3
_____15. What is the pH of a saturated solution of Cu(OH)2 (Ksp = 2.6 x 10-19)?
A. 8.82
B. 7.91
C. 7.31
D. 8.42
_____16. Which of the following reactions is associated with the most negative change in
entropy?
A. 4 NH3(g) + 7 O2(g)
C. CH4(g) + 4 Cl2(g)
4 NO2(g) + 6 H2O(g)
CCl4(l) + 4 HCl(g)
B. N2(g) + O2(g)
2 NO(g)
D. CO(g) + 2 H2(g)
CH3OH(l)
_____17. Calculate the entropy change (J/mole.K) of the reaction. The molar entropies [So] are
given in brackets after each substance.
Bi2O3(s) [151.5] + 3 CO(g) [197.7]
A. +35.2
2 Bi(s) [56.7] + 3 CO2(g) [213.7]
B. +9.9
C. +79.5
D. -79.5
_____18. Calculate the pH of a 0.10 M solution of Ca(OH)2 .
A. 0.500
B. 2.50
C. 1.20
D. none of these
_____19. Which of the following equation is correct for the equilibrium constant, K a?
A. Ka = (Kw/Kb) B. Ka = Kw . Kb
C. Ka = 1/Kb
D. Ka = (Kb/Kw)
_____ 20. Which statement below is correct about the third law of thermodynamics?
A.the change of entropy of the universe is always positive
B.the entropy of perfect crystal at zero degree K is zero
C.the entropy of universe is constant
D.every spontaneous reaction is non-spontaneous in opposite direction
10
PART II- ( 8 points each) Please show all your work .
21. The Kp = 9.66 x 10-16 at 25oC for the following reaction. Calculate
(R = 8.314 J/mol.K)
N2(g) + O2(g) ↔ 2 NO(g)
Go (kJ).
22. What is the molar solubility of Zn(OH)2 at pH = 8 ? (Ksp = 2.1 x 10-16)
23. Given
S = -266.3 and the listed [So values] calculate So for Fe3O4(s).
4 Fe3O4(s) [__?__] + O2(g) [205.1]
6 Fe2O3(s) [87.4]
24. What is the pH of a solution of 0.81 M acid and 0.35 M of its conjugate base if the acid
dissociation ionization constant, Ka is 5.45 x 10-8?
25. A 20.0 ml sample of 0.100 M CH3COOH is titrated with 0.100 M NaOH. Calculate the pH of
the solution at points where 15.00 ml of NaOH have been added. (Ka for CH3 COOH =
1.8x10-5)
Bonus Question- (10 points)- show all your work .
Ka for hypochlorous acid, HClO, is 3.0x10-8. Calculate the pH after addition of 15.0 , 20.0 , 40.0 ,
and 50.0 ml of 0.100 M NaOH to 40.0 ml of 0.100 M HClO. Identify the pH at equilibrium and
half equilibrium.
11
CHEM 1412 Exam # 2B (Chapters 15,16, and 18) ( Answers)
PART I
1. C
2. D
3. C
4. C
HSO4 - + H+
H2SO4
-5
pH = - log (3.5x10 ) = 4.30
pOH = -log (3.5x10-11) = 10.46  pH = 14 – pOH = 14 – 10.46 = 3.54
[H3O+] = 10 – pH = 10 –3. 85 = 1.4x10 – 4
5. D
6. B
7. C
8. B
[H+] = √ MaKa = √ (0.15)(2.95x10-8) = 6.65x10 –5 M  pH = -log (6.65x10 –5 ) = 4.18
Na2CO3 (aq) + H2O(l)  2NaOH(aq) (strong base) + H2CO3 (aq) (weak acid)
Kb = ( Kw / Ka) = ( 1.0x10 –14 / 6.2x10 –10) = 1.6x10 –5
Ka = 10 –5.51 = 3.1x10-6  [H+] = ( MaKa / Mb) = ( 0.01x3.1x10-6 / 0.36) = 3.96x10 –6 M
pH = -log (3.95x10 –6)= 5.40
OR: pH = pKa + log [base/acid] = 5.51 + log [0.36/0.46] = 5.51 – 0.11 = 5.40
9. D less oxygen and less electronegativity of central element
10. C [H+] = √ MaKa = √ (0.010)(4.4x10-4) = 2.10x10 –3 M
% i = ([H+] x100/ Ma) = (2.10x10 –3 x 100 / 0.01) ≈ 21%
11. B 10.0 ml is half of 20.0 ml and is at half equilibrium  pH = pKa = - log(1.8x10 –5) = 4.74
12. B strong acid and strong base are at equilibrium point, therefore pH =7.00
13. D Ag3PO4(s)
3 Ag+ (aq) + PO4 3-(aq)
3X
X
+
3
33
Ksp = [Ag ] [ PO4 ] = (3X) (X) = 27 X4  X4 = Ksp/ 27  X = (Ksp/ 27)1/ 4
14. D MgC2O4(s)
Mg2+ (aq) + C2O4 2 -(aq)
X
X + 0.02
Ksp = [Mg2+] [C2O4 2 -] = (X)(X + 0.02) = X(0.02)  X = Ksp / 0.02 = (8.5x10 –5 /0.02)
X = 4.25x10 –3 M
15. B
Cu (OH)2 (s)
4 X3 = 2.6x10 –19
Cu2+(aq) + 2 OH –(aq)
 Ksp = [Cu2+ ] [OH –]2 = (X)(2X)2
X
2X
 X = (Ksp /4) 1/ 3 = (2.6x10 –19 /4) 1/ 3 = 4.02x10 –7 M
[OH-] = 2X = 2(4.02x10 –7) = 8.04x10 –7 M  pOH = -log (8.04x10 –7) = 6.09  pH =7.91
16. D (gases change to liquid)
17. B
S = [3(213.7) + 2(56.7)] – [1(151.5) + 3(197.7)] = + 9.9 J/K
18. D [OH-] = nbMb = (2 OH-)(0.10) = 020 M , pOH = -log(0.20) = 0.70  pH = 14 – 0.70 = 13.3
19. A Ka.Kb = Kw  Ka = Kw / Kb
20. B
12
PART II
21.
22.
Go = - RT ln Kp = - (8.314x10 –3) (25+273) ln (9.66x10 –16) = + 85.7 kJ
pH = 8  pOH = 14 –8 = 6  [OH -] = 10 –pOH = 10 – 6 = 1.0x10 –6 M
Zn(OH)2 (s)
Zn2+(aq) + 2 OH X
1.0x10 -6
Ksp = 2.1x10 –16 = [Zn2+] [OH –]2 = (X) (1.0x10 –6)2  X = [Zn2+] = 2.1x10 –4 M
23.
S = [6 So (Fe2O3) ] – [So (O2) + 4 So ( Fe3O4) ] = [ 6(87.40] – [ (205.10 + 4 So ( Fe3O4)]
-166.3 = 524.4 –205.1 – 4 So (Fe3O4)  So (Fe3O4) = 146.4 J/mol.K
24.
[H+] = MaKa /Ms = (0.81x5.45x10-8 /0.35) = 1.26x10 –7  pH = 6.90
OR
pH = pKa + log [base/acid] = -log(5.45x10-8) + log (0.35/0.81) = 7.26 – 0.36  pH = 6.90
25. Ma = [acid] = (0.1x20/ 20+15) = 0.057 M
Mb = [base] = (0.1x15 /15+20) = 0.043 M
pH = pKa + log [Mb / Ma -Mb] = -log(1.8x10-5) + log ( 0.043 / 0.057-0.043) = 4.74 + 0.48
 pH = 5.22
BONUS
pKa = -log Ka = - log(3.0x10 –8) = 7.52
(a) 15 ml
Ma = [acid] = (0.1x40/ 40+15) = 0.0727 M
Mb = [base] = (0.1x15 /15+40) = 0.0273 M
pH = pKa + log [Mb / Ma -Mb] = 7.52 + log ( 0.0273 / 0.0727-0.0273) = 7.52 - 0.221 = 7.30
(b) 20.0 ml (half of 40 ml = ½ equilibrium point)  pH = pKa = 7.52
(c) 40 ml (at equilibrium point) – strong base is dominating the solution.
Ma = [base] = (0.x40/ 40+40) = 0.05 M
[OH -] = √ MbKb = √ MbKw/Ka = √ 0.05x 1.0x10 –14 /3.0x10 –8
= 1.29x10 –4 M
–4
pOH = -log [1.29x10 ] = 3.89  pH = 14 –pOH = 14 – 3.89  pH = 10.11
(d) 50 ml – solution is basic
Ma = [acid] = (0.1x40/ 40+50) = 0.0444 M
Mb = [base] = (0.1x50 /50+40) = 0.0556 M (more basis)
[OH -] = Mb - Ma = 0.0556 – 0.0444 = 0.112 M  pOH = -log (0.0112) = 1.95  pH = 14 – pOH
pH = 14 – 1.95  pH = 12.05
13