Chemistry 1412
EXAM # 2A Sample
1
CHEM 1412 Exam #2A Sample
(Chapters 13-15)
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.
Consider the following equilibrium:
PCl3(g) + Cl2(g)
PCl5(g
H = –92 kJ
The concentration of PCl3 at equilibrium may be increased by
a.
b.
c.
d.
increasing the pressure.
adding Cl2 to the system.
decreasing the temperature.
the addition of PCl5.
_____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 correctly describes the equilibrium constant for the gas-phase
reaction between H2 and O2 to form gaseous H2O?
A)
B)
Kc =
[H 2 O]
[H 2 ][O 2 ]
[H 2 O]2
Kc =
[H 2 ][O 2 ]
C)
[H 2O]2
Kc =
[H 2 ]2 [O2 ]
D)
[H 2 ][O 2 ]
Kc =
[H 2 O]
2
_____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. For the system CaO(s) + CO2(g)
A. [CO2]
B. 1 / [CO2]
 CaCO3(s) the equilibrium constant expression is
C. [CaO] [CO2] / [CaCO3]
D. [CaCO3] / [CaO] [CO2]
_____ 9. The value of Kc for the reaction C(s) + CO2(g)  2CO(g) is 1.6. What is the equilibrium
concentration of CO if the equilibrium concentration of CO2 is 0.50 M?
A. 0.31
B. 0.80
C. 0.89
D.
0.75
____ 10. Consider the reaction 2HI(g)  H2(g) + I2(g). What is the value of the equilibrium
constant, Kc, if at
equilibrium , [H2] = 6.50 x 10-7 M, [I2] = 1.06 x 10-5 M, and [HI] = 1.87 x 10-5 M?
A. 3.68 x 10-7
B. 1.97 x 10-2
C. 1.29 x 10-16
D. 50.8
____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. Rank H3PO4, H2PO4–, and HPO42– in order of increasing acid strength.
A) H3PO4 < H2PO4– < HPO42–
B) H2PO4– < HPO42– < H3PO4
C) HPO42– < H2PO4– < H3PO4
D) H2PO4– < H3PO4 < HPO42–
____ 14. What is the pH of a solution prepared by dissolving 0.172 L of HCl(g), measured at STP, in enough
water such that the total volume of the solution is 9.00 L? (R = 0.0821 L · atm/(K · mol)
3
A)
B)
C)
D)
11.885
3.069
10.931
7.000
____ 15. What will happen if a small amount of sodium hydroxide is added to a 0.1 M solution of ammonia?
A)
B)
C)
D)
Kb for ammonia will increase.
Kb for ammonia will decrease.
The percent ionization of ammonia will increase.
The percent ionization of ammonia will decrease.
____16. Hypobromous acid, HOBr, has an acid-ionization constant of 2.5 × 10-9 at 25ºC. What is the
hydronium-ion concentration in a 0.64 M HOBr solution?
A) 1.6 × 10-9 M
B) 3.9 × 10-8 M
C) 4.0 × 10-5 M
D) 5.0 × 10-4 M
____ 17. For which of the following equilibria does Kc correspond to an acid-ionization constant, Ka?
A) NH3(aq) + H3O+(aq)
B) NH4+(aq) + H2O(l)
C) NH4+(aq) + OH–(aq)
–
D) HF(aq) + OH (aq)
NH4+(aq) + H2O(l)
NH3(aq) + H3O+(aq)
NH3(aq) + H2O(l)
H2O(l) + F–(aq)
____18. A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration was titrated with 0.145 M
KOH. The equivalence point was reached when 41.54 mL of base had been added. What is the concentration of
the propionate ion at the equivalence point?
A)
B)
C)
D)
0.145 M
0.0905 M
0.241 M
0.128 M
____19. Which of the following statements is incorrect?
A)
B)
C)
D)
One reason why HCl is a stronger acid than HF is that Cl has a larger atomic radius than F.
One reason why HCl is a stronger acid than HF is that the H–Cl bond is weaker than the H–F bond.
One reason why HCl is a stronger acid than HF is that Cl is more electronegative than F.
The acids HBr and HI both appear equally strong in water.
4
____20. A sample of ammonia (Kb = 1.8 x 10–5) is titrated with 0.1 M HCl. At the equivalence point, what is
the approximate pH of the solution?
A)
B)
C)
D)
1
5
7
9
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. A mixture of 0.100 mol of NO, 0.0500 mol of H2, and 0.100 mol of H2O is placed in a 1.00-L vessel. The
following equilibrium is established:
2NO(g) + 2H2 (g)  N2(g) + 2H2O(g)
At equilibrium [NO] = 0.0620 M. Calculate the equilibrium concentrations of H2, N2, and H2O
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?
BONUS QUESTION - (10 points)
5
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
6
CHEM 1412 Exam # 2A (Chapters 13-15) ( Answers)
PART - I
1. C
4. D
7. D
[H2] [I2]
(6.50x10-7)(1.06x10-5)
Kc = ------------- = ------------------------------- = 1.97x10-2
[HI]
(1.87x10-5)2
10. B
2. C
3. D
13. C
16.C
19.C
5. C
8. B
11. B
14.B
17. B
20. B
6. C
9. C for reverse reaction n = - 1  K′ = (Kc)n = (Kc)-1 = ( 1.52)-1 = (1/1.52) = 0.658
12. C
15. D
18. 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.
[NO] = 0.100 mol/1L = 0.100 M , [H2O] = 0.100 mol/1L = 0.100 M , [H2] = 0.050 mol/1L =
0.050 M
2 NO + 2 H2
 N2
(0.100-2x) (0.050-2x)
+x
+
2 H2O
(0.100+2x)
0.100 –2x = 0.0620  2x = -0.062 – 0.100 = 0.038  x = 0.019 M
[H2] = 0.050 – 2x = 0.050 – 2(0.019) = 0.012 M , [H2O] = 0.100 + 2x = 0.100 + 2(0.0190 = 0.138 M
[N2] = x = 0.019 M
23. pka = 4.82 ;
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)
=
Ms
24.
= 3.0 x 10 - 5 M, % I = 0.0300%
(0.05)
(0.50)(50.0)
[ H + ] = Ma =
= 0.33 M (MORE ACIDIC), pKa = 4.75
(50.0 + 25.0)
7
(0.150) (25.0)
[OH
-]
= Mb =
= 0.05 M
(50.0 + 25.0)
Mb
0.05
pH =4.75 +log
= 4.75 + log
Ma - Mb
= 4.75 - 0.748 = 4.00
0.33 - 0.05
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)
8
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
Chemistry 1412
EXAM # 2B Sample
9
CHEM 1412 Exam # 2B Sample
(Chapters 13 -15)
Name: _____________________
Score:
PART I - (3.5 points each) - Please write your correct answer next to each question
number. DO NOT CIRCLE
_____ 1. For the hypothetical reaction A + 3 B  2C, the rate of appearance of C,  [C]/  t
may also be expressed as
a)  [C]/  t = - [A] /t
c)  [C]/  t = -2/3 [B]/ t
b)  [C]/  t = - 3/2 [B] /t
d)  [C]/  t = -1/2 [A]/ t
_____ 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
10
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. For the reaction CaCO3(s)
CaO(s) + CO2 (g) , increasing the pressure on the
system at equilibrium causes
a) increased amount of CaCO3 and CaO
b) decreased the amount of CaO and CO2
c) increased the amount of CO2 and CaCO3
d) increased the amount of CaCO3 and CO2
_____ 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
_____ 9. Equilibrium is reached in all reversible reaction when the
a) forward reaction stops
b) reversed reaction stops
c) concentrations of reactants and the products become equal
d) rates of the opposing reactions become equal
_____10. The value of Kc for the following reaction is 1.60. C(s) + CO2(g)  2 CO(g)
What is the equilibrium concentration of CO if the equilibrium concentration of
CO2 is 0.50 M?
a) 0.79
b) 0.40
c) 0.894
d) 2.24
_____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. Phosgene, COCl2 , a poisonous gas decomposes according to the following equation;
COCl2 (g)  CO(g) + Cl2 (g)
11
If Kc = 0.083 at 900 oC, what is the value of Kp?
a) 0.125
b) 8.0
c) 6.1
d) 0.16
_____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. Consider the two gaseous equilibria;
SO2(g) + ½ O2 (g) SO3 (g)
2SO3 (g)  2 SO2 (g) + O2 (g)
, K1
, K2
The value of the equilibrium constant s are related by
a) K2 = K1
b) K2 = (K1) -1
c) K2 = (K1) -2
d) K2 = (K1)2
_____15. Consider the reaction N2g) + O2(g)
2 NO(g) , Kc = 0.10 at 200 oC.
Starting with initial concentration of 0.04 mol/L of N2 and 0.040 mol/L of O2, calculate
the equilibrium concentration on NO in mol/L.
a) 5.4x10-3
b) 0.0096
c) 0.013
d) 1.6x10-4
_____16. 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
_____17. Which of the following equation is correct for the equilibrium constant, Ka?
A. Ka = (Kw/Kb) B. Ka = Kw . Kb
C. Ka = 1/Kb
D. Ka = (Kb/Kw)
PART II- ( 8 points each) Please show all your work .
21. What is the molar solubility of Zn(OH)2 at pH = 8 ? (Ksp = 2.1 x 10-16)
22. 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?
12
23. The equilibrium constant for the following reaction is 10.5 at 500 K. A system at equilibrium
has [CO] = 0.250 M and [H2] = 0.120 M. What is the concentration of [CH3OH]?
CO (g) + 2 H2 (g)  CH3OH (g)
24. What is the equilibrium concentration of H2C2O4 in a 0.220 M oxalic acid, H2C2O4, solution?
For oxalic acid, Ka1 = 5.6 10–2 and Ka2 = 5.1 10–5. (hint: quadratic formula)
25. HCl(g) initially at a partial pressure of 0.445 atm; is reacting with I2(s) ;
2 HCl(g) + I2 (s)  2 HI (g) + Cl2 (g) , Keq = 3.9x10 -33 , at 25 0C .
Calculate the final partial pressures at equilibrium.
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.
CHEM 1412 Exam # 2B (Chapters 13-15) ( Answers)
PART I
1. C  [C]/  t = -1/3 [B]/ t = - ½ [A]/ t   [C]/  t = -2/3 [B]/ t
2. D pH = - log (3.5x10-5) = 4.30
3. C pOH = -log (3.5x10-11) = 10.46  pH = 14 – pOH = 14 – 10.46 = 3.54
4. C [H3O+] = 10 – pH = 10 –3. 85 = 1.4x10 – 4
5. D [H+] = √ MaKa = √ (0.15)(2.95x10-8) = 6.65x10 –5 M  pH = -log (6.65x10 –5 ) = 4.18
6. B reaction will shift to left, decreases the amounts of CO2 and CaO
7. C Kb = ( Kw / Ka) = ( 1.0x10 –14 / 6.2x10 –10) = 1.6x10 –5
8. B 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
13
OR: pH = pKa + log [base/acid] = 5.51 + log [0.36/0.46] = 5.51 – 0.11 = 5.40
9. D
10. C
Kc = [CO]2 / [CO2]  1.60 = [CO]2 / 0.50  [CO]2 = 0.80  [CO] = 0.894 M
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
Δ n = 2-1 = 1 , T = 900 + 273 = 1173 K
Kp = Kc (RT) Δ n = (0.083)(0.0821x 1173)1 = 8.0
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. C
Reverse and double K2 = ( K1)-2
15. C Kc = [NO]2 / [N2][O2]  0.1 = [NO]2/ (0.04)(0.040)  solve for [NO] = 0.013 mol/L
[OH-] = 2X = 2(4.02x10 –7) = 8.04x10 –7 M  pOH = -log (8.04x10 –7) = 6.09  pH =7.91
16. D [OH-] = nbMb = (2 OH-)(0.10) = 020 M , pOH = -log(0.20) = 0.70  pH = 14 – 0.70 = 13.3
17. A Ka.Kb = Kw  Ka = Kw / Kb
PART II
21
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
–16
2+
Ksp = 2.1x10
= [Zn ] [OH –]2 = (X) (1.0x10 –6)2  X = [Zn2+] = 2.1x10 –4 M
22.
[CH3OH]
Keq =
[CO][ H2]2
[CH3OH]
 10.5 =
 [CH3OH] = 0.0373 M
(
0.250)(0.120)2
23. 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
24. 1.3  10–1 M
14
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
pOH = -log [1.29x10 –4] = 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
15