Honors Chemistry
Name _____________________________________
Chapter 14/15: Acids/Bases Review Worksheet II
Date _____/_____/_____
Period _____
1. What volume of 0.050 M NaOH(aq) will exactly neutralize 100.0 mL of 0.075 M HBr solution?
a. 100.0 mL
c. 6.6 x 10-3 mL
e. 120 mL
b. 7.5 mL
d. 150 mL
NaOH + HBr  NaBr + HOH
100.0 mL
1L
1000 mL
0.075 mole HBr
1L
1 mole NaOH
1 mole HBr
1L
0.050 mole NaOH
1000 mL
1L
= 150 mL
2. What volume of 0.200 M H2SO4(aq) would exactly neutralize 2.00 L of 1.0 x 10-3 M Al(OH)3?
a. 6.7 mL
b. 10 mL
c. 15 mL
d. 1.00 L
e. 75 mL
3 H2SO4 + 2 Al(OH)3  6 HOH + Al2(SO4)3
2.00 L
1.0 x 10-3 mole Al(OH)3
1L
3 mole H2SO4
2 mole Al(OH)3
1 LH2SO4
0.200 mole H2SO4
1000 mL
1L
= 15 mL
3. What is the hydroxide concentration in a solution which results from pouring 100 mL of 0.010 M HCl
together with 200 mL of 0.030M Ca(OH)2(aq) solution? Assume that the liquid volumes are additive.
a. 0.020 M
c. 0.0367 M
e. 0.030 M
b. 0.0183 M
d. 0.060 M
2 HCl + Ca(OH)2  2 HOH + CaCl2
0.1 L HCl
0.010 mole HCl
1L
0.2 L Ca(OH)2
1 mole H+1
1 mole HCl
2 mole (OH)-1
1 mole Ca(OH)2
0.030 Ca(OH)2
1L
= 0.001 mole H+1
= 0.012 mole (OH)-1
0.012 mole (OH)-1
- 0.001 mole H+1
0.011 mole (OH)-1
0.011 mole (OH)-1 = 0.0367 M
0.3 L
4. Write the conjugate acid for each of the following:
a. HPO4-2
b. NO2-1
c. HS-1
H2PO4-1
HNO2
H2S
d. S-2
HS-1
5. In standardizing an HCl solution of unknown concentration, 21.50 mL of HCl solution are found to
reach the endpoint in a titration with 4.086 g of Na2CO3. (MM = 106.0 g/mol) What is the molarity of
the HCl solution?
a. 1.79 M
b. 3.58 M
c. 0.896 M
d. 3.6 x 10-3 M
2 HCl + Na2CO3  2 NaCl + H2CO3
4.086 g Na2CO3
1 mole Na2CO3
106.0 g Na2CO3
2 mole HCl
1 mole Na2CO3
= 0.07709 mole Na2CO3
0.0215 L
= 3.58 M HCl
6. What is the equivalent weight of Ca(OH)2?
a. 24.7 g/eq
b. 74.1 g/eq
74.1 g Ca(OH)2
1 mole
c. 148.2 g/eq
1 mole
2 eq
= 37 g/eq
7. How many equivalents are in 1 mole of H3PO4?
a. 0.33
b. 1
c. 3
1 mole H3PO4
d. 37 g/eq
d. 98
3 eq
1 mole
e. 33
= 3 eq
8. How many equivalents are in 1.00 gram of H2SO4? (MM = 98 g/mol)
a. 0.020
b. 0.010
c. 0.0051
1.00 g H2SO4
1 mole
98 g
9. What is the normality of 12.0 M H2SO4?
a. 3.0 N
b. 6.0 N
2 eq
1 mole
10. What is the molarity of 1.0 N HI?
a. 7.8 x 10-3
b. 128
c. 1.0
N = #eq X M
M= N/eq
e. 49
= 0.020 eq
c. 12.0 N
N = 2 X 12.0 M 
N = #eq X M
d. 98
d. 4.0 N
e. 24.0 N
N = 24.0 N
d. 2.0
M= 1.0N/1eq 
M = 1.0M
11. Suppose 9.8 grams of H2SO4 (MW = 98 g/mol) are dissolved in 250.0 mL of aqueous solution. What
is the Normality of H2SO4 in the solution?
a. 19.6 N
b. 0.40 N
c. 0.20 N
d. 4.9 N
e. 0.80 N
9.8 g H2SO4
1 mole H2SO4
98 g H2SO4
= 0.10 mole H2SO4
0.250 L
= 0.40 M x 2 eq = 0.80 N H2SO4
12. What volume of 3.0 N H3PO4 would exactly neutralize 250.0 mL of aqueous 2.0 N NaOH?
a. 375 mL
b. 166.7 mL
c. 250.0 mL
d. 45 mL
H3PO4 + 3 NaOH  3 HOH + Na3PO4
H3PO4 = 3.0 N/ 3 eq= 1.0 M
250.0 mL
1L
2.0 mole NaOH
1000 mL
1L
1 mole
H3PO4
3 mole
NaOH
1L
1000 mL
1.0 mole
H3PO4
1L
13. What is the hydronium ion concentration in a solution which is 0.10 M HNO3(aq)?
a. 0.30 M
c. 6.7 x 10-3 M
e. 2.1 x 10-2 M
-7
b. 0.10 M
d. 1.0 x 10 M
0.10 mole HNO3
1L
1 mole H3O+
1 mole HNO3
= 0.10 M H3O+
= 166.7 mL
H3PO4
14. What is the hydroxide ion concentration in 1.0 M HBr?
a. 1.0 M
b. 1.0 x 10-13 M
c. 1.0 x 10-14 M
Kw= [H3O+][OH-]
[OH-] = 1.0x10-14 / 1.0  1.0x 10-14
15. What is the pH of 1.0 x 10-3 M aqueous HClO4?
a. 10-7
b. -3.0
c. 0.0
pH = -log [H3O+]
d. 1.0 x 10-7 M
d. 3.0
e. 1.0
pH = -log[1.0x10-3]  pH = 3.0
16. What is the hydronium ion concentration in a solution which has [OH-1] = 1.0 x 10- 2 M?
a. 1.0 x 10-12 M
c. 1.0 x 10-16 M
e. 1.0 x 10-7 M
-2
b. 1.0 x 10 M
d. 12
Kw= [H3O+][OH-]
[H3O+]= 1.0x10-14 / 1.0x10-2 1.0x10-12
17. Identify the INCORRECT statement:
a. As the pH increases the hydroxide ion decreases.
b. As the pH increases the hydronium ion concentration decreases.
c. As the pH increases the Kw of water remains the same.
d. As the pH increases the product [H3O+][OH-] remains constant.
e. As the pH increases the solution becomes less acidic and more alkaline.
18. Write the conjugate base for each of the following:
a. H2S
HSc. NH3
-1
b. HS
S-2
d. H2SO3
NH2HSO3-
19. Which acid is the weakest acid? Refer to the table 6 on p 485 of your textbook.
a. H2S
c. HC2H3O2
e. HF
b. HNO3
d. HClO
20. What is the Ka of HNO2(aq)? Refer to the table of acids given.
a. -0.525
c. 0.525
e. 1.8 x 10-5
b. 3.35
d. 4.5 x 10-4
21. What is the pH of a 0.10 M solution of HF in water? The Ka = 7.2 x 10- 4.
a. -2.07
c. 0.10
e. 2.1
b. 8.5 x 10-3
d. 3.14
HF + H2O  H3O+ + F0.10 M
0
0
0.10-x
x
x
K= [P]/[R]  7.2x10-4 = x2/0.10
x = 0.0085 M
pH = -log[0.0085]  pH= 2.1
22. What is the percent ionization of the acid in Problem #21?
a. 100%
c. 0.11%
b. 8.5%
d. 0.0011%
e. 0.08%
7.2x10-4/0.0085 = 0.085 x 100% = 8.5%
23. What is the pH of a 0.20 M aqueous solution of the base methylamine, CH3NH2? Kb = 5.0 x 10-4.
a. 2.0
b. 12.0
c. 16.0
d. 10.0
e. 0.01
CH3NH2 + H2O  CH3NH3+ + OH0.20 M
0
0
0.20 – x
x
x
K= [P]/[R]  5.0x10-4 = x2/0.20
x = 0.010 M
pH = -log[0.010]  pOH = 2.00  pH + pOH = 14 pH = 14 – 2 = 12.00
24. Using your knowledge of the Brønsted-Lowry theory of acids and bases, write equations for the
following acid-base reactions and indicate each conjugate acid-base pair:
a. HNO3 + (OH)-1  HOH(l) + NO-(aq)
A
B
CA
CB
b. CH3NH2 + H2O  HCH3NH2-(aq) + OH-(aq)
B
A
CA
CB
c. (OH)-1 + HPO4-2  HOH(l) + PO4-3(aq)
B
A
CA
CB
25. The compound NaOH is a base by both of the theories we discussed in class. However, both of the
theories describe what a base is in different terms. Use your knowledge of these theories to
describe NaOH as an Arrhenius base and a Brønsted-Lowry base.
NaOH is an Arrhenius base because it produces an (OH)-1 ion when dissolved in solution.
NaOH is a Brønsted-Lowry base because it is a proton acceptor. The (OH)-1 portion of the
NaOH reacts to remove a hydrogen ion from another compound.
26. When hydrogen chloride reacts with ammonia, ammonium chloride is formed. Write the
equation for this process, and indicate which of the reagents is the acid and which is the
base.
HCl(aq) + NH3(aq)  (NH4)Cl(aq) NH3 is a Lewis base because it uses its lone pair electrons
to pull a hydrogen atom from hydrochloric acid. HCl is a Lewis acid because it accepts
electrons from NH3 when the H is transferred.
27. Write an equation for the reaction each of the following reactions and indicate what each
reaction produces:
a. Carbon dioxide and water
CO2(g) + H2O()  H2CO3(aq)
b. Sodium oxide and water
Na2O(s) + H2O()  2 NaOH(aq)
c. Dichlorine trioxide and water
Cl2O3(g) + H2O()  2 HClO2(aq)
d. Beryllium oxide and water
BeO(s) + H2O()  Be(OH)2(aq)
28. Write the names for the following acids and bases:
a.
b.
c.
d.
e.
KOH _____potassium hydroxide_____________________
H2S ______hydrosulfuric acid_______________________
HC2H3O2 ____acetic acid___________________________
Fe(OH)2 _____iron (II) hydroxide_____________________
HCN _____hydrocyanic acid________________________
29. Calculate the pH, pOH and [OH-1] of a solution in which the [H+1] = 1.2 x 10-3. Calculate each
significantly!
pH = -log [H+]  pH = -log[1.2x10-3] 
pH = 2.92
pH + pOH = 14  pOH = 14 – 2.92 =
11.08 pOH
kw = ka x kb  kb = 1.0x10-14 / 1.2x10*3 
kb = [OH-] = 8.33x10-12 or: [OH-] = 10-11.08 = 8.33x10-12 M
30. Calculate the pH, pOH, and [H+1] of a solution in which the [OH-1] = 2.34 x 10-5. Calculate each
significantly!
pOH = -log[2.34 x 10-5]  pOH = 4.631
pH = 14 - 4.631
pH = 9.369
kw = ka x kb  ka = 1.0x10-14 / 2.34 x 10-5
ka = [H+] = 4.27x10-10 or: [H+] = 10-9.369 = 4.27x10-10 M
31. What is the pH and pOH of a solution that was made by adding 400 mL of water to 350 mL
of 5.0 x 10-3 M NaOH solution?
NaOH Na+ + OH(5.0x10 M)(0.350L) = 0.0017 mole NaOH
0.0017 mole NaOH x 1 mole OH-/1 mole NaOH = 0.0017 mole OH0.0017 mole OH- / 0.750 L = 0.0023 M OHpOH = -log[0.0023] 2.63
pH = 14 – 2.63 = 11.37
-3
32. What is the pH and pOH of a solution with a volume of 5.4 L that contains 15 grams of hydrochloric
acid and 25 grams of nitric acid?
15 g HCl
25 g HNO3
1 mole
36.46 g
1 mole
63.02 g
= 0.41 mole HCl
= 0.40 mole HNO3
0.41 + 0.40 = 0.81 mole total / 5.4 L = 0.15 M H+
pH = -log[0.15] 
pH = 0.82
pH + pOH = 14  14 - 0.82 
pOH = 13.18
33. What is an electrolyte? Which of the following will produce stronger electrolyte solutions, strong
acids and bases or weak acids and weak bases? If one was comparing two strong acids or two
strong bases, how would they determine which solution would produce a stronger electrolyte
solution?
An electrolyte is a substance that produces a large number of ions when dissolved in
solution. Strong acids and bases would be stronger electrolytes because they produce
more ions in solution than weak acids and bases. To compare two strong acids or two
strong bases, one would compare the Ka and Kb values to determine which would be a
stronger electrolyte. The larger the K value the stronger the electrolyte.
34. Write the dissociation reactions for the polyprotic acid, Arsenic acid (H3AsO4). Using the following k
values, label the k values for each reaction: (ka1 = 5 x 10-3, ka2 = 8 x 10-8, ka3 = 6 x 10-10)
H3AsO4(aq)  H2AsO4-1(aq) + H3O+1(aq)
H2AsO4-1(aq)  HAsO4-2(aq) + H3O+1(aq)
HAsO4-2(aq)  AsO4-3(aq) + H3O+1(aq)
ka1 = 5 x 10-3
ka2 = 8 x 10-8
ka3 = 6 x 10-10
35. Using the following Ka values, place the following acids in order of increasing acid strength.
HClO4
CH3OOH
ka = 1 x 107
ka = 1.76 x 10-5
HCN < CH3OOH < HF < HClO4
HCN
HF
Ka = 4.93 x 10-10
Ka = 3.53 x 10-4
36. Using the Ka values from above, calculate the Kb values and place the following bases in order of
increasing strength. (Kw = Ka x Kb)
Kb = Kw/Ka
CN-1, F-1, CH3OO-1, ClO4-1
CN- : kb = 1.0x10-14 /4.93x10-10  kb = 2.0x10-5
F- : kb = 1.0x10-14 / 3.53x10.4  Kb = 2.8x10.11
CH3OO- : kb = 1.0x10-14 / 1.76x10-5  kb = 5.7x10=10
ClO4- : kb = 1.0x10-14 / 1x107  kb = 1x10-21
ClO4-1 <F-1 < CH3OO-1 < CN-1
37. Draw titration curves for the following titrations and label the x-axis, y-axis, and equivalence point:
a. 0.100 M NaOH is titrated into a solution of 0.200 M HNO3.
b. 1.0 M HCl is titrated into a solution of 0.50 M NaOH.
38. Indicate whether the following salts will produce an acidic, basic, or neutral solution when dissolved
in water.
a. NaNO3
SB-SA  WCA-WCB  neutral
b. NH4ClO4 WB-SA  SCA-WCB  acidic
c. NaCO3
SB-WA  WCA-SCB  basic
39. Which of the following acids have relatively weak conjugate bases?
a. HF
b. HCl
c. H2SO4
d. HC2H3O2
40. Which of the following will have relatively strong conjugate acids?
a. SO4-2
b. Br-1
c. CN-1
d. C2H3O2-1