Section 10.3: Acid–Base Stoichiometry
Tutorial 1 Questions, page 481
1. It is necessary to keep the volume of indicator used to a minimum because many acid–base
indicators are weak acids. Some of the base used in a titration reacts with the indicator.
2. It is necessary to rinse with titrant after rinsing with water so that the inner walls of the burette
are lined with titrant rather than with water. If water were present, it could dilute the titrant.
3. The air bubble occupies a certain volume in the burette. The presence of an air bubble would
make the volume of titrant used seem larger than it actually is.
4. Rinsing the sides of a titration flask near the end of the titration washes any unreacted titrant
into the bottom of the flask.
5. The solution turned pink momentarily because the sodium hydroxide solution was briefly
concentrated in one spot. Then, as the flask is swirled, the sodium hydroxide spreads throughout
the flask where it reacts with excess acid, changing the indicator to colourless.
Tutorial 2 Practice, page 484
1. (a) Given: three volumes of acid
Required: average volume of acid used, VHNO
3
Solution: Calculate the average of titrant (acid) used by subtracting the final from the initial
burette volume for each trial.
8.00 mL + 8.06 mL + 8.12 mL
VHNO (average) =
3
3
VHNO (average) = 8.06 mL
3
Statement: The average volume of nitric acid used is 8.06 mL.
(b) Given: concentration of acid, cHNO = 0.500 mol/L
3
volume of base, VNaOH = 25.00 mL
average volume of acid, VHNO
3 (average)
= 8.06 mL
Required: amount concentration of base, cNaOH
Analysis: c =
n
V
Solution:
Step 1. Convert the volume of solution to litres.
1L
VHNO (average) = 8.06 mL !
3
1000 mL
VHNO
3 (average)
= 8.06 ! 10 "3 L
VNaOH = 25.00 mL !
VNaOH = 2.500 ! 10
"2
1L
1000 mL
L
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Chapter 10: Acids and Bases
10.3-1
Step 2. Write a balanced equation for the reaction listing the given values.
NaOH(aq)
+
HNO3(aq)
→
NaNO3
+
H2O
2.5 × 10–2 L
8.06 × 10–3 L
cNaOH
0.500 mol/L
Step 3. Use the concentration equation to determine the amount of acid.
nHNO = cHNO VHNO (average)
3
3
=
3
0.500 mol
! 8.06 ! 10 "3 L
1L
nHNO = 4.03 ! 10 "3 mol
3
Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the
amount of base.
1 mol NaOH
nNaOH = 4.03 ! 10 "3 mol HNO !
3
1 mol HNO
3
"3
nNaOH = 4.03 ! 10 mol NaOH
Step 5. Use the concentration equation to determine the concentration of base.
n
cNaOH = NaOH
VNaOH
=
4.03 ! 10 "3 mol
2.500 ! 10 "2 L
mol
cNaOH = 0.161
L
Statement: The amount concentration of the sodium hydroxide solution is 0.161 mol/L.
2. Given: mass of potassium carbonate, mK CO = 1.00 g
2
3
volume of acid, VHNO = 24.20 mL
3
Required: amount concentration of acid, cHNO
Analysis: c =
3
n
V
Solution:
Step 1. Convert the volume of titrant used to litres.
1L
VHNO = 24.20 mL !
3
1000 mL
VHNO = 2.420 ! 10 "2 L
3
Step 2. Write the balanced equation listing the given values.
K2CO3(s)
+ 2 HNO3(aq)
→ 2 KNO3 + H2CO3
–2
1.00 g
2.420 × 10 L
138.21 g/mol
cHNO
3
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Chapter 10: Acids and Bases
10.3-2
Step 3. Convert the mass of potassium carbonate to the amount.
1 mol
nK CO = 1.00 g !
2
3
138.21 g
nK CO = 7.2354 ! 10 "3 mol [2 extra digits carried]
2
3
Step 4. Use the amount and the mole ratio in the balanced equation to determine the amount of
the acid.
2 mol HNO
3
nHNO = 7.2354 ! 10 "3 mol K CO !
3
2
3
1 mol K CO
2
3
nHNO = 1.4471 ! 10 "2 mol HNO [2 extra digits carried]
3
3
Step 5. Use the concentration equation to determine the concentration of acid.
nHNO
3
cHNO =
3
VHNO
3
=
1.4471 ! 10
"2
mol
"2
2.420 ! 10 L
mol
cHNO = 0.598
3
L
Statement: The amount concentration of nitric acid is 0.598 mol/L.
3. Given: concentration of acid, cHCl = 0.100 mol/L
volume of acid, VHCl = 15.52 mL
volume of base, VBa(OH) = 25.00 mL
2
Required: amount concentration of base, cBa(OH)
Analysis: c =
2
n
V
Solution:
Step 1. Convert the volume of the solutions to litres.
1L
VHCl = 15.52 mL !
1000 mL
VHCl = 1.552 ! 10 "2 L
VBa(OH) = 25.00 mL !
2
VBa(OH) = 2.500 ! 10
2
"2
1L
1000 mL
L
Copyright © 2011 Nelson Education Ltd.
Chapter 10: Acids and Bases
10.3-3
Step 2. Write a balanced equation for the reaction listing the given values.
2 HCl(aq)
+
Ba(OH)2(aq) →
BaCl2
+
2 H2O
1.552 × 10–2 L
2.500 × 10–2 L
0.100 mol/L
cBa(OH)
2
Step 3. Use the concentration equation to determine the amount of acid.
nHCl = cHClVHCl
0.100 mol
! 1.552 ! 10 "2 L
1L
=
"3
nHCl = 1.552 ! 10 mol
Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the
amount of base.
1 mol Ba(OH)
2
nBa(OH) = 1.552 ! 10 "3 mol HCl !
2
2 mol HCl
nBa(OH) = 7.760 ! 10 "4 mol Ba(OH)
2
2
Step 5. Use the concentration equation to determine the concentration of base.
nBa(OH)
2
cBa(OH) =
2
VBa(OH)
2
=
"4
7.760 ! 10 mol
2.500 ! 10 "2 L
mol
cBa(OH) = 3.10 ! 10 "2
2
L
Statement: The amount concentration of the barium hydroxide solution is 3.10 × 10–2 mol/L.
Section 10.3 Questions, page 485
1. The purpose of an acid–base indicator in a titration is to signify when the endpoint of the
titration is reached.
2. The equivalence point is the point during a titration when neutralization is complete. The
endpoint of a titration is a sudden change in a property of the solution such as a change in pH or
conductivity. Ideally, the endpoint occurs at a pH very close to that of the equivalence point.
3. (a) Given: mass of sodium hydroxide, mNaOH = 4.00 g
volume of solution, VNaOH = 100.0 mL
Required: amount concentration of base, cNaOH
Analysis: cNaOH =
nNaOH
VNaOH
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Chapter 10: Acids and Bases
10.3-4
Solution:
Step 1. Convert the volume of solution to litres.
1L
VNaOH = 100.0 mL !
1000 mL
VNaOH = 0.1000 L
Step 2. Convert mass of the sodium hydroxide to the amount.
1 mol
nNaOH = 4.00 g !
40.00 g
nNaOH = 0.100 mol
Step 3. Use the concentration equation to determine the concentration of the solution.
n
cNaOH = NaOH
VNaOH
0.100 mol
0.100 L
mol
cNaOH = 1.00
L
Statement: The amount concentration of sodium hydroxide solution is 1.00 mol/L.
(b) The actual concentration differs from the predicted concentration because, over time,
contaminants may enter the solution and react with sodium hydroxide. Carbon dioxide, for
example, is known to react with sodium hydroxide to form sodium carbonate. This decreases the
amount of sodium hydroxide present. Assuming the volume of solution remains constant, carbon
dioxide contamination decreases the concentration of the solution.
4. It is necessary to re-standardize a solution after it has been stored for some time because the
concentration of the solution may vary over time. This may be caused by several factors
including reactions with contaminants and changes in the volume of the solution.
5. Storage bottles of solutions are often filled right to the top to eliminate the amount of air that
would otherwise occupy the space above the solution. Components in air, such as oxygen and
carbon dioxide, may react with the solution.
6. Drying the samples in an oven before using them to standardize a base is necessary to remove
any water that may have been absorbed by the solid. If the mass were not dried, the mass of a
sample contaminated with water would be larger than a dried sample. As a result, the mass of
potassium hydrogen phthalate recorded would be larger than it actually is, and the calculated
amount and concentration of base would be incorrect.
7. A pH meter can be used to monitor the pH changes during a titration. Near the equivalence
point, a sharp rise or drop should be observed. For an acid–base indicator to be suitable for a
titration, the colour change should occur when the equivalence point is reached.
8. (a) The two acids have different initial pH because Acid 1 is stronger than Acid 2 and, thus, it
has a lower pH.
(b) The volume that is required to neutralize these acids is 50 mL. The same volume is required
for both acids because the acids are of the same concentration. This also assumes that both acids
contain the same number of hydrogen atoms that ionize.
(c) An acid–base indicator that could be used for one of the acids, but not the other is methyl red.
=
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Chapter 10: Acids and Bases
10.3-5
9. (a) CaCO3(s) + 2 HCl(aq) → CO2(g) + 2 H2O(l) + CaCl2(aq)
(b) Given: volume of hydrochloric acid, VHCl = 25.20 mL
amount concentration of hydrochloric acid, cHCl = 0.50 mol/L
Required: mass of calcium carbonate, mCaCO
Analysis: c =
3
n
V
Solution:
Step 1. Convert the volume of solution to litres.
1L
VHCl = 25.20 mL !
1000 mL
VHCl = 0.025 20 L
Step 2. Rearrange the concentration equation to determine the amount of hydrochloric acid.
nHCl = cHCl ! VHCl
mol
! 0.025 20 L
L
nHCl = 0.0126 mol
Step 3. Use the amount of acid and the mole ratio in the balanced equation to determine the
amount of calcium carbonate.
CaCO3(s) + 2 HCl(aq) → CO2(g) + 2 H2O(l) + CaCl2(aq)
1 molCaCO
3
nCaCO = 0.0126 mol HCl !
3
2 mol HCl
= 0.50
nCaCO = 0.006 30 molCaCO
3
3
Step 4. Calculate the mass of calcium carbonate.
100.09 g
mCaCO = 0.006 30 mol !
3
1 mol
mCaCO = 0.63 g
3
Statement: The mass of calcium carbonate required is 0.63 g.
10. (a) 2 NH4OH(aq) + H2SO4(aq) → 2 H2O(l) + (NH4)2SO4(aq)
(b) Given: three volumes of acid
Required: average volume of acid used, VH SO
2
4
Solution: Calculate the average of titrant (acid) used by subtracting the final from the initial
burette volume for each trial.
8.46 mL + 8.50 mL + 8.66 mL
VH SO (average) =
2
4
3
VH SO (average) = 8.54 mL
2
4
Statement: The average volume of sulfuric acid used is 8.54 mL.
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Chapter 10: Acids and Bases
10.3-6
(c) Given: concentration of acid, cH SO = 0.25 mol/L
2
4
volume of base, VNH OH = 10.00 mL
4
average volume of acid, VH SO
2
4 (average)
= 8.54 mL
Required: amount concentration of base, cNH OH
4
Analysis: c =
n
V
Solution:
Step 1. Convert the volume of solution to litres.
1L
VNH OH = 10.00 mL !
4
1000 mL
VNH OH = 0.010 00 L
4
VH SO
4 (average)
VH SO
4 (average)
2
2
= 8.54 mL !
1L
1000 mL
= 0.00854 L
Step 2. Write a balanced equation for the reaction listing the given values.
2 NH4OH(aq) +
H2SO4(aq) → 2 H2O(l) + (NH4)2SO4(aq)
0.010 00 L
0.008 54 L
0.25 mol/L
cNH OH
4
Step 3. Use the concentration equation to determine the amount of acid.
nH SO = cH SO VH SO (average)
2
4
2
4
2
4
0.25 mol
! 0.008 54 L
1L
= 0.002 135 mol
=
nH SO
2
4
Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the
amount of base.
2 mol NH OH
4
nNH OH = 0.002 135 mol H SO !
4
2
4
1 mol H SO
2
4
nNH OH = 0.004 27 mol NH OH
4
4
Copyright © 2011 Nelson Education Ltd.
Chapter 10: Acids and Bases
10.3-7
Step 5. Use the concentration equation to determine the concentration of base.
nNH OH
4
2
cNH OH =
4
2
VNH OH
4
2
0.004 27 mol
0.0100 L
mol
cNH OH = 0.43
4
2
L
Statement: The amount concentration of ammonium hydroxide solution is 0.43 mol/L.
(d) It is necessary to conduct more than one trial when standardizing a solution to show that the
data is reproducible and reliable.
11. Given: concentration of base, cNaOH = 0.100 mol/L
=
average volume of base, VNaOH (average) = 1.60 mL
volume of acid, Vmilk = 10.00 mL
Required: amount concentration of lactic acid in milk, cHC H O
3
Analysis: c =
5
2
n
V
Solution:
Step 1. Convert the volume of solutions to litres.
1L
VNaOH(average) = 1.60 mL !
1000 mL
VNaOH(average) = 0.001 60 L
Vmilk = 10.00 mL !
1L
1000 mL
Vmilk = 0.0100 L
Step 2. Write the balanced equation for the reaction listing the given values.
HC3H5O2(aq) + NaOH(aq) → H2O(l) + NaC3H5O2(aq)
0.00160 L
cHC H O
3
5
2
0.100 mol/L
Step 3. Use the concentration equation to determine the amount of base.
nNaOH = cNaOHVNaOH(average)
=
0.100 mol
! 0.00160 L L
1L
nNaOH = 1.60 ! 10 –4 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 10: Acids and Bases
10.3-8
Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the
amount of base.
1 mol HC H O
3 5 2
nHC H O = 1.60 ! 10 –4 mol NaOH !
3 5 2
1 mol NaOH
nHC H O = 1.60 ! 10 –4 mol HC H O
3
5
2
3
5
2
Step 5. Use the concentration equation to determine the concentration of base.
nHC H O
3 5 2
cHC H O =
3 5 2
VHC H O
3
5
2
1.60 ! 10 –4
0.0100 L
mol
cHC H O = 0.0160
3 5 2
L
Statement: The amount concentration of lactic acid in milk is 0.0160 mol/L.
12. (a) KH(IO3)2(aq) + KOH(aq) → H2O(l) + 2 KIO3(aq)
(b) Given: mass of potassium hydrogen iodate, mKH(IO ) = 1.50 g
=
3 2
average volume of base, VKOH = 25.42 mL
Required: amount concentration of potassium hydroxide, cKOH
Solution:
Step 1. Convert the volume of titrant used to litres.
1L
VKOH = 25.42 mL !
1000 mL
VKOH = 2.542 ! 10 "2 L
Step 2. Write the balanced equation for the reaction listing the given values.
KH(IO3)2(aq) + KOH(aq) → H2O(l) + 2 KIO3(aq)
1.50 g
2.420 × 10–2 L
389.91 g/mol cKOH
Step 3. Convert the mass of potassium hydrogen iodate to the amount.
1 mol
nKH(IO ) = 1.50 g !
3 2
389.91 g
nKH(IO
3 )2
= 3.847 ! 10 "3 mol
Step 4. Use the amount and the mole ratio in the balanced equation to determine the amount of
the base.
1 mol KOH
nKOH = 3.847 ! 10 "3 mol KH(IO ) !
3 2
1 mol KH(IO )
3 2
nKOH = 3.847 ! 10 "3 mol KOH
Copyright © 2011 Nelson Education Ltd.
Chapter 10: Acids and Bases
10.3-9
Step 5. Use the concentration equation to determine the concentration of base.
n
cKOH = KOH
VKOH
=
3.847 ! 10 "3 mol
2.542 ! 10 "2 L
mol
cKOH = 0.151
L
Statement: The amount concentration of potassium hydroxide is 0.151 mol/L.
13. Given: mass of potassium carbonate, mK CO = 1.00 g
2
3
average volume of acid, VHCl(average) = 24.20 mL
volume of base, VK CO = 100 mL
2
3
Required: amount concentration of acid, cHCl
Analysis: c =
n
V
Solution:
Step 1. Convert the volume of solutions to litres.
1L
VHCl(average) = 24.20 mL !
1000 mL
VHCl(average) = 2.420 ! 10 "2 L
VK CO = 100 mL !
2
3
1L
1000 mL
VK CO = 0.100 L
2
3
Step 2. Write a balanced equation for the reaction listing the values.
K2CO3(aq) + 2 HCl(aq) → 2 H2O(l) + CO2(g) + 2 KCl(aq)
1.00 g
2.420 × 10–2 L
138.21 g/mol cHCl
Step 3. Convert the mass of potassium carbonate to the amount.
1 mol
nK CO = 1.00 g !
2
3
138.21 g
nK CO = 7.235 ! 10 "3 mol
2
3
Step 4. Use the amount of potassium carbonate and the mole ratio in the balanced equation to
determine the amount of acid.
2 mol HCl
nHCl = 7.235 ! 10 "3 mol K CO !
2
3
1 mol K CO
2
3
nHCl = 1.447 ! 10 "2 mol HCl
Copyright © 2011 Nelson Education Ltd.
Chapter 10: Acids and Bases
10.3-10
Step 5. Use the concentration equation to determine the concentration of acid.
n
cHCl = HCl
VHCl
=
1.447 ! 10 "2 mol
2.420 ! 10 "2 L
mol
cHCl = 0.598
L
Statement: The amount concentration of the hydrochloric acid solution is 0.598 mol/L.
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Chapter 10: Acids and Bases
10.3-11