Chemistry 11 Solutions
Chapter 10
Acids and Bases
Section 10.2 Neutralization Reactions and Acid-base Titrations
Solutions for Practice Problems
Student Edition page 466
1. Practice Problem (page 466)
Hydrochloric acid was slowly added to an Erlenmeyer flask that contained
50.0 mL of 1.50 mol/L sodium hydroxide, NaOH(aq), and a pH meter. The pH
meter read 7.0 after the addition of 35.3 mL of hydrochloric acid. Calculate the
concentration of the hydrochloric acid.
What Is Required?
You need to find the concentration in moles per litre of a hydrochloric acid
solution.
What Is Given?
You know the volume and concentration of the NaOH(aq): V = 50.0 mL;
c = 1.50 mol/L
You know the volume of the HCl(aq): V = 35.3 mL
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volumes of the reactants to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of NaOH(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
HCl(aq).
n
Calculate the concentration of HCl(aq) using the relationship c = .
V
Act on Your Strategy
Balanced chemical equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(ℓ)
Mole ratio:
1 mole
1 mole
1 mole
1 mole
Volume of NaOH(aq):
V = 50.0 mL × 1 × 10 –3 L/ mL
= 0.0500 L
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Chemistry 11 Solutions
Volume of HCl(aq):
–3
V = 35.3 mL × 1 × 10 L/ mL
= 0.0353 L
Amount in moles, n, of NaOH(aq):
nNaOH = cV
= 1.50 mol/ L × 0.0500 L
= 0.0750 mol
Amount in moles, n, of HCl(aq):
1 mol NaOH 0.0750 mol NaOH
=
1 mol HCl
nHCl
0.0750 mol NaOH × 1 mol HCl
1 mol NaOH
= 0.0750 mol
nHCl =
Concentration of HCl(aq):
n
c=
V
0.0750 mol
=
0.0353 L
= 2.12 mol/L
The concentration of HCl(aq) is 2.12 mol/L.
Check Your Solution
For the mole ratio of 1:1 between NaOH(aq) and HCl(aq), it is reasonable that
the HCl(aq) would have a slightly higher concentration than NaOH(aq) since
slightly less HCl(aq) was used in the titration. The answer correctly shows
three significant digits.
2. Practice Problem (page 466)
What volume of 0.400 mol/L sodium hydroxide, NaOH(aq), is needed to
neutralize 26.8 mL of 0.504 mol/L sulfuric acid, H2SO4(aq), completely? Hint:
Sulfuric acid loses two hydrogen ions during this neutralization reaction.
What Is Required?
You need to find the volume of a sodium hydroxide solution.
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Chemistry 11 Solutions
What Is Given?
You know the volume and concentration of the H2SO4(aq): V = 26.8 mL;
c = 0.504 mol/L
You know the concentration of the NaOH(aq): c = 0.400 mol/L
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volume of H2SO4(aq) from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of H2SO4(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
NaOH(aq).
n
Calculate the volume of NaOH(aq) using the relationship V = .
c
Act on Your Strategy
Balanced chemical equation:
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ)
Mole ratio:
2 moles
1 mole
1 mole
1 mole
Volume of H2SO4(aq):
V = 26.8 mL × 1 × 10 –3 L/ mL
= 0.0268 L
Amount in moles, n, of H2SO4(aq):
nH2SO4 = cV
= 0.504 mol/ L × 0.0268 L
= 0.0135 mol
Amount in moles, n, of NaOH(aq):
nNaOH
2 mol NaOH
=
1 mol H 2SO 4 0.0135 mol H 2SO 4
nNaOH =
2 mol NaOH × 0.0135 mol H 2SO 4
1 mol H 2SO 4
= 0.0270 mol
Volume of NaOH(aq):
n
V=
c
0.0270 mol
=
0.400 mol/L
= 0.0675 L
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Chemistry 11 Solutions
The volume of NaOH(aq) is 0.0675 L.
Check Your Solution
Since the mole ratio between NaOH(aq) and H2SO4(aq) is 2:1, the volume of
2
NaOH required will be times more than the volume of H2SO4(aq). Since
1
the concentration of NaOH(aq) is less than that of H2SO4(aq), the volume of
0.5
more than the volume of H2SO4(aq).
NaOH(aq) required will be
0.4
The volume, V, of NaOH(aq) required can be estimated:
V ≈ volume of H2SO4(aq) × concentration factor × mole ratio factor
0.5
2
× = 0.067 L
≈ 0.027 L ×
0.4
1
The calculated answer of 0.0675 L seems reasonable and correctly shows three
significant digits.
3. Practice Problem (page 466)
A 25.00 mL sample of a nitric acid solution, HNO3(aq), is neutralized by
18.55 mL of a 0.1750 mol/L sodium hydroxide solution, NaOH(aq). What is
the concentration of the nitric acid solution?
What Is Required?
You need to find the concentration in moles per litre of a nitric acid solution.
What Is Given?
You know the volume and concentration of the NaOH(aq): V = 18.55 mL;
c = 0.1750 mol/L
You know the volume of the HNO3( aq): V = 25.00 mL
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volumes of the reactants to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of NaOH(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
HNO3(aq).
n
Calculate the concentration of HNO3(aq) using the relationship c = .
V
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Chemistry 11 Solutions
Act on Your Strategy
Balanced chemical equation: NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(ℓ)
Mole ratio:
1 mole
1 mole
1 mole
1 mole
Volume of NaOH(aq):
V = 18.55 mL × 1 × 10 –3 L/ mL
= 0.01855 L
Volume of HNO3(aq):
V = 25.00 mL × 1 × 10 –3 L/ mL
= 0.02500 L
Amount in moles, n, of NaOH(aq):
nNaOH = cV
= 0.1750 mol/ L × 0.01855 L
= 0.003246 mol
Amount in moles, n, of HNO3(aq):
nHNO3
1 mol HNO3
=
1 mol NaOH 0.003246 mol NaOH
1 mol HNO3 × 0.003246 mol NaOH
nHNO3 =
1 mol NaOH
= 0.003246 mol
Concentration of HNO3(aq):
n
c=
V
0.003246 mol
=
0.02500 L
= 0.1298 mol / L
The concentration of HNO3(aq) is 0.1298 mol/L.
Check Your Solution
For the mole ratio of 1:1 between NaOH(aq) and HNO3(aq), it is reasonable
that the concentration of HNO3(aq) would be slightly less than the
concentration of NaOH(aq) since the volume of HNO3(aq) used is slightly
greater than the volume of NaOH(aq). The answer correctly shows four
significant digits.
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Chemistry 11 Solutions
4. Practice Problem (page 466)
What volume of 1.25 mol/L hydrobromic acid, HBr(aq), will neutralize
75.0 mL of 0.895 mol/L magnesium hydroxide, Mg(OH)2(aq)?
What Is Required?
You need to find the volume of a hydrobromic acid solution.
What Is Given?
You know the volume and concentration of the Mg(OH)2(aq): V = 75.0 mL;
c = 0.895 mol/L
You know the concentration of the HBr(aq): c = 1.25 mol/L
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volume of Mg(OH)2(aq) to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of Mg(OH)2(aq) using the relationship n = cV.
Use the mole ratio in the balanced equation to calculate the amount in moles of
HBr(aq).
n
Calculate the volume of HBr(aq) using the relationship V = .
c
Act on Your Strategy
Balanced chemical equation:
2HBr(aq) + Mg(OH)2(aq) → MgBr2(aq) + 2H2O(ℓ)
Mole ratio:
2 moles
1 mole
1 mole
2 moles
Volume of Mg(OH)2(aq):
V = 75.0 mL × 1 × 10 –3 L/ mL
= 0.0750 L
Amount in moles, n, of Mg(OH)2(aq):
nMg( OH ) = cV
2
= 0.895 mol/ L × 0.0750 L
= 0.067125 mol
Amount in moles, n, of HBr(aq):
2 mol HBr
nHBr
=
1 mol Mg(OH)2 0.067125 mol Mg(OH) 2
nHBr =
2 mol HBr × 0.067125 mol Mg(OH) 2
1 mol Mg(OH) 2
= 0.13425 mol
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Chemistry 11 Solutions
Volume of HBr(aq):
n
V=
c
0.13425 mol
=
1.25 mol /L
= 0.1074 L
= 0.107 L
The volume of HBr(aq) is 0.107 L.
Check Your Solution
Since the mole ratio between Mg(OH)2(aq) and HBr(aq) is 1:2, the volume of
2
times more than the volume of Mg(OH)2(aq). Since
HBr required will be
1
the concentration of HBr is more than that of Mg(OH)2(aq), the volume of HBr
0.9
the volume of Mg(OH)2(aq).
will be about
1.25
The volume, V, of HBr(aq) can be estimated:
V ≈ volume Mg(OH)2(aq) × concentration ratio factor × mole ratio factor
0.9
2
×
= 0.108 L
≈ 0.0750 L ×
1.25
1
The calculated answer of 0.107 L seems reasonable and correctly shows three
significant digits.
5. Practice Problem (page 466)
A solution of sodium hydroxide was prepared by dissolving 4.0 g of sodium
hydroxide, NaOH(s), in 250 mL of water. It was found that 20.0 mL of the
sodium hydroxide solution neutralizes 25.0 mL of vinegar. Determine the
concentration of acetic acid, CH3COOH(aq), in the sample of vinegar. Assume
that acetic acid is the only acidic substance in the vinegar.
What Is Required?
You need to find the concentration in moles per litre of acetic acid in a vinegar
solution.
What Is Given?
You know the mass of the NaOH(s): 4.0 g
You know the volume of the NaOH(aq) stock solution: 250 mL
You know the volume of the NaOH(aq) used in the neutralization = 20.0 mL
You know the volume of the vinegar containing CH3COOH(aq): 25.0 mL
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Plan Your Strategy
Use the periodic table to determine the molar mass of NaOH.
Calculate the amount in moles of NaOH(aq) in the stock solution using the
relationship n =
m
.
M
Convert the volume of NaOH(aq) from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the concentration of NaOH(aq).
Write the balanced chemical equation for the reaction.
Calculate the amount in moles of NaOH(aq) used in the neutralization using
the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
CH3COOH(aq).
Convert the volume of CH3COOH(aq) from millilitres to litres: 1 mL = 1 × 10–3 L
n
Calculate the concentration of CH3COOH(aq) using the relationship c = .
V
Act on Your Strategy
Molar mass of NaOH(aq):
M NaOH = 1M Na + 1M O + 1M H
=1(22.99 g/mol) + 1(16.00 g/mol) + 1(1.01g/mol)
= 40.00 g/mol
Amount in moles, n, of NaOH(aq) in the stock solution:
m
nNaOH =
M
4.0 g
=
40.00 g /mol
= 0.10 mol
Volume of NaOH(aq):
V = 250 mL × 1 × 10−3 L/ mL
= 0.250 L
Concentration of NaOH(aq):
n
c=
V
0.10 mol
=
0.250 L
= 0.40 mol/L
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Balanced chemical equation:
NaOH(aq) + CH3COOH(aq) → NaCH3COO(aq) + H2O(ℓ)
Mole ratio:
1 mole
1 mole
1 mole
1 mole
Volume of NaOH(aq) used in the neutralization:
V = 0.0 mL × 1 × 10 –3 L/ mL
= 0.0200 L
Amount in moles, n, of NaOH(aq) in the 20.0 mL sample:
nNaOH = cV
= 0.40 mol/ L × 0.0200 L
= 0.00800 mol
Amount in moles, n, of CH3COOH(aq):
nCH3COOH
1 mol CH 3COOH
=
1 mol NaOH
0.00800 mol NaOH
1 mol CH 3COOH × 0.00800 mol NaOH
nCH3COOH =
1 mol NaOH
= 0.00800 mol
Volume of CH3COOH(aq):
V = 25.00 mL × 1 × 10 –3 L/ mL
= 0.0250 L
Concentration of CH3COOH(aq) in the vinegar:
n
c=
V
0.008000 mol
=
0.0250 L
= 0.32 mol/L
The concentration of CH3COOH(aq) in the vinegar is 0.32 mol/L.
Check Your Solution
The acetic acid and sodium hydroxide react in a mole ratio of 1:1. The volume
of acetic acid solution used was a little more than the volume of sodium
hydroxide solution. It is reasonable to expect the concentration of acetic acid to
be a little less than the concentration of the sodium hydroxide. The calculated
concentration of acetic acid seems reasonable and the answer correctly shows
two significant digits.
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Chemistry 11 Solutions
6. Practice Problem (page 466)
Phosphoric acid, H3PO4(aq), is a triprotic acid. If 15.0 mL of phosphoric acid
completely neutralizes 38.5 mL of 0.150 mol/L sodium hydroxide, NaOH(aq),
what is the concentration of the phosphoric acid?
What Is Required?
You need to find the concentration in moles per litre of a phosphoric acid
solution.
What Is Given?
You know the volume and concentration of the NaOH(aq): V = 38.5 mL;
c = 0.105 mol/L
You know the volume of the H3PO4(aq): V = 15.0 mL
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volumes of the reactants from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of NaOH(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
H3PO4(aq).
n
Calculate the concentration of H3PO4(aq) using the relationship c = .
V
Act on Your Strategy
Balanced chemical equation:
3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3H2O(ℓ)
Mole ratio:
3 moles
1 mole
1 mole
3 moles
Volume of NaOH(aq):
V = 38.5 mL ×1 × 10 –3 L/ mL
= 0.0385 L
Volume of H3PO4(aq):
V = 15.00 mL × 1 × 10 –3 L/ mL
= 0.01500 L
Amount in moles, n, of NaOH(aq):
nNaOH = cV
= 0.150 mol/ L × 0.0385 L
= 0.005775 mol
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Chemistry 11 Solutions
Amount in moles, n, of H3PO4(aq):
nH3PO4
1 mol H 3PO 4
=
3 mol NaOH 0.005775 mol NaOH
1 mol H 3PO 4 × 0.005775 mol NaOH
nH3PO4 =
3 mol NaOH
= 0.001925 mol
Concentration of H3PO4(aq):
n
c=
V
0.001925 mol
=
0.0150 L
= 0.1283 mol/L
= 0.128 mol/L
The concentration of H3PO4(aq) is 0.128 mol/L.
Check Your Solution
Since the mole ratio between NaOH(aq) and H3PO4(aq) is 3:1, the
1
concentration of H3PO4(aq) will be
the concentration of NaOH(aq). Since
3
40
more than the volume of H3PO4(aq), the
the volume of NaOH used is about
15
40
concentration of the H3PO4(aq) should be
greater than the concentration of
15
NaOH(aq).
The concentration, c, of H3PO4(aq) can be estimated:
c ≈ concentration of NaOH(aq) × volume factor × mole ratio factor
40
1
≈ 0.15 mol/L ×
× = 0.13 mol/L
15
3
This estimate is in the range of the calculated answer. The calculated answer of
0.128 mol/L seems reasonable and correctly shows three significant digits.
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Chemistry 11 Solutions
7. Practice Problem (page 466)
The acidity of a water sample can be measured by a neutralization reaction
with a solution of sodium hydroxide, NaOH(aq). What is the concentration of
hydrogen ions in a water sample if 100 mL of the sample is neutralized by the
addition of 8.0 mL of 2.5 ×10–3 mol/L sodium hydroxide?
What Is Required?
You need to find the concentration of hydrogen ions, H+(aq), in moles per litre.
What Is Given?
You know the volume and concentration of the NaOH(aq): V = 8.0 mL;
c = 2.5 × 10–3 mol/L
You know the volume of the acid water sample: V = 100 mL
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volumes of the reactants from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of NaOH(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
H+(aq).
n
Calculate the concentration of H+(aq) using the relationship c = .
V
Act on Your Strategy
Balanced chemical equation: NaOH(aq) + H+(aq) → Na+(aq) + H2O(ℓ)
Mole ratio:
1 mole
1 mole
1 mole
1 mole
Volume of NaOH(aq):
V = 8.0 mL × 1 × 10 –3 L/ mL
= 0.0080 L
Volume of H+(aq):
V = 100 mL × 1 × 10 –3 L/ mL
= 0.100 L
Amount in moles, n, of NaOH(aq):
nNaOH = cV
= 2.5 × 10 –3 mol/ L × 0.0080 L
= 2.0 × 10 –5 mol
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Amount in moles, n, of H+(aq):
nH+
1 mol H +
=
–5
1 mol NaOH 2.0 × 10 mol NaOH
1 mol H + × 2.0 × 10 –5 mol NaOH
nH+ =
1 mol NaOH
= 2.0 × 10 –5 mol
Concentration of H+(aq):
n
c=
V
2.0 × 10 –5 mol
=
0.100 L
= 2 × 10 –4 mol/L
The concentration of H+(aq) is 2 × 10–4 mol/L.
Check Your Solution
The mole ratio between NaOH(aq) and H+(aq) is 1:1, and the volume of acid
solution used is about 10 times more than the volume of NaOH(aq) used. The
1
the concentration
concentration of the acid would be expected to be about
10
of the NaOH(aq). The answer seems reasonable and has one significant digit.
8. Practice Problem (page 466)
Citric acid, H3C6H5O7(aq), is a weak triprotic acid that occurs naturally in
many fruits and vegetables, especially the citrus fruits from which it gets its
name. What volume of 0.165 mol/L sodium hydroxide, NaOH(aq), will
completely react with 40.0 mL of 0.120 mol/L citric acid? For this calculation,
assume that all the hydrogen ions are released by the citric acid.
What Is Required?
You need to find the volume of a sodium hydroxide solution.
What Is Given?
You know the volume and concentration of the H3C6H5O7(aq): V = 40.0 mL;
c = 0.120 mol/L
You know the concentration of the NaOH(aq): c = 0.165 mol/L
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volume of H3C6H5O7(aq) from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of H3C6H5O7(aq) using the relationship n = cV .
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Chemistry 11 Solutions
Use the mole ratio in the balanced equation to calculate the amount in moles of
NaOH(aq).
n
Calculate the volume of NaOH(aq) using the relationship V = .
c
Act on Your Strategy
Balanced chemical equation:
3NaOH(aq) + H3C6H5O7(aq) → Na3C6H5O7(aq) + 3H2O(ℓ)
Mole ratio:
3 moles
1 mole
1 mole
3 moles
Volume of H3C6H5O7(aq):
V = 40.0 mL × 1 × 10 –3 L/ mL
= 0.0400 L
Amount in moles, n, of H3C6H5O7(aq):
nH3C6 H5O7 = cV
= 0.120 mol/ L × 0.0400 L
= 4.80 × 10−3 mol
Amount in moles, n, of NaOH(aq):
nNaOH
3 mol NaOH
=
–3
1 mol H 3C6 H 5O7 4.80 × 10 mol H 3C6 H 5O7
nNaOH =
3 mol NaOH × 4.80 × 10 –3 mol H 3C6 H 5O 7
1 mol H 3C6 H 5O 7
= 1.44 × 10−2 mol
Volume of NaOH(aq):
n
V=
c
1.44 × 10 –2 mol
=
0.165 mol /L
= 8.7272 × 10 –2 L
= 8.73 × 10 –2 L
The volume of NaOH(aq) is 8.73 × 10–2 L.
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Chemistry 11 Solutions
Check Your Solution
0.12
less than the
0.16
volume of citric acid because the ratio of the concentrations of NaOH(aq) to
0.12
3
. The volume of NaOH(aq) will be times greater
H3C6H5O7(aq) is
0.16
1
than the volume of H3C6H5O7(aq) because of the 3:1 mole ratio between
NaOH(aq) and H3C6H5O7(aq).
Estimating the answer, the volume of NaOH(aq) will be
The volume, V, of NaOH(aq) can be estimated:
V ≈ volume of H3C6H5O7(aq) × concentration ratio factor × mole ratio factor
0.12 3
× = 0.09 L
≈ 0.04 L ×
0.16 1
The calculated answer of 8.73 × 10–2 L seems reasonable and has three
significant digits.
9. Practice Problem (page 466)
Phosphoric acid, H3PO4(aq), is a weak triprotic acid. When phosphoric acid
reacts with a base, different salts can be prepared, depending on how many
hydrogen ions are replaced by cations.
For example, potassium hydrogen phosphate, K2HPO4 (aq), can be prepared in
an aqueous solution by adding just enough potassium hydroxide, KOH(aq), to
replace two hydrogen ions:
H3PO4(aq) + 2KOH(aq) → K2HPO4(aq) + 2H2O(ℓ)
What volume of 0.185 mol/L potassium hydroxide should be added to 80.0 mL
of 0.137 mol/L phosphoric acid, H3PO4(aq), to form a solution of potassium
hydrogen phosphate?
What Is Required?
You need to find the volume of a potassium hydroxide solution required to
produce potassium hydrogen phosphate.
What Is Given?
You know the volume and concentration of the H3PO4(aq): V = 80.0 mL;
c = 0.137 mol/L
You know the concentration of the KOH(aq): c = 0.185 mol/L
You know the balanced chemical equation for the reaction.
Plan Your Strategy
Convert the given volume of H3PO4(aq) from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of H3PO4(aq) using the relationship n = cV .
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Use the mole ratio in the balanced equation to calculate the amount in moles of
KOH(aq).
n
Calculate the volume of KOH(aq) using the relationship V = .
c
Act on Your Strategy
Balanced chemical equation:
H3PO4(aq) + 2KOH(aq) → K2HPO4(aq) + 2H2O(ℓ)
Mole ratio:
1 mole
2 moles
1 mole
2 moles
Volume of H3PO4(aq):
V = 80.0 mL × 1 × 10 –3 L/ mL
= 0.0800 L
Amount in moles, n, of H3PO4(aq):
nH3PO4 = cV
= 0.137 mol/ L × 0.0800 L
= 0.01096 mol
Amount in moles, n, of KOH(aq):
nKOH
2 mol KOH
=
1 mol H 3 PO 4 0.01096 mol H 3 PO 4
nKOH =
2 mol KOH × 0.01096 mol H 3 PO 4
1 mol H 3 PO 4
= 2.1920 × 10 –2 mol
Volume of KOH(aq):
n
V=
c
2.1920 × 10 –2 mol
=
0.185 mol /L
= 0.118 L
The volume of KOH(aq) is 0.118 L.
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Check Your Solution
2
times that of
1
phosphoric acid because the mole ratio between KOH(aq) and H3PO4(aq) is
0.14
2:1. The volume will be approximately
the volume of phosphoric acid
0.18
because the ratio of the concentrations of H3PO4(aq) to KOH(aq) is 0.14:0.18.
Estimating the answer, the volume of KOH(aq) will be
The volume, V, of KOH(aq) can be estimated:
V ≈ volume of H3PO4(aq) × concentration factor × mole ratio factor
0.14
2
≈ 0.080 L ×
× = 0.12 L
0.18
1
The calculated answer of 0.118 L seems reasonable and has three significant
digits.
10. Practice Problem (page 466)
What volume of 0.150 mol/L calcium hydroxide, Ca(OH)2(aq), is needed to
completely neutralize 20 mL of 0.185 mol/L sulfuric acid, H2SO4(aq)?
What Is Required?
You need to find the volume of a calcium hydroxide solution.
What Is Given?
You know the volume and concentration of the H2SO4(aq) : V = 20 mL;
c = 0.185 mol/L
You know the concentration of the Ca(OH)2(aq): c = 0.150 mol/L
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the given volumes of H2SO4(aq) and Ca(OH)2(aq) from millilitres to
litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of H2SO4(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
Ca(OH)2(aq).
n
Calculate the volume of Ca(OH)2(aq) using the relationship V = .
c
Act on Your Strategy
Balanced chemical equation:
H2SO4(aq) + Ca(OH)2 (aq) → CaSO4(s) + 2H2O(ℓ)
Mole ratio:
1 mole
1 mole
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Chemistry 11 Solutions
Volume of H2SO4(aq):
V = 20 mL × 1 × 10 –3 L/ mL
= 0.020 L
Amount in moles, n, of H2SO4(aq):
nH 2SO4 = cV
= 0.185 mol/ L × 0.020 L
= 0.3.70 × 10 –3 mol
Amount in moles, n, of Ca(OH)2(aq):
nCa (OH)2
1 mol Ca(OH) 2
=
1 mol H 2SO 4
3.70 × 10 –3 mol H 2SO 4
nCa (OH)2 =
1 mol Ca(OH) 2 × 3.70 × 10 –3 mol H 2SO 4
1 mol H 2SO 4
= 3.7 × 10 –3 mol
Volume of Ca(OH)2 (aq):
n
V=
c
3.70 × 10 –3 mol
=
0.150 mol /L
= 2.466 × 10 –2 L
= 2.5 × 10 –2 L
The volume of Ca(OH)2 (aq) is 2.5 × 10–2 L.
Check Your Solution
The mole ratio between H2SO4(aq) and Ca(OH)2 (aq) is 1:1. Since the
concentrations are almost the same, the volumes used in this neutralization
reaction should be about the same. The answer seems reasonable and has two
significant digits.
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Chemistry 11 Solutions
Section 10.2 Neutralization Reactions and Acid-base Titrations
Solutions for Selected Review Questions
Student Edition page 470
8. Review Question (page 470)
What amount of calcium hydroxide, Ca(OH)2(aq), will be neutralized by 1 mol
of hydrochloric acid?
Balanced chemical equation: Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(ℓ)
Mole ratio:
1 mole
2 moles
1 mole
2 moles
Amount in moles, n, of Ca(OH)2(aq):
nCa (OH)2
1 mol Ca(OH) 2
=
2 mol HCl
1 mol HCl
1 mol Ca(OH)2 × 1 mol HCl
nCa (OH)2 =
2 mol HCl
= 0.5 mol
Therefore, 0.5 mol of Ca(OH)2(aq) will be neutralized by 1 mol of HCl.
9. Review Question (page 470)
Heartburn is a condition that is caused when fluid from the stomach moves up
into the esophagus, causing irritation. Some people use milk of magnesia,
which contains magnesium hydroxide, Mg(OH)2(s), to relieve the symptoms of
heartburn. Explain why this medicine works.
The digestive fluid in the stomach contains hydrochloric acid. Magnesium
hydroxide is a base that will neutralize the hydrochloric acid in the stomach.
The balanced equation for this reaction is as follows:
Balanced chemical equation: 2HCl(aq) + Mg(OH)2 → MgCl2(aq) + 2H2O(ℓ)
Mole ratio:
2 moles
1 mole
1 mole
2 moles
Note that every mole of Mg(OH)2(aq) will neutralize 2 mol of HCl(aq). This
makes milk of magnesia a good choice for an antacid.
10. Review Question (page 470)
What volume of 0.996 mol/L barium hydroxide, Ba(OH)2(aq), is needed to
neutralize 25.0 mL of 1.70 mol/L nitric acid, HNO3(aq)?
What Is Required?
You need to find the volume of a barium hydroxide solution.
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Chemistry 11 Solutions
What Is Given?
You know the volume and concentration of the HNO3(aq): V = 25.0 mL;
c = 1.70 mol/L
You know the concentration of the Ba(OH)2(aq): c = 0.996 mol/L
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volume of HNO3(aq) from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of HNO3(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
Ba(OH)2(aq).
n
Calculate the volume of Ba(OH)2(aq) using the relationship V = .
c
Act on Your Strategy
Balanced chemical equation:
Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(ℓ)
Mole ratio:
1 mole
2 moles
1 mole
2 moles
Volume of HNO3(aq):
V = 25.0 mL × 1 × 10 –3 L/ mL
= 0.0250 L
Amount in moles, n, of HNO3(aq):
nHNO3 = cV
= 1.70 mol/ L × 0.0250 L
= 4.25 × 10−2 mol
Amount in moles, n, of Ba(OH)2(aq):
nBa (OH)2
1 mol Ba(OH)2
=
2 mol HNO3
4.25 × 10 –2 mol HNO3
nBa (OH)2 =
1 mol Ba(OH) 2 × 4.25 × 10 –2 mol HNO3
2 mol HNO3
= 2.125 × 10−2 mol
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Chemistry 11 Solutions
Volume of Ba(OH)2(aq):
n
V=
c
2.125 × 10 –2 mol
=
0.996 mol /L
= 2.1335 × 10 –2 L
= 2.13 × 10 –2 L
The volume of Ba(OH)2(aq) is 2.13 × 10–2 L.
Check Your Solution
1
the volume of
2
HNO3(aq) because the mole ratio between HNO3(aq) and Ba(OH)2(aq) is 1:2.
The volume of Ba(OH)2(aq) will be 1.7 times the volume of HNO3(aq) because
1.7
the ratio of the concentrations of the reactants is about
.
1
The volume, V, of Ba(OH)2(aq) can be estimated:
Estimating the answer, the volume of Ba(OH)2(aq) will be
V ≈ volume of HNO3(aq) × concentration factor × mole ratio factor
1.7
1
≈ 0.025 L ×
×
1
2
≈ 0.021 L
The calculated answer of 2.13 × 10–2 L seems reasonable and has three
significant digits.
12. Review Question (page 470)
Describe how you would design and perform a titration in which you use 0.250
mol/L sulfuric acid, H2SO4(aq), to determine the concentration of a strontium
hydroxide solution, Sr(OH)2(aq). Include the balanced equation for the
reaction, as well as an outline of the calculations you would make.
Materials
•
standard solution of 0.250 mol/L H2SO4(aq)
•
solution of unknown concentration of Sr(OH)2(aq)
•
bromothymol blue indicator
•
25.00 mL volumetric or graduated pipette
•
burette
•
250 mL Erlenmeyer flask
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Chemistry 11 Solutions
Preparation
1. Using distilled water, wash and rinse the required glassware: 25.00 mL
pipette, burette, 250 mL Erlenmeyer flask
2. Rinse the burette twice with the titrant, 0.250 mol/L H2SO4(aq).
3. Rinse the pipette twice with the solution of unknown concentration,
Sr(OH)2(aq).
Procedure
1. Pipette a 25.00 mL sample of Sr(OH)2(aq) into the Erlenmeyer flask.
2. Add 2–3 drops of bromothymol blue indicator. The solution will now have
the characteristic blue colour of this indicator in a base.
3. Fill the burette with the titrant, 0.250 mol/L H2SO4(aq). Record the initial
volume of this standard solution.
4. While gently swirling the flask, add the titrant, quickly at first until the first
signs appear that the blue colour is beginning to turn green or a greenishyellow. Continue to add the titrant drop-wise until the solution is the neutral
green colour of the bromothymol blue indicator.
5. Record the final burette reading.
6. Subtract the two burette readings to obtain the volume of H2SO4(aq) used.
7. Repeat steps 1 through 6 two more times or until three precise results are
obtained.
8. Average the three volume readings.
9. Proceed with the calculation to determine the concentration of Sr(OH)2(aq).
Data table:
Final burette reading (mL)
Initial burette reading
Volume of H2SO4(aq)
Trial 1
15.20 mL
0.00 mL
15.20 mL
Trial 2
30.39 mL
15.20 mL
15.19 mL
Trial 3
45.60 mL
30.39 mL
15.21 mL
Calculation:
Average volume of H2SO4(aq):
15.20 mL + 15.19 mL + 15.21 mL
V=
3
= 15.20 mL × 1 × 10 –3 L/ mL
= 0.01520 L
Balanced chemical equation: H2SO4(aq) + Sr(OH)2 → SrSO4(aq) + 2H2O(ℓ)
Mole ratio:
1 mole
1 mole
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Chemistry 11 Solutions
Amount in moles, n, of H2SO4(aq):
nH2SO4 = cV
= 0.250 mol/ L × 0.01520 L
= 3.80 × 10 –3 mol
Amount in moles, n, of Sr(OH)2(aq):
nSr(OH)2
1 mol Sr(OH) 2
=
1 mol H 2SO 4
3.80 × 10 –3 mol H 2SO 4
nSr(OH)2 =
1 mol Sr(OH) 2 × 3.80 × 10 –3 mol H 2SO 4
1 mol H 2SO 4
= 3.80 × 10−3 mol
Volume of Sr(OH)2(aq):
V = 25.00 mL × 1 × 10 –3 L/ mL
= 0.02500 L
Concentration of Sr(OH)2(aq):
n
c=
V
3.80 × 10 –3 mol
=
0.02500 L
= 0.152 mol/L
The concentration of the Sr(OH)2(aq) is 0.152 mol/L.
13. Review Question (page 470)
Methanoic acid, HCOOH(aq), is a weak monoprotic acid. A 25.00 mL sample
of methanoic acid was titrated with a standard solution of 0.1004 mol/L
sodium hydroxide, NaOH(aq). Three trials were conducted. The average
volume of sodium hydroxide solution that was required to reach the endpoint
was 16.32 mL. What is the concentration of the methanoic acid solution?
What Is Required?
You need to find the concentration in moles per litre of a methanoic acid
solution.
What Is Given?
You know the volume and concentration of the NaOH(aq): V = 16.32 mL;
c = 0.1004 mol/L
You know the volume of the HCOOH(aq): V = 25.00 mL
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Chemistry 11 Solutions
Plan Your Strategy
Write the balanced chemical equation for the reaction.
Convert the volumes of the reactants from millilitres to litres: 1 mL = 1 × 10–3 L
Calculate the amount in moles of NaOH(aq) using the relationship n = cV .
Use the mole ratio in the balanced equation to calculate the amount in moles of
HCOOH(aq).
n
Calculate the concentration of HCOOH(aq) using the relationship c = .
V
Act on Your Strategy
Balanced chemical equation:
NaOH(aq) + HCOOH(aq) → NaHCOO(aq) + H2O(ℓ)
Mole ratio:
1 mole
1 mole
1 mole
1 mole
Average volume of NaOH(aq):
V = 16.32 mL ×1 × 10 –3 L/ mL
= 0.01632 L
Volume of HCOOH(aq):
V = 25.00 mL × 1 × 10 –3 L/ mL
= 0.02500 L
Amount in moles, n, of NaOH(aq):
nNaOH = cV
= 0.1004 mol/ L × 0.01632 L
= 1.6385 × 10 –3 mol
Amount in moles, n, of HCOOH(aq):
nHCOOH
1 mol HCOOH
=
1 mol NaOH 1.6385 × 10 –3 mol NaOH
1 mol HCOOH × 1.6385 × 10 –3 mol NaOH
nHCOOH =
1 mol NaOH
= 1.6385 × 10 –3 mol
Concentration of HCOOH(aq):
n
c=
V
1.6385 × 10 –3 mol
=
0.02500 L
= 6.554 × 10 –2 mol/L
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Chemistry 11 Solutions
The concentration of HCOOH(aq) is 6.554 × 10–2 mol/L.
Check Your Solution
The mole ratio between HCOOH(aq) and NaOH(aq) is 1:1. Since 25.00 mL of
HCOOH(aq) was used compared with 16.32 mL of NaOH(aq), the
16
the concentration of
concentration of HCOOH will be approximately
25
NaOH(aq).
The concentration, c, of HCOOH(aq) can be estimated:
c ≈ concentration of NaOH(aq) × volume ratio factor × mole ratio factor
16 1
≈ 0.10 mol/L × × = 6.4 × 10–2 mol/L
25 1
The calculated answer of 6.554 × 10–2 mol/L is reasonable and has four
significant digits.
14. Review Question (page 470)
Suppose that you titrate 25.0 mL of 0.100 mol/sodium hydroxide, NaOH(aq),
with two different acids. In the first titration, you use 0.150 mol/L hydrochloric
acid, HCl(aq), which is a strong acid. In the second titration, you use
0.150 mol/L acetic acid, CH3COOH(aq), which is a weak acid. How will the
volume of acid used in each titration compare? Explain your answer.
Both HCl(aq) and CH3COOH(aq) have one replaceable hydrogen ion, H+(aq).
The same volume of each acid will be required for each titration. A difference
in the two trials will be in the pH of the solution at the endpoint. Using
HCl(aq) will result in a neutral solution of pH 7 whereas CH3COOH(aq) will
result in a solution of ~pH 9.
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