Section 10.3: Acid–Base...

Copyright © 2011 Nelson Education Ltd. Chapter 10: Acids and Bases 10.3-1

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,

V

HNO3

Solution: Calculate the average of titrant (acid) used by subtracting the final from the initial burette volume for each trial.

VHNO

3(average)

=8.00 mL + 8.06 mL + 8.12 mL

3

VHNO

3(average)

= 8.06 mL

Statement: The average volume of nitric acid used is 8.06 mL. (b) Given: concentration of acid,

c

HNO3

= 0.500 mol/L

volume of base, V

NaOH= 25.00 mL

average volume of acid, V

HNO3

(average)= 8.06 mL

Required: amount concentration of base, c

NaOH

Analysis:

c =n

V

Solution: Step 1. Convert the volume of solution to litres.

VHNO

3(average)

= 8.06 mL !1 L

1000 mL

VHNO

3(average)

= 8.06 !10"3

L

VNaOH

= 25.00 mL !1 L

1000 mL

VNaOH

= 2.500 !10"2

L


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

c

NaOH 0.500 mol/L

Step 3. Use the concentration equation to determine the amount of acid.

nHNO

3

= cHNO

3

VHNO

3(average)

=0.500 mol

1 L! 8.06 !10

"3 L

nHNO

3

= 4.03!10"3

mol

Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the amount of base.

nNaOH

= 4.03!10"3

molHNO

3

!1 mol

NaOH

1molHNO

3

nNaOH

= 4.03!10"3

molNaOH

Step 5. Use the concentration equation to determine the concentration of base.

cNaOH

=n

NaOH

VNaOH

=4.03!10

"3 mol

2.500 !10"2

L

cNaOH

= 0.161 mol

L

Statement: The amount concentration of the sodium hydroxide solution is 0.161 mol/L. 2. Given: mass of potassium carbonate,

m

K2CO

3

= 1.00 g

volume of acid, V

HNO3

= 24.20 mL

Required: amount concentration of acid, c

HNO3

Analysis:

c =n

V

Solution: Step 1. Convert the volume of titrant used to litres.

VHNO

3

= 24.20 mL !1 L

1000 mL

VHNO

3

= 2.420 !10"2

L

Step 2. Write the balanced equation listing the given values. K2CO3(s) + 2 HNO3(aq) → 2 KNO3 + H2CO3 1.00 g 2.420 × 10–2 L 138.21 g/mol

c

HNO3


Step 3. Convert the mass of potassium carbonate to the amount.

nK

2CO

3

= 1.00 g !1 mol

138.21 g

nK

2CO

3

= 7.2354 !10"3

mol [2 extra digits carried]

Step 4. Use the amount and the mole ratio in the balanced equation to determine the amount of the acid.

nHNO

3

= 7.2354 !10"3

molK

2CO

3

!

2 molHNO

3

1molK

2CO

3

nHNO

3

= 1.4471!10"2

molHNO

3

[2 extra digits carried]

Step 5. Use the concentration equation to determine the concentration of acid.

cHNO

3

=

nHNO

3

VHNO

3

=1.4471!10

"2 mol

2.420 !10"2

L

cHNO

3

= 0.598 mol

L

Statement: The amount concentration of nitric acid is 0.598 mol/L. 3. Given: concentration of acid,

c

HCl= 0.100 mol/L

volume of acid, V

HCl= 15.52 mL

volume of base, V

Ba(OH)2

= 25.00 mL


Ba(OH)2

Analysis:

c =n

V

Solution: Step 1. Convert the volume of the solutions to litres.

VHCl

= 15.52 mL !1 L

1000 mL

VHCl

= 1.552 !10"2

L

VBa(OH)

2

= 25.00 mL !1 L

1000 mL

VBa(OH)

2

= 2.500 !10"2

L


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

c

Ba(OH)2


nHCl

= cHCl

VHCl

=0.100 mol

1 L!1.552 !10

"2 L

nHCl

= 1.552 !10"3

mol


nBa(OH)

2

= 1.552 !10"3

molHCl

!

1 molBa(OH)

2

2 molHCl

nBa(OH)

2

= 7.760 !10"4

molBa(OH)

2


cBa(OH)

2

=

nBa(OH)

2

VBa(OH)

2

=7.760 !10

"4mol

2.500 !10"2

L

cBa(OH)

2

= 3.10 !10"2

mol

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,

m

NaOH= 4.00 g

volume of solution, V

NaOH= 100.0 mL


NaOH

Analysis:

cNaOH

=n

NaOH

VNaOH



VNaOH

= 100.0 mL !1 L

1000 mL

VNaOH

= 0.1000 L

Step 2. Convert mass of the sodium hydroxide to the amount.

nNaOH

= 4.00 g !1 mol

40.00 g

nNaOH

= 0.100 mol

Step 3. Use the concentration equation to determine the concentration of the solution.

cNaOH

=n

NaOH

VNaOH

=0.100 mol

0.100 L

cNaOH

= 1.00 mol

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.


9. (a) CaCO3(s) + 2 HCl(aq) → CO2(g) + 2 H2O(l) + CaCl2(aq) (b) Given: volume of hydrochloric acid,

V

HCl= 25.20 mL

amount concentration of hydrochloric acid, c

HCl= 0.50 mol/L

Required: mass of calcium carbonate, m

CaCO3

Analysis:

c =n

V


VHCl

= 25.20 mL !1 L

1000 mL

VHCl

= 0.025 20 L

Step 2. Rearrange the concentration equation to determine the amount of hydrochloric acid.

nHCl

= cHCl

!VHCl

= 0.50mol

L! 0.025 20 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)

nCaCO

3

= 0.0126 molHCl

!

1 molCaCO

3

2 molHCl

nCaCO

3

= 0.006 30 molCaCO

3

Step 4. Calculate the mass of calcium carbonate.

mCaCO

3

= 0.006 30 mol !100.09 g

1 mol

mCaCO

3

= 0.63 g

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,

V

H2SO

4

Solution: Calculate the average of titrant (acid) used by subtracting the final from the initial burette volume for each trial.

VH

2SO

4(average)

=8.46 mL + 8.50 mL + 8.66 mL

3

VH

2SO

4(average)

= 8.54 mL

Statement: The average volume of sulfuric acid used is 8.54 mL.


(c) Given: concentration of acid, c

H2SO

4

= 0.25 mol/L

volume of base, V

NH4OH

= 10.00 mL


H2SO

4(average)

= 8.54 mL


NH4OH

Analysis:

c =n

V


VNH

4OH

= 10.00 mL !1 L

1000 mL

VNH

4OH

= 0.010 00 L

VH

2SO

4(average)

= 8.54 mL !1 L

1000 mL

VH

2SO

4(average)

= 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

c

NH4OH

0.25 mol/L


nH

2SO

4

= cH

2SO

4

VH

2SO

4(average)

=0.25 mol

1 L! 0.008 54 L

nH

2SO

4

= 0.002 135 mol


nNH

4OH

= 0.002 135 molH

2SO

4

!

2 molNH

4OH

1molH

2SO

4

nNH

4OH

= 0.004 27 molNH

4OH



cNH

4OH

2

=

nNH

4OH

2

VNH

4OH

2

=0.004 27 mol

0.0100 L

cNH

4OH

2

= 0.43 mol

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,

c

NaOH= 0.100 mol/L

average volume of base, V

NaOH (average)= 1.60 mL

volume of acid, V

milk= 10.00 mL

Required: amount concentration of lactic acid in milk, c

HC3H

5O

2

Analysis:

c =n

V

Solution: Step 1. Convert the volume of solutions to litres.

VNaOH(average)

= 1.60 mL !1 L

1000 mL

VNaOH(average)

= 0.001 60 L

Vmilk

= 10.00 mL !1 L

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)

c

HC3H

5O

2

0.00160 L

0.100 mol/L Step 3. Use the concentration equation to determine the amount of base.

nNaOH

= cNaOH

VNaOH(average)

=0.100 mol

1 L! 0.00160 L L

nNaOH

= 1.60 !10–4

mol



nHC

3H

5O

2

= 1.60 !10–4

molNaOH

!

1 molHC

3H

5O

2

1molNaOH

nHC

3H

5O

2

= 1.60 !10–4

molHC

3H

5O

2


cHC

3H

5O

2

=

nHC

3H

5O

2

VHC

3H

5O

2

=1.60 !10

–4

0.0100 L

cHC

3H

5O

2

= 0.0160 mol

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,

m

KH(IO3

)2

= 1.50 g

average volume of base, V

KOH= 25.42 mL

Required: amount concentration of potassium hydroxide, c

KOH

Solution: Step 1. Convert the volume of titrant used to litres.

VKOH

= 25.42 mL !1 L

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

c

KOH

Step 3. Convert the mass of potassium hydrogen iodate to the amount.

nKH(IO

3)2

= 1.50 g !1 mol

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.

nKOH

= 3.847 !10"3

molKH(IO

3)2

!1 mol

KOH

1molKH(IO

3)2

nKOH

= 3.847 !10"3

molKOH



cKOH

=n

KOH

VKOH

=3.847 !10

"3 mol

2.542 !10"2

L

cKOH

= 0.151 mol

L

Statement: The amount concentration of potassium hydroxide is 0.151 mol/L. 13. Given: mass of potassium carbonate,

m

K2CO

3

= 1.00 g


HCl(average)= 24.20 mL

volume of base, V

K2CO

3

= 100 mL

Required: amount concentration of acid, c

HCl

Analysis:

c =n

V

Solution: Step 1. Convert the volume of solutions to litres.

VHCl(average)

= 24.20 mL !1 L

1000 mL

VHCl(average)

= 2.420 !10"2

L

VK

2CO

3

= 100 mL !1 L

1000 mL

VK

2CO

3

= 0.100 L

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

c

HCl

Step 3. Convert the mass of potassium carbonate to the amount.

nK

2CO

3

= 1.00 g !1 mol

138.21 g

nK

2CO

3

= 7.235!10"3

mol

Step 4. Use the amount of potassium carbonate and the mole ratio in the balanced equation to determine the amount of acid.

nHCl

= 7.235!10"3

molK

2CO

3

!2 mol

HCl

1molK

2CO

3

nHCl

= 1.447 !10"2

molHCl


Step 5. Use the concentration equation to determine the concentration of acid.

cHCl

=n

HCl

VHCl

=1.447 !10

"2mol

2.420 !10"2

L

cHCl

= 0.598 mol

L

Statement: The amount concentration of the hydrochloric acid solution is 0.598 mol/L.

Section 10.3: Acid–Base...

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