Chapter 15
Acid-Base Equilibria
and Reactions
Note to teacher: You will notice that there are two different formats for the Sample
Problems in the student textbook. Where appropriate, the Sample Problem contains the
full set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act on
Your Strategy, and Check Your Solution. Where a shorter solution is appropriate, the
Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check
Your Solution step is also included in the shorter Sample Problems.
Solutions for Practice Problems
Student Textbook page 586
1. Problem
Calculate the concentration of hydronium ions in each solution.
4.5 mol/L HCl(aq)
30.0 mL of 4.50 mol/L HBr(aq) diluted to 100.0 mL
18.6 mL of 2.60 mol/L HClO4(aq) added to 24.8 mL of 1.92 mol/L NaOH(aq)
17.9 mL of 0.175 mol/L HNO3(aq) added to 35.4 mL of 0.0160 mol/L
Ca(OH)2(aq)
(a)
(b)
(c)
(d)
What Is Required?
You must calculate the [H3O+] in each solution.
(a)
What Is Given?
[HCl] = 4.5 mol/L
Solution
HCl is a strong acid. Therefore, [HCl] = [H3O+].
[H3O+] = 4.5 mol/L
Check Your Solution
The [H3O+] is equal to the concentration of a strong acid, provided it is not very
dilute.
(b)
What Is Given?
[HBr] = 4.50 mol/L and 30.0 mL is diluted to 100.0 mL
Solution
Calculate the concentration of HBr after dilution. HBr is a strong acid.
Therefore, [HBr] = [H3O+].
Amount of HBr = 4.50 mol/L × 0.0300 L
= 0.135 mol
After dilution, the volume is 0.100 L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
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[HBr] after dilution = 0.135 mol
0.100 L
= 1.35 mol/L
[H3O+] = 1.35 mol/L
Check Your Solution
The acid is diluted a little more than three times, so the concentration of the diluted
solution should be a little less than one-third the original concentration. The [H3O+]
is equal to the concentration of a strong acid, provided it is not very dilute.
(c)
What Is Given?
Volume of perchloric acid = 18.6 mL
[HClO4] = 2.60 mol/L
Volume of sodium hydroxide = 24.8 mL
[NaOH] = 1.92 mol/L
Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equation:
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reagent.
Step 4 The reagent in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Act on Your Strategy
HClO4(aq) + NaOH(aq) → NaClO4(aq) + H2O()
Amount of HClO4 = 2.60 mol/L × 0.0186 L = 0.04836 mol
Amount of NaOH = 1.92 mol/L × 0.0248 L = 0.04762 mol
Step 3 The reactants combine in a 1:1 ratio. The amount of NaOH is less, so this
reactant must be the limiting reactant.
Step 4 Amount of excess
HClO4 = 0.04836 mol − 0.04762 mol (carrying one more digit)
= 7.40 × 10-4 mol
Therefore, the amount of H3O+(aq) = 7.40 × 10-4 mol
Step 5 Total volume of solution = 18.6 mL + 24.8 mL = 43.4 mL
7.40 × 10−4
[H 3O+ ] =
0.0434 L
= 0.0170 mol/L
Step 1
Step 2
Thus, [H3O+] = 0.0170 mol/L
Check Your Solution
The chemical equation has a 1:1 ratio between reactants. The amount of acid is
greater than the amount of base, so the resulting solution should be acidic.
Subtracting the amount of base from the amount of acid gives an answer with one
decimal place. The final answer should have one significant digit.
(d)
What Is Given?
Volume of nitric acid = 17.9 mL
[HNO3] = 0.175 mol/L
Volume of calcium hydroxide = 35.4 mL
[Ca(OH)2] = 0.0160 mol/L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
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Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equation.
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reagent.
Step 4 The reagent in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Act on Your Strategy
Step 1 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O()
Step 2 Amount of HNO3 = 0.175 mol/L × 0.0179 L = 0.00313 mol
Amount of Ca(OH)2 = 0.0160 mol/L × 0.0354 L = 0.000566 mol
Step 3 The reactants combine in a 2:1 ratio. The amount of HNO3 that will react
with 0.000566 mol of Ca(OH)2 is:
Amount HNO3 = 2 × 0.000566 mol = 0.00113 mol
The amount of acid is greater, therefore Ca(OH)2 must be the limiting
reactant.
Step 4 Amount of excess
HNO3 = 0.00313 − 0.00113 mol (carrying one more digit)
= 0.00200 mol
Therefore, the amount of H3O+(aq) is 0.00200 mol.
Step 5 Total volume of solution = 17.9 mL + 35.4 mL = 53.3 mL
[H3O+] = 0.00200 mol
0.0533 L
= 0.0375 mol/L
Check Your Solution
The chemical equation has a 2:1 ratio between reactants. The amount of acid is
greater than the amount of base that reacts, so the resulting solution should be acidic.
2. Problem
Calculate the concentration of hydroxide ions in each solution.
3.1 mol/L KOH(aq)
21.0 mL of 3.1 mol/L KOH diluted to 75.0 mL
23.2 mL of 1.58 mol/L HCl(aq) added to 18.9 mL of 3.50 mol/L NaOH
16.5 mL of 1.50 mol/L H2SO4(aq) added to 12.7 mL of 5.50 mol/L NaOH
(a)
(b)
(c)
(d)
What Is Required?
You must calculate [OH−] in each solution.
(a)
What Is Given?
[KOH] = 3.1 mol/L
Solution
KOH is a strong base. Therefore, [KOH] = [OH−].
[OH−] = 3.1 mol/L
Check Your Solution
The [OH−] is equal to the concentration of a strong base, provided it is not very
dilute.
Chapter 15 Acid-Base Equilibria and Reactions • MHR
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(b)
What Is Given?
[KOH] = 3.1 mol/L
21.0 mL is diluted to 75.0 mL
Solution
Calculate the concentration of KOH after dilution. KOH is a strong base.
Therefore, [KOH] = [OH−].
Amount of KOH = 3.1 mol/L × 0.0210 L = 0.0651 mol
After dilution, the volume is 0.0750 L.
[KOH] after dilution = 0.00651 mol
0.075 L
= 0.868 mol/L
[OH−] = 0.87 mol/L
Check Your Solution
The [OH−] is equal to the concentration of a strong base, provided it is not very
dilute. The base is diluted to almost a quarter the original concentration. Therefore,
the [OH−] should be just greater than one-quarter the original concentration of the
base.
(c)
What Is Given?
Volume of hydrochloric acid = 23.2 mL
[HCl] = 1.58 mol/L
Volume of sodium hydroxide = 18.9 mL
[NaOH] = 3.50 mol/L
Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equation:
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reagent.
Step 4 The reagent in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Act on Your Strategy
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O()
Amount of HCl = 1.58 mol/L × 0.0232 L = 0.0367 mol
Amount of NaOH = 3.50 mol/L × 0.0189 L = 0.0662 mol
Step 3 The reactants combine in a 1:1 ratio. The amount of HCl is less, so this
reactant must be the limiting reactant.
Step 4 Amount of excess NaOH = 0.0662 mol − 0.0367 mol
= 0.0295 mol
Therefore, the amount of OH−(aq) is 0.02949 mol.
Step 5 Total volume of solution = 23.2 mL + 18.9 mL = 42.1 mL
0.0295 mol
[OH − ] =
0.0421 L
= 0.701 mol/L
Step 1
Step 2
Check Your Solution
The chemical equation has a 1:1 ratio between reactants. The amount of base is
greater than the amount of acid, so the resulting solution should be basic.
Chapter 15 Acid-Base Equilibria and Reactions • MHR
259
(d)
What Is Given?
Volume of sulfuric acid = 16.5 mL
[H2SO4] = 1.50 mol/L
Volume of sodium hydroxide = 12.7 mL
[NaOH] = 5.50 mol/L
Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equation:
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reagent.
Step 4 The reagent in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Act on Your Strategy
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O()
Amount of H2SO4 = 1.50 mol/L × 0.0165 L = 0.0248 mol
Amount of NaOH = 5.50 mol/L × 0.0127 L = 0.0698 mol
Step 3 The reactants combine in a 1:2 ratio. The amount of NaOH that will react
with 0.02475 mol of H2SO4 is:
Amount of NaOH = 2 × 2.0248 = 0.0496 mol
The amount of base is greater, therefore H2SO4 must be the limiting
reactant.
Step 4 Amount of excess NaOH = 0.0698 − 0.0496 mol = 0.0202 mol
Therefore, the amount of OH−(aq) is 0.0202 mol.
Step 5 Total volume of solution = 16.5 mL + 12.7 mL = 29.2 mL
0.0202 mol
[OH − ] =
0.0292
= 0.692 mol/L
Check Your Solution
The chemical equation has a 1:2 ratio between reactants. The amount of base is
greater than the amount of acid that reacts, so the resulting solution should be basic.
Step 1
Step 2
3. Problem
Determine whether reacting each pair of solutions results in an acidic or basic
solution. Then calculate the concentration of the ion that causes the solution to be
acidic or basic. (Assume that the volumes in part (a) are additive. Assume that the
volumes in part (b) stay the same.)
(a) 31.9 mL of 2.75 mol/L HCl(aq) added to 125 mL of 0.0500 mol/L Mg(OH)2(aq)
(b) 4.87 g of NaOH(s) added to 80.0 mL of 3.50 mol/L HBr(aq)
What Is Required?
You must determine the ion in excess, its concentration and the pH of the final
solution.
(a)
What Is Given?
Volume of hydrochloric acid = 31.9 mL
[HCl] = 2.75 mol/L
Volume of magnesium hydroxide = 125 mL
[Mg(OH)2] = 0.0500 mol/L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
260
Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equation:
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reactant.
Step 4 The reactant in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Step 6 Calculate the pH of the final solution.
Act on Your Strategy
2HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2H2O()
Amount of HCl = 2.75 mol/L × 0.0319 L = 0.0877 mol
Amount of Mg(OH)2 = 0.0500 mol/L × 0.125 L = 0.00625 mol
Step 3 The reactants combine in a 2:1 ratio. The amount of HCl that will react
with 0.00625 mol of Mg(OH)2 is:
Amount of HCl = 2 × 0.00625 = 0.0125 mol
The amount of acid is greater, therefore Mg(OH)2 must be the limiting reactant.
Step 4 Amount of excess HCl = 0.0877 − 0.0125 mol = 0.0752 mol
Therefore, the amount of H3O+(aq) = 0.0752 mol
Step 5 Total volume of solution = 31.9 mL + 125 mL = 157 mL
[H3O+] = 0.0752 mol
0.157 L
= 0.479 mol/L
Step 6 pH = −log 0.479 = 0.32
Step 1
Step 2
Check Your Solution
The chemical equation has a 2:1 ratio between reactants. The amount of acid is
greater than the amount of base that reacts, so the resulting solution should be acidic.
(b)
What Is Given?
Mass of sodium hydroxide = 4.87 g
Volume of hydrobromic acid = 80.0 mL
[HBr] = 3.50 mol/L
Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equations:
Mass (g)
Amount (mol) =
Molar mass (g/mol)
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reactant.
Step 4 The reactant in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Step 6 Calculate the pH of the final solution.
Act on Your Strategy
HBr(aq) + NaOH(aq) → NaBr(aq) + H2O()
Amount of HBr = 3.50 mol/L × 0.0800 L = 0.280 mol
Molar mass of NaOH = 40.0 g/mol
4.87 g
Amount of NaOH =
40.0 g/mol
= 0.122 mol
Step 1
Step 2
Chapter 15 Acid-Base Equilibria and Reactions • MHR
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The reactants combine in a 1:1 ratio. The amount of acid is greater, therefore
NaOH must be the limiting reactant.
Step 4 Amount of excess HBr = 0.280 − 0.122 mol = 0.158 mol
Therefore, the amount of H3O+(aq) = 0.158 mol
0.158 mol
Step 5 [H3O+] =
0.0800 L
= 1.98 mol/L
Step 6 pH = −log 1.98 = 0.297
Step 3
Check Your Solution
The chemical equation has a 1:1 ratio between reactants. The amount of acid is
greater than the amount of base that reacts, so the resulting solution should be acidic.
4. Problem
2.75 g of MgO(s) is added to 70.0 mL of 2.40 mol/L HNO3(aq). Is the solution that
results from the reaction acidic or basic? What is the concentration of the ion that is
responsible for the character of the solution and determine the pH of the final solution?
What Is Given?
Mass of magnesium oxide = 2.75 g
Volume of nitric acid = 70.0 mL
[HNO3] = 2.40 mol/L
Plan Your Strategy
Step 1 Write the chemical equation for the reaction.
Step 2 Calculate the amount of each reactant using the following equations:
Mass (g)
Amount (mol) =
Molar mass (g/mol)
Amount (mol) = Concentration (mol/L) × Volume (L)
Step 3 Determine the limiting reactant.
Step 4 The reactant in excess is a strong acid or base. Thus, the excess amount
results in the same amount of H3O+ or OH−.
Step 5 Calculate the concentration of the excess ion using the amount in excess and
the total volume of solution.
Step 6 Calculate the pH of the final solution.
Act on Your Strategy
2HNO3(aq) + MgO(s) → Mg(NO3)2(aq) + H2O()
Amount of HNO3 = 2.40 mol/L × 0.0700 L = 0.168 mol
Molar mass of MgO = 40.3 g/mol
2.75 g
Amount of MgO =
40.3 g/mol
= 0.0682 mol
Step 3 The reactants combine in a 2:1 ratio. The amount of acid that will combine
with 0.0682 mol MgO is:
Amount of HNO3 = 2 × 0.0682 mol = 0.136 mol
This is less than the amount of acid. Therefore, MgO is the limiting
reactant.
Step 4 Amount of excess HNO3 = 0.168 − 0.136 mol = 0.032 mol
(Note: the subtraction gives 3 decimal places, but only 2 significant digits.)
Therefore, the amount of H3O+(aq) is 0.032 mol.
0.032 mol
Step 5 [H3O+] =
0.0700 L
= 0.46 mol/L
Step 6 pH = −log 0.46 = 0.34
Step 1
Step 2
Chapter 15 Acid-Base Equilibria and Reactions • MHR
262
Check Your Solution
The chemical equation has a 2:1 ratio between reactants. The amount of acid is
greater than the amount of base that reacts, so the resulting solution should be acidic.
Solutions for Practice Problems
Student Textbook pages 591–592
5. Problem
Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid. What is
the percent dissociation of the vinegar?
What Is Required?
You need to calculate the pH of the solution and the percent dissociation for acetic acid.
What Is Given?
Initial [CH3COOH] = 0.83 mol/L
Acid dissociation constant for acetic acid (listed in Table 8.2):
Ka = 1.8 × 10−5
Plan Your Strategy
[CH3COOH]
to see whether or not the amount that
Step 1 Check the value of
Ka
dissociates is negligible compared with the initial concentration of the acid.
Write the equation for the dissociation equilibrium of acetic acid in water.
Then set up an ICE table.
Step 3 Write the equation for the acid dissociation constant. Substitute equilibrium
concentrations into the equilibrium expression. Solve the equilibrium
expression for x. If the amount that dissociates is not negligible compared
with the initial concentration of acid, you will need to solve a quadratic
equation.
Step 4 pH = − log[H3O+]
Step 5 Calculate the percent dissociation by expressing the fraction of molecules
that dissociate out of 100.
Step 2
Act on Your Strategy
[CH3COOH]
0.83
=
Step 1
Ka
1.8 × 10−5
= 4.6 × 104
Since this value is greater than 500, the amount that dissociates can be
neglected compared to the initial concentration of the acid.
Step 2
Concentration (mol/L)
Initial
Change
Equilibrium
Step 3
CH3COOH(aq)
+
H2O()
CH3COO−(aq)
+
H3O+(aq)
0.83
0
~0
−x
+x
+x
0.83 − x ≈ 0.83
+x
+x
[CH3COO−][H3O+]
[CH3COOH]
(x)(x)
1.8 × 10−5 =
0.83
x = 1.49 × 10−5
Ka =
= ±3.9 × 10−3
Only the positive root is chemically important.
Therefore, [H3O+] = 3.9 × 10−3 mol/L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
263
pH = − log 3.9 × 10−3
= 2.41
3.9 × 10−3 mol/L × 100
Step 5 Percent dissociation =
0.83 mol/L
= 0.47%
Check Your Solution
The pH indicates an acidic solution, as expected. Data given in the problem has
two significant digits, and the pH has two digits following the decimal place.
The assumption that x is negligible compared with the initial concentration is
valid because, using the rules for subtracting measured quantities,
0.83 − 3.9 × 10−3 = 0.83
Next, check the equilibrium values:
(3.9 × 10−3)2 = 1.8 × 10−5
0.83
This is equal to Ka.
Step 4
6. Problem
In low doses, barbiturates act as sedatives. Barbiturates are made from barbituric acid,
a weak monoprotic acid that was first prepared by the German Chemist Adolph von
Baeyer in 1864. The formula of barbituric acid is C4H4N2O3. A chemist prepares a
0.10 mol/L solution of barbituric acid. The chemist finds the pH of the solution to
be 2.50. What is the acid dissociation constant for barbituric acid? What percent of
its molecules dissociate?
What Is Required?
You need to find Ka and the percent dissociation for barbituric acid.
What Is Given?
Initial [C4H4N2O3] = 0.10 mol/L
pH = 2.50
Plan Your Strategy
Step 1 Write the equation for the dissociation equilibrium of barbituric acid in
water. Then set up an ICE table.
Step 2 Write the equation for the acid dissociation constant. Substitute equilibrium
terms into the equation.
Step 3 Calculate [H3O+] using [H3O+] = 10−pH
Step 4 Use the stoichiometry of the equation and [H3O+] to substitute for the
unknown term, x, and calculate Ka.
Step 5 Calculate the percent dissociation by expressing the fraction of molecules
that dissociate out of 100.
Act on Your Strategy
The equation for the dissociation equilibrium of propanoic acid in water is
given in the ICE table.
Step 1
Concentration (mol/L)
C4H4N2O3(aq)
Initial
Step 2
H2O()
C4H3N2O3−(aq)
+
H3O+(aq)
0.10
0
~0
−x
+x
+x
0.10 − x
x
x
Change
Equilibrium
+
[C4H3N2O3−][H3O+]
[C4H4N2O3]
(x)(x)
=
(0.10 − x)
Ka =
Chapter 15 Acid-Base Equilibria and Reactions • MHR
264
Step 3
The value of x is equal to [H3O+] and [C4H3N2O3−].
[H3O+] = 10−2.50
= 3.16 × 10−3 mol/L
(3.16 × 10−3)2
0.10 − (3.16 × 10−3)
= 1.0 × 10−4
3.16 × 10−3 mol/L × 100
Step 5 Percent dissociation =
0.10 mol/L
= 3.2%
Ka =
Step 4
Check Your Solution
The value of Ka and the percent dissociation are reasonable for a weak acid.
7. Problem
A solution of hydrofluoric acid has a molar concentration of 0.0100 mol/L. What is
the pH of this solution?
What Is Required?
You need to calculate the pH of the solution.
What Is Given?
Initial [HF] = 0.0100 mol/L
The acid dissociation constant for hydrofluoric acid (listed in Table 8.2) is:
Ka = 6.6 × 10−4
Plan Your Strategy
[HF]
Step 1 Check the value of
to see whether or not the amount that dissociates
Ka
is negligible compared with the initial concentration of the acid.
Step 2 Write the equation for the dissociation equilibrium of hydrofluoric acid in
water. Then set up an ICE table.
Step 3 Write the equation for the acid dissociation constant. Substitute equilibrium
concentrations into the equilibrium expression. Solve the equilibrium
expression for x. If the amount that dissociates is not negligible compared
with the initial concentration of acid, you will need to solve a quadratic
equation.
Step 4 Calculate the pH using pH = − log[H3O+].
Act on Your Strategy
HF = 0.0100
Step 1
Ka
6.6 × 10−4
= 152
Since this value is less than 500, the amount that dissociates cannot be
neglected compared to the initial concentration of the acid.
Step 2
Concentration (mol/L)
Initial
Change
Equilibrium
Step 3
HF(aq)
+
H2O()
F−(aq)
+
H3O+(aq)
0.0100
0
~0
−x
+x
+x
0.0100 − x
x
x
[F−][H3O+]
[HF]
(x)(x)
−4
6.6 × 10 =
0.0100 − x
This equation must be rearranged into a quadratic equation.
x 2 + (6.6 × 10−4)x − (6.6 × 10−6) = 0
Ka =
Chapter 15 Acid-Base Equilibria and Reactions • MHR
265
Recall that a quadratic equation of the form ax 2 + bx + c = 0 has the
solution
√
−b
±
b2 − 4ac
x =
2a
√
−4
−7
−5
x = −6.6 × 10 ± 4.36 × 10 − 2.64 × 10
2
x = −2.92 × 10−3 and x = 2.26 × 10−3
The value x = −2.92 × 10−3 is not physically possible. It would result in
negative concentrations of F−(aq) and H3O+ at equilibrium.
Therefore, [H3O+] = 2.26 × 10−3 mol/L
Step 4 pH = − log 2.26 × 10−3
= 2.65
Check Your Solution
The pH indicates an acidic solution, as expected. The value for Ka has two significant
digits, and the pH has two digits following the decimal place. Next, check the
equilibrium values:
(2.26 × 10−3)2
= 6.6 × 10−4
0.0100 − 2.26 × 10−3
This is equal to Ka.
8. Problem
Hypochlorous acid, HOCl, is used as a bleach, and a germ-killer. A chemist finds
that 0.027% of hypochlorous acid molecules are dissociated in a 0.40 mol/L solution
of the acid. What is the value of Ka for the acid?
What Is Required?
You need to calculate the value of Ka for hypochlorous acid.
What Is Given?
You know that 0.027% of HOCl molecules dissociate in a 0.40 mol/L solution.
Plan Your Strategy
Step 1 Calculate [H3O+] using the following equation:
[H3O+]
× 100
Percent dissociation =
[HOCl]
Step 2 Write the equation for the dissociation equilibrium of hypochlorous acid in
water. Then set up an ICE table.
Step 3 Write the equation for the acid dissociation constant. Substitute equilibrium
terms into the equation and calculate Ka.
Act on Your Strategy
[H3O+]
× 100
Step 1 0.027 =
0.40
Therefore, [H3O+] = 1.08 × 10−4
= [OCl−]
Step 2
Concentration (mol/L)
Equilibrium
Step 3
HOCl(aq) + H2O()
(0.40 − 1.08 × 10−4) ≈ 0.40
OCl−(aq)
1.08 × 10−4
+
H3O+(aq)
1.08 × 10−4
[OCl−][H3O+]
[HOCl]
−4 2
= (1.08 × 10 )
0.40
= 2.9 × 10−8
Ka =
Chapter 15 Acid-Base Equilibria and Reactions • MHR
266
Check Your Solution
The value of Ka is reasonable for a weak acid.
9. Problem
The word “butter” comes from the Greek butyros. Butanoic acid (common name:
butyric acid) gives rancid butter its distinctive odour. Calculate the pH of a
1.00 × 10−2 mol/L solution of butanoic acid (Ka = 1.51 × 10−5).
What Is Required?
You need to calculate the pH of the solution.
What Is Given?
Initial [CH3CH2CH2COOH] = 1.00 × 10−2 mol/L
The acid dissociation constant for butanoic acid is 1.51 × 10−5.
Plan Your Strategy
Butanoic acid is monoprotic, with the formula CH3CH2CH2COOH.
2COOH]
to see whether or not the amount
Check the value of [CH3CH2CH
Ka
that dissociates is negligible compared with the initial concentration of the
acid.
Step 2 Write the equation for the dissociation equilibrium of butanoic acid in water.
Then set up an ICE table.
Step 3 Write the equation for the acid dissociation constant. Substitute equilibrium
concentrations into the equilibrium expression. Solve the equilibrium
expression for x.
Step 4 Use the equation pH = − log[H3O+].
Step 1
Act on Your Strategy
−2
[CH3CH2CH2COOH]
= 1.00 × 10−5
Step 1
Ka
1.51 × 10
= 662
Since this value is greater than 500, the amount that dissociates can be
neglected compared to the initial concentration of the acid.
Step 2
Concentration (mol/L)
Initial
CH3CH2CH2COOH(aq)
H2O()
CH3CH2CH2COO−(aq)
+
H3O+(aq)
1.00 × 10−2
0
~0
−x
+x
+x
−x≈
x
x
Change
−2
1.00 × 10
Equilibrium
+
1.00 × 10−2
Step 3
[CH3CH2CH2COO−][H3O+]
[CH3CH2CH2COOH]
(x)(x)
−5
1.51 × 10 =
1.00 × 10−2
√
x = 1.51 × 10−7
Ka =
x = −3.89 × 10−4 and x = 3.89 × 10−4
The value x = −3.89 × 10−4 is not physically possible. It would result in
negative concentrations of CH3CH2CH2COO−(aq) and H3O+ at
equilibrium.
Therefore, [H3O+] = 3.89 × 10−4 mol/L
Step 4 pH = − log 3.89 × 10−4
= 3.411
Chapter 15 Acid-Base Equilibria and Reactions • MHR
267
Check Your Solution
The given data has three significant digits, and the pH has three digits following the
decimal place. The assumption that the amount of acid that dissociates can be
neglected is valid because:
1.00 × 10−2 − 3.89 × 10−4 = 9.61 × 10−3
The error introduced by the assumption is 4 × 10−4, or 4%. This is less than the
typical experimental error in Ka values, which is 5%. Next, check the equilibrium
values:
(3.98 × 10−4)2 = 1.51 × 10−5
1.00 × 10−2
This is equal to Ka.
10. Problem
Caproic acid, C5H11COOH, occurs naturally in coconut and palm oil. It is a weak
monoprotic acid, with Ka = 1.3 × 10−5. A certain aqueous solution of caproic acid
has pH = 2.94. How much acid was dissolved to make 100 mL of this solution?
What Is Required?
You need to calculate how much caproic acid was dissolved to make 100 mL of
solution.
What Is Given?
The acid dissociation constant for caproic acid is 1.3 × 10−5.
pH = 2.94
Plan Your Strategy
Step 1 Calculate [H3O+] using the equation [H3O+] = 10−pH.
Step 2 Write the equation for the dissociation equilibrium of caproic acid in water.
Then show the equilibrium concentration below each chemical.
Step 3 Write the equation for the acid dissociation constant. Substitute equilibrium
concentrations into the equilibrium expression. Solve the equilibrium
expression for [C5H11COOH].
Step 4 Calculate the amount of caproic acid dissolved in 100 mL of solution using
the following equation:
C H COOH (mol)
[C5H11COOH] = 5 11
Volume (L)
Step 5 Calculate the mass of caproic acid using the following equation:
Mass (g)
Amount (mol) =
Molar mass (g/mol)
Act on Your Strategy
[H3O+] = 10−2.94
= 1.15 × 10−3 mol/L
Step 1
Step 2
Concentration (mol/L)
C5H11COOH(aq)
Equilibrium
[C5H11COOH] −
−3
1.15 × 10
+
H2O()
C5H11COO−(aq)
1.15 × 10−3
+
H3O+(aq)
1.15 × 10−3
≈
[C5H11COOH]
Step 3
[C5H11COO−][H3O+]
[C5H11COOH]
−3 2
= (1.15 × 10 )
[C5H11COOH]
Ka =
1.3 × 10−5
Therefore, [C5H11COOH] = 0.102 mol/L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
268
C5H11COOH (mol)
0.100 L
Therefore, the amount of C5H11COOH (mol) = 1.02 × 10−2 mol
Step 5 Molar mass of C5H11COOH (mol) = 116 g/mol
Mass (g)
1.02 × 10−2 mol =
116 g/mol
Therefore, mass = 1.2 g
Step 4
0.102 mol/L =
Check Your Solution
The data given in the problem has two significant digits, and the final answer has two
significant digits.
Solutions for Practice Problems
Student Textbook page 595
11. Problem
An aqueous solution of household ammonia has a molar concentration of
0.105 mol/L. Calculate the pH of the solution.
What Is Required?
You need to find the pH of the solution.
What Is Given?
From Table 8.3, Kb = 1.8 × 10−5
Concentration of ammonia solution = 0.105 mol/L
Plan Your Strategy
[NH3]
to see if the amount that dissociates can be
Step 1 Check the value of
Kb
neglected compared to the initial [NH3].
Step 2 Write the equation for the equilibrium reaction of ammonia in water. Then
set up an ICE table.
Step 3 Write the equation for the base dissociation constant. Substitute equilibrium
terms into the equation and solve for x.
Step 4 Use the following equations:
pOH = − log[OH−]
pH = 14.00 − pOH
Act on Your Strategy
0.105 = 5.8 × 104
Step 1
1.8 × 10−5
Since this value is greater than 500, the amount that dissociates can be
neglected compared to the initial concentration of ammonia.
Step 2
Concentration (mol/L)
Initial
Change
Equilibrium
Step 3
NH3(aq)
+
H2O()
NH4+(aq) + OH−(aq)
0.105
0
~0
−x
+x
+x
0.105 − x ≈ 0.105
x
x
+
−
Kb = [NH4 ][OH ]
[NH3]
2
x
−5
1.8 × 10 =
0.105
x = 1.89 × 10−6
x = ±1.37 × 10−3
The negative root is not reasonable.
Therefore, x = 1.37 × 10−3 mol/L = [OH−]
Chapter 15 Acid-Base Equilibria and Reactions • MHR
269
Step 5
pOH = − log 1.37 × 10−3
= 2.86
pH = 14.00 − 2.86
= 11.14
Check Your Solution
The pH of the solution is greater than 7, as expected for a basic solution.
12. Problem
Hydrazine, N2H4, has been used as a rocket fuel. The concentration of an aqueous
solution of hydrazine is 5.9 × 10−2 mol/L. Calculate the pH of the solution.
What Is Required?
You need to find the pH of the solution.
What Is Given?
From Table 8.3, Kb = 1.7 × 10−6.
[N2H4] = 5.9 × 10−2 mol/L
Plan Your Strategy
Step 1 Check the value of [NK2H4] to see if the amount that dissociates can be
b
neglected compared to the initial [N2H4].
Step 2 Write the equation for the equilibrium reaction of hydrazine in water.
Then set up an ICE table.
Step 3 Write the equation for the base dissociation constant. Substitute equilibrium
terms into the equation and solve for x.
Step 4 Use the following equations:
pOH = − log[OH−]
pH = 14.00 − pOH
Act on Your Strategy
5.9 × 10−2 = 3.5 × 104
Step 1
1.7 × 10−6
Since this value is greater than 500, the amount that dissociates can be
neglected compared to the initial concentration of hydrazine.
Step 2
Concentration (mol/L)
Initial
N2H4(aq)
H2O()
N2H5+(aq) +
OH−(aq)
5.9 × 10−2
0
~0
−x
+x
+x
5.9 × 10−2 − x ≈
x
x
Change
Equilibrium
+
5.9 × 10−2
Step 3
[N2H5+][OH−]
[N2H4]
x2
1.7 × 10−6 =
5.9 × 10−2
√
x = 1.00 × 10−7
Kb =
x = ±3.17 × 10−4
The negative root is not reasonable.
Therefore, x = 3.17 × 10−4 mol/L = [OH−]
Step 5 pOH = − log 3.17 × 10−4
= 3.50
pH = 14.00 − 3.50
= 10.50
Chapter 15 Acid-Base Equilibria and Reactions • MHR
270
Check Your Solution
The pH of the solution is greater than 7, as expected for a basic solution.
13. Problem
Morphine, C17H19NO3, is a naturally occurring base that is used to control pain.
A 4.5 × 10−3 mol/L solution has a pH of 9.93. Calculate Kb for morphine.
What Is Required?
You need to find Kb.
What Is Given?
[C17H19NO3] = 4.5 × 10−3 mol/L
pH = 9.93
Plan Your Strategy
Step 1 Write the equation for the equilibrium reaction of morphine in water. Then
set up an ICE table.
Step 2 Write the equation for the base dissociation constant. Substitute equilibrium
terms into the equation.
Step 3 Use the equation pOH = 14.00 − pH.
Step 4 Use the equation [OH−] = 10−pOH.
Step 5 Substitute for x into the equilibrium equation. Calculate the value of Kb.
Act on Your Strategy
Concentration (mol/L) C17H19NO3(aq) + H2O()
Step 1
Initial
4.5 × 10−3
0
~0
−x
+x
+x
−x
x
x
Change
Equilibrium
C17H19NO3H+(aq) + OH−(aq)
4.5 × 10
−3
[C17H19NO3H+][OH−]
[C17H19NO3]
(x)(x)
=
(4.5 × 10−3 − x)
Step 2
Kb =
Step 3
Step 4
pOH = 14.00 − 9.93 = 4.07
[OH−] = 10−4.07
= 8.5 × 10−5 mol/L
Step 5
4.5 × 10−3 − 8.5 × 10−5 = 4.4 × 10−3
−5 2
Kb = (8.5 × 10 −3)
4.4 × 10
= 1.6 × 10−6
Check Your Solution
The value for Kb is reasonable for a weak organic base. The final answer has two
significant digits, consistent with the two decimal places in the given pH.
14. Problem
Methylamine, CH3NH2, is a fishy-smelling gas at room temperature. It is used to
manufacture several prescription drugs, including methamphetamine. Calculate
[OH−] and pOH of a 0.25 mol/L aqueous solution.
What Is Required?
You need to find [OH−] and the pOH of the solution.
What Is Given?
From Table 8.3, Kb = 4.4 × 10−4
[CH3NH2] = 0.25 mol/L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
271
Plan Your Strategy
[CH3NH2]
to see if the amount that dissociates can
Step 1 Check the value of
Kb
be neglected compared to the initial [CH3NH2].
Step 2 Write the equation for the equilibrium reaction of methylamine in water.
Then set up an ICE table.
Step 3 Write the equation for the base dissociation constant. Substitute equilibrium
terms into the equation and solve for x.
Step 4 Use the equation pOH = − log[OH−].
Act on Your Strategy
0.25
= 568
Step 1
4.4 × 10−4
Since this value is greater than 500, the amount that dissociates can probably
be neglected compared to the initial concentration of methylamine.
Step 2
Concentration (mol/L)
CH3NH2(aq) + H2O()
0.25
0
~0
−x
+x
+x
0.25 − x ≈ 0.25
x
x
Initial
Change
Equilibrium
Step 3
CH3NH3+(aq) + OH−(aq)
[CH3NH3+][OH−]
[CH3NH2]
2
4.4 × 10−4 = x
0.25
√
x = 1.1 × 10−4
Kb =
x = ±1.0 × 10−2
The negative root is not reasonable.
Therefore, x = 1.0 × 10−2 mol/L = [OH−]
Step 4 pOH = − log 1.0 × 10−2
= 1.98
Check Your Solution
The pOH of the solution is less than 7, as expected for a basic solution.
15. Problem
At room temperature, trimethylamine, (CH3)3N, is a gas with a strong ammonia-like
odour. Calculate [OH−] and the percent of trimethylamine molecules that react with
water in a 0.22 mol/L aqueous solution.
What Is Required?
You need to find [OH−] and the percent dissociation.
What Is Given?
From Table 8.3, Kb = 6.5 × 10−5
[(CH3)3N] = 0.22 mol/L
Plan Your Strategy
[(CH3)3N]
to see if the amount that dissociates can
Step 1 Check the value of
Kb
be neglected compared to the initial [(CH3)3N].
Step 2 Write the equation for the equilibrium reaction of trimethylamine in water.
Then set up an ICE table.
Step 3 Write the equation for the base dissociation constant. Substitute equilibrium
terms into the equation and solve for x.
Step 4 Calculate the percent of molecules that dissociated.
Chapter 15 Acid-Base Equilibria and Reactions • MHR
272
Act on Your Strategy
0.22
= 3.4 × 103
Step 1
6.5 × 10−5
Since this value is greater than 500, the amount that dissociates can be
neglected compared to the initial concentration of trimethylamine.
Concentration (mol/L)
Step 2
0
~0
−x
+x
+x
0.22 − x ≈ 0.22
x
x
Change
Step 3
(CH3)3NH+(aq) + OH−(aq)
0.22
Initial
Equilibrium
(CH3)3N(aq) + H2O()
[(CH3)3NH+][OH−]
[(CH3)3N]
2
6.5 × 10−5 = x
0.22
x = 1.43 × 10−5
Kb =
x = ±3.8 × 10−3
The negative root is not reasonable.
Therefore, [OH−] = 3.8 × 10−3 mol/L
3.8 × 10−3 × 100
Step 4 Percent dissociation =
0.22
= 1.7%
Check Your Solution
The [OH−] of the solution is greater than 1.0 × 10−7, as expected for a basic
solution. The percent dissociation is small, consistent with trimethylamine being
a weak base.
16. Problem
An aqueous solution of ammonia has a pH of 10.85. What is the concentration
of the solution?
What Is Required?
You need to find [NH3].
What Is Given?
You know the pH of the solution is 10.85. Kb for aqueous ammonia = 1.8 × 10−5.
Plan Your Strategy
Step 1 Calculate [OH−] using the following equations:
pOH = 14.00 − pH
[OH−] = 10−pOH
Step 2 Write the equation for the equilibrium reaction of ammonia in water. Then
write down the equilibrium concentrations.
Step 3 Write the equation for the base dissociation constant. Substitute equilibrium
terms into the equation and solve for [NH3].
Act on Your Strategy
pOH = 14.00 − 10.85
= 3.15
[OH−] = 10−3.15
= 7.1 × 10−4 mol/L
Step 2
Concentration (mol/L)
Equilibrium
NH3(aq) + H2O()
[NH3] − 7.1 × 10−4 ≈ [NH3]
NH4+(aq) +
7.1 × 10−4
OH−(aq)
7.1 × 10−4
Chapter 15 Acid-Base Equilibria and Reactions • MHR
273
Step 3
+
−
Kb = [NH4 ][OH ]
[NH3]
−4 2
1.8 × 10−5 = (7.1 × 10 )
[NH3]
Therefore, [NH3] = 2.8 × 10−2 mol/L
Check Your Solution
The assumption that the amount of NH3 that dissociates is negligible compared
with the initial concentration is valid because:
2.8 × 10−2 − 7.1 × 10−4 = 2.7 × 10−2 (an error of 4%)
Check the equilibrium values:
(7.1 × 10−4)2 = 1.8 × 10−5
2.8 × 10−2
This is equal to Ka for aqueous ammonia.
Solutions for Practice Problems
Student Textbook page 602
17. Problem
17.85 mL of nitric acid neutralizes 25.00 mL of 0.150 mol/L NaOH(aq). What is the
concentration of the nitric acid?
What Is Required?
You have to find the concentration of nitric acid needed to neutralize the given
amount of NaOH.
What Is Given?
Volume of HNO3 = 17.85 mL = 0.01785 L
Volume of NaOH = 25.00 mL = 0.025 L
Concentration of NaOH = 0.150 mol/L
Plan Your Strategy
Step 1 Write the balanced equation for the reaction.
Step 2 Calculate the number of moles of NaOH used by multiplying its concentration by the given volume.
Step 3 From the mole ratio of acid:base in the balanced equation and the number of
moles of NaOH from step 2, equate and solve for the number of moles of
HNO3 reacted.
Step 4 Divide the number of moles of HNO3 by the given volume to get its concentration.
Act on Your Strategy
Step 1 HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
Step 2 Number of moles of NaOH = 0.15 mol/L × 0.025 L = 0.00375 mol
Step 3 HNO3 reacts with NaOH in a 1:1 ratio so there must be 0.00375 mol
HNO3
Step 4 [HNO3] = 0.00375 mol / 0.01785 L = 0.210 mol/L
Therefore, the concentration of HNO3 is 0.210 mol/L.
Check Your Solution
The concentration of nitric acid is greater than that of sodium hydroxide, so one
would expect the volume of acid needed for the neutralization to be less than the volume of the base. This is the case with the volumes given in the question.
Chapter 15 Acid-Base Equilibria and Reactions • MHR
274
18. Problem
What volume of 1.015 mol/L magnesium hydroxide is needed to neutralize 40.0 mL
of 1.60 mol/L hydrochloric acid?
What Is Required?
You have to find the volume of magnesium hydroxide needed to neutralize the given
amounts of hydrochloric acid.
What Is Given?
Volume of hydrochloric acid = 40.0 mL = 0.04 L
Concentration of hydrochloric acid = 1.60 mol/L
Concentration of magnesium chloride = 1.015 mol/L
Plan Your Strategy
Step 1 Write the balanced equation for the reaction.
Step 2 Calculate the number of moles of hydrochloric acid used by multiplying its
concentration by the given volume.
Step 3 From the mole ratio of acid:base in the balanced equation and the number of
moles of HCl from step 2, equate and solve for the number of moles of
magnesium hydroxide reacted.
Step 4 Divide the number of moles of magnesium hydroxide by the given concentration to get its volume in L.
Act on Your Strategy
Step 1 2HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2H2O(l)
Step 2 Number of moles of HCl = 1.6 mol/L × 0.04 L = 0.064 mol
Step 3 HCl reacts with Mg(OH)2 in a 2:1 ratio. Therefore, the number of moles of
Mg(OH)2 needed is
1 mol Mg(OH)2
× 0.064 mol HCl = 0.032 mol Mg(OH)2
2 mol HCl
0.032 mol
= 0.0315 L = 31.5 mL
Step 4 Volume of Mg(OH)2 = 1.015
mol/L
Therefore, the volume of magnesium hydroxide used is 31.5 mL.
Check Your Solution
The mole ratio of HCl to Mg(OH)2 is 2:1, meaning twice as much acid is needed to
neutralize the base. Although the given concentration of HCl is higher than
Mg(OH)2, it is still nowhere near twice that of the base, so one would expect the
volume of acid needed for the neutralization to still be more than the volume of the
base. This is the case with the volume of Mg(OH)2 calculated. Your answer is
reasonable.
19. Problem
What volume of 0.150 mol/L hydrochloric acid is needed to neutralize each solution
below?
(a) 25.0 mL of 0.135 mol/L sodium hydroxide
(b) 20.0 mL of 0.185 mol/L ammonia solution
(c) 80 mL of 0.0045 mol/L calcium hydroxide
What Is Required?
You need to calculate the volume of hydrochloric acid needed to neutralize the given
amounts of bases.
What Is Given?
Concentration of hydrochloric acid = 0.150 mol/L
(a) Volume of sodium hydroxide = 25.0 mL = 0.025 L;
concentration = 0.135 mol/L
Chapter 15 Acid-Base Equilibria and Reactions • MHR
275
(b) Volume of ammonia solution = 20.0 mL = 0.020 L;
concentration = 0.185 mol/L
(c) Volume of calcium hydroxide = 80 mL = 0.080 L;
concentration = 0.0045 mol/L
Plan Your Strategy
For each case, do the following steps.
Step 1 Write the balanced equation for the reaction.
Step 2 Calculate the number of moles of base used by multiplying its concentration
by the given volume.
Step 3 From the mole ratio of acid:base in the balanced equation and the number of
moles of base from step 2, equate and solve for the number of moles of
hydrochloric acid reacted.
Step 4 Divide the number of moles of hydrochloric acid by the given concentration
to get its volume in L.
Note: An aqueous ammonia solution, i.e., NH3 + H2O, is represented as
NH4(OH)(aq).
Act on Your Strategy
(a) Step 1 HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2 Number of moles of NaOH = 0.135 mol/L × 0.025 L = 0.003375 mol
Step 3 NaOH reacts with HCl in a 1:1 ratio. Therefore, 0.003375 mol NaOH
reacts with 0.003375 mol HCl
mol
= 0.0225 L = 22.5 mL
Step 4 Volume ofHCl = 0.003375
0.150 mol/L
(b) Step 1 HCl(aq) + NH4OH(aq) → NH4Cl(aq) + H2O(l)
Step 2 Number of moles of NH4OH = 0.185 mol/L × 0.020 L = 0.0037 mol
Step 3 NH4OH reacts with HCl in a 1:1 ratio. Therefore, 0.0037 mol NH4OH
reacts with 0.0037 mol HCl
0.0037 mol
= 0.02466 L = 24.7 mL
Step 4 Volume of HCl = 0.150
mol/L
(c) Step 1 2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
Step 2 Number of moles of Ca(OH)2 = 0.0045 mol/L × 0.080 L = 0.00036 mol
Step 3 Ca(OH)2 reacts with HCl in a 1:2 ratio. Therefore, 0.00036 mol
Ca(OH)2 reacts with 0.00072 mol HCl
mol
= 0.0048 L = 4.8 mL
Step 4 Volume of HCl = 0.00072
0.150 mol/L
Check your Solution
(a) The mole ratio of HCl to NaOH is 1:1. The given concentration of HCl is higher
than NaOH, so its volume should be less than that of NaOH. This is the case
with the volume of 22.5 mL calculated. Your answer is reasonable.
(b) The mole ratio of HCl to NH4(OH) is 1:1. The given concentration of HCl is
lower than NH4(OH), so its volume should be more than that of NH4(OH).
This is the case with the volume of 24.7 mL calculated. Your answer is reasonable.
(c) The mole ratio of HCl to Ca(OH)2 is 2:1 meaning twice as much acid is needed
to neutralize the base. However, the given concentration of HCl is already almost
three times that of Ca(OH)2, so its volume should be less than that of Ca(OH)2.
This is the case with the volume of 4.8 mL calculated. Your answer is reasonable.
20. Problem
What concentration of sodium hydroxide solution is needed for each neutralization
reaction?
(a) 37.82 mL of sodium hydroxide neutralizes 15.00 mL of 0.250 mol/L hydrofluoric
acid.
Chapter 15 Acid-Base Equilibria and Reactions • MHR
276
(b) 21.56 mL of sodium hydroxide neutralizes 20.00 mL of 0.145 mol/L sulfuric
acid.
(c) 14.27 mL of sodium hydroxide neutralizes 25.00 mL of 0.105 mol/L phosphoric
acid.
What Is Required?
You need to calculate the concentration of sodium hydroxide used to neutralize the
given amounts of acids.
What Is Given?
(a) Volume of sodium hydroxide = 37.82 mL = 0.03782 L
[hydrofluoric acid] = 0.250 mol/L; volume = 15.00 mL = 0.015 L
(b) Volume of sodium hydroxide = 21.56 mL = 0.02156 L
[sulfuric acid] = 0.145 mol/L; volume = 20.00 mL = 0.020 L
(c) Volume of sodium hydroxide = 14.27 mL = 0.01427 L
[phosphoric acid] = 0.105 mol/L; volume = 25.00 mL = 0.025 L
Plan Your Strategy
For each case, do the following steps.
Step 1 Write the balanced equation for the reaction.
Step 2 Calculate the number of moles of acid used by multiplying its concentration
by the given volume.
Step 3 From the mole ratio of acid:base in the balanced equation and the number of
moles of acid from step 2, equate and solve for the number of moles of
sodium hydroxide used.
Step 4 Divide the number of moles of sodium hydroxide by the given volume to get
its concentration in mol/L.
Act on Your Strategy
(a) Step 1 HF(aq) + NaOH(aq) → NaF(aq) + H2O(l)
Step 2 Number of moles of HF = 0.250 mol/L × 0.015 L = 3.75 × 10−3 mol
Step 3 HF reacts with NaOH in a 1:1 ratio so there must be 3.75 × 10−3 mol
NaOH
× 10−3 mol
= 0.09915 mol/L
Step 4 Concentration of NaOH = 3.750.03782
L
(b) Step 1 H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Step 2 Number of moles of H2SO4 = 0.145 mol/L × 0.02 L = 2.9 × 10−3 mol
Step 3 H2SO4 reacts with NaOH in a 1:2 ratio so there must be 5.8 × 10−3 mol
NaOH
× 10−3 mol
= 0.2690 mol/L
Step 4 Concentration of NaOH = 5.80.02156
L
(c) Step 1 H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
Step 2 Number of moles of H3PO4 = 0.105 mol/L × 0.025 L = 2.625 × 10−3
mol
Step 3 H3PO4 reacts with NaOH in a 1:3 ratio so there must be 7.875 × 10−3
mol NaOH
× 10−3 mol
= 0.5519 mol/L
Step 4 Concentratio’n of NaOH = 7.875
0.01427 L
Check your Solution
(a) The mole ratio of HF to NaOH is 1:1. The given volume of HF is lower than
that of NaOH, so its given concentration must be more than that of NaOH. This
is the case with the [NaOH] of 0.09915 mol/L calculated. Your answer is
reasonable.
(b) The mole ratio of H2SO4 to NaOH is 1:2, meaning twice as much base is needed
to neutralize the acid. The given volume of H2SO4 is only slightly lower than that
of NaOH, so its given concentration should be almost half of that of NaOH. This
Chapter 15 Acid-Base Equilibria and Reactions • MHR
277
is the case with the [NaOH] of 0.2690 mol/L calculated. Your answer is
reasonable.
(c) The mole ratio of H3PO4 to NaOH is 1:3, meaning three times as much base is
needed to neutralize the acid. However, the given volume of NaOH is only a little
over one half that of H2PO4, so its concentration should be very much higher
than that of the acid. This is the case with the [NaOH] of 0.5519 mol/L
calculated. Your answer is reasonable.
Solutions for Practice Problems
Student Textbook page 611
21. Problem
Write balanced stepwise net ionic equation and the overall equation for the reaction
between the following substances in aqueous solution.
(a) sodium thiosulfate and excess hydronium ions
(b) carbonic acid and excess hydroxide ions
What is Required?
You must write the balanced equation for each step and the overall equation for each
reaction.
What is Given?
The word equation is given for each reaction.
Plan Your Strategy
(a) Determine the correct chemical formula for sodium thiosulfate. Sodium will be a
spectator ion in the process and can be omitted from the equations. Since the
thiosulfate ion is dibasic, it will accept two protons and there sill be two steps in
the process. For each step, transfer a proton from H3O+(aq) to the base and write
the conjugate products.
(b) Carbonic acid is a diprotic acid, that is, it contains 2 hydrogen ions, H+(aq).
Carbonic acid will react in 2 steps, transferring a H+(aq) in each step to the base.
For each step, write the conjugate products.
The overall reaction is obtained by adding the two steps in the process.
Act on Your Strategy
–
–
(a) Step 1 S2O32 (aq) + H3O+(aq) → HS2O31 (aq) + H2O(l)
–
Step 2 HS2O31 (aq) + H3O+(aq) → H2S2O3(aq) + H2O(l)
2–
Overall: S2O3 (aq) + 2H3O+(aq) → H2S2O3(aq) + 2H2O(l)
–
(b) Step 1 H2CO3(aq) + OH–(aq) → HCO3 (aq) + H2O(l)
–
–
Step 2 HCO3 (aq) + OH–(aq) → CO32 (aq) + H2O(l)
–
Overall H2CO3(aq) + 2OH–(aq) → CO32 (aq) + 2H2O(l)
Check Your Solution
For the overall reaction in (a), the product is a neutral substance and two hyrdonium
ions were added to the dibasic species. In reaction (b) the product is a dibasic ion
–
(CO32 (aq)) that was produced in a two step process from a diprotic acid, H2CO3(aq).
22. Problem
Sodium carbonate solution, Na2CO3(aq), is a common primary standard for acid
titrants. Write reactions showing the quantitative transfer of protons from the hydronium ions of a strong cid to the sodium carbonate, and the overall reaction for the
titration.
Chapter 15 Acid-Base Equilibria and Reactions • MHR
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What is Required?
You must write the balanced equation for each step and the overall equation for the
reaction.
What is Given?
You know the reactant species and the chemical formula for sodium carbonate.
Plan Your Strategy
The sodium ion is a spectator ion and can be omitted from the equations. The
–
carbonate ion, CO32 (aq), is dibasic and can accept 2 hyrdronium ions, H3O+(aq). This
will be a two step process. The overall reaction is obtained by adding the two steps in
the process.
Act on Your Strategy
–
Step 1 CO32 (aq) + H3O+(aq) → HCO3–(aq) + H2O(l)
Step 2 HCO3–(aq) + H3O+(aq) → H2CO3(aq) + H2O(l)
–
overall CO32 (aq) + 2H3O+(aq) → H2CO3(aq) + H2O(l)
Check Your Solution
The product is a diprotic acid, H2CO3(aq) the was produced from a dibasic ion
–
(CO32 (aq)).
23. Problem
Write the stepwise reactions that occur for the proton transfers between sulfurous
acid and sodium hydroxide, and write the overall neutralization reaction.
What is Required?
You must write the balanced equation for each step and the overall equation for the
reaction.
What is Given?
The names of the reactants are given.
Plan Your Strategy
Sulfurous acid is a diprotic acid, H2SO3(aq) and sodium hydroxide is a strong acid and
a source of hydroxide ions, OH–(aq). The sodium ion is a spectator ion and can be
omitted from the equations. Since there are two hyrdogen ions, H+(aq), in sulfurous
acid, this reaction will be a two step process in which one H+(aq) is removed in each
step. The overall reaction is obtained by adding the two steps in the process.
Act on Your Strategy
Step 1 H2SO3(aq) + OH–(aq) → HSO3–(aq) + H2O(l)
–
Step 2 HSO3–(aq) + OH–(aq) → SO32 (aq) + H2O(l)
–
Overall H2SO3–(aq) + 2OH–(aq) → SO32 (aq) + 2H2O(l)
Check Your Solution
–
The product is a dibasic species, SO32 (aq), that was produced from a diprotic acid,
H2SO3(aq).
Chapter 15 Acid-Base Equilibria and Reactions • MHR
279