CHEMISTRY 11
Chapter 10
Acids and Bases
Solutions for Practice Problems
Student Textbook page 378
1. Problem
Hydrogen cyanide is a poisonous gas at room temperature. When this gas dissolved in
water, the following reaction occurs:
HCN(aq) + H2O(l) → H3O+(aq) + CN−(aq)
Identify the conjugate acid-base pairs.
What Is Required?
You have to identity the conjugate acid-base pairs in the hydrogen cyanide reaction.
What Is Given?
The balanced chemical equation is given.
Plan Your Strategy
Step 1 Identity the proton donor on the left side of the equation. This is the acid.
The other reactant is the proton receiver, which is the base.
Step 2 Identity the proton receiver as a product on the right side of the equation. It
has the same basic formula as the base, plus one extra proton. This is the
conjugate acid of the base reactant.
Step 3 The other product is, by default, the conjugate base of the acid reactant. It has
one less proton than the acid reactant.
Act on Your Strategy
Conjugate Acid-Base Pairs:
HCN(aq) / CN−(aq)
H2O(l) / H3O+(aq)
Check Your Solution
The formulas of the conjugate acid-base pairs differ by only one proton, as expected.
2. Problem
Sodium acetate is a good electrolyte. In water, the acetate ion reacts as follows:
CH3COO−(aq) + H2O(l) → CH3COOH(aq) + OH−(aq)
Identify the conjugate acid-base pairs.
What Is Required?
You have to identity the conjugate acid-base pairs in the sodium acetate reaction.
What Is Given?
The balanced chemical equation is given.
Plan Your Strategy
Step 1 Identity the proton donor on the left side of the equation. This is the acid.
The other reactant is the proton receiver, which is the base.
Step 2 Identity the proton receiver as a product on the right side of the equation.
It has the same basic formula as the base, plus one extra proton. This is the
conjugate acid of the base reactant.
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Step 3 The other product is, by default, the conjugate base of the acid reactant. It has
one less proton than the acid reactant.
Act on Your Strategy
Conjugate Acid-Base Pairs:
CH3COOH(aq) / CH3COO−(aq)
H3O+(aq) / H2O(l)
Check Your Solution
The formulas of the conjugate acid-base pairs differ by only one proton, as expected.
3. Problem
Write equations to show how the hydrogen sulfide ion, HS− , can react with water.
First show the ion acting as an acid, then show the ion acting as a base.
What Is Required?
You have to write balanced equations for the reaction of the hydrogen sulfide ion
with water.
What Is Given?
The reactants are HS− and H2O. HS− can be both an acid and a base.
Plan Your Strategy
HS− as a base: The water is, by default, the acid.
Step 1 Write the formula for the conjugate acid. This will be the same formula as the
base HS ion, plus one proton, making it H2S.
Step 2 Write the formula for the conjugate base. This will be the same formula as the
acid H2O, minus one proton, making it the negative OH− ion.
Step 3 Identify the states of the reactants and products. Water is liquid and the HS
ion is aqueous. The products will be aqueous too.
Step 4 Put the equation together and check that the number of atoms on the left
equal the atoms on the right.
−
HS as an acid: The water is, by default, the base.
Step 1 Write the formula for the conjugate base. This will be the same formula as the
acid HS ion, minus one proton, making it S2− .
Step 2 Write the formula for the conjugate acid. This will be the same formula as the
base H2O, plus one proton, making it the positive H3O+ ion.
Repeat Steps 3 and 4 above.
Act on Your Strategy
As a base: HS−(aq) + H2O(l) → H2S(aq) + OH−(aq)
As an acid: HS−(aq) + H2O(l) → S2−(aq) + H3O+(aq)
Check Your Solution
The formulas of the conjugate acid-base pairs differ by only one proton, as expected.
The number of H, S, and O atoms on the left equal their numbers on the right.
Solutions for Practice Problems
Student Textbook page 384
4. Problem
(a) Write the chemical formula for hydrobromic acid. Then write the name and
formula for the anion that it forms.
(b) Hydrosulfuric acid, H2S, forms two anions. Name them and write their formulas.
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What Is Required?
(a) You need to give the chemical formula of hydrobromic acid and the name and
formula of its anion.
(b) You need to give the chemical formula of hydrosulfuric acid and the name and
formula of its two anions.
What Is Given?
The formula for hydrosulfuric acid is H2S. Hydrobromic acid has one anion, while
hydrosulfuric acid has two anions.
Plan Your Strategy
To determine the chemical formula, we use the naming rules for acids and binary
compounds. Hydrobromic acid is hydrogen bromide, while hydrosulfuric acid is
hydrogen disulfide. We can apply the same steps as used for Practice Problems 16–18
in Chapter 3 to determine the chemical formula.
The anion of the acid is essentially its conjugate base in an acid-base reaction. That is,
it has the acid formula minus one proton, and acquires a negative charge. If the conjugate base is able to dissociate further, it becomes an acid in the next reaction; then
its own conjugate base is its same formula minus one proton, plus an extra negative
charge. The charge of a conjugate base becomes one negative charge for every proton
that is lost.
Act on Your Strategy
(a) HBr(aq) gives the Bromide ion, Br−
(b) H2S(aq) gives the hydrogen sulfide ion, HS−(aq) and the sulfide ion, S2−(aq)
Check Your Solution
The formulas of the ions (conjugate bases) differ by only one proton and one
negative charge to their parent acid formula, as expected.
5. Problem
Write the chemical formula for the following acids. Then name and write the formulas for the oxoanions that form from each acid. Refer to Chapter 3, Table 3.5, Names
and Valences of Some Common Polyatomic Ions, as necessary.
(a) nitric acid
(b) nitrous acid
(c) hyponitrous acid
(d) phosphoric acid
(e) phosphorus acid
(f) periodic acid
What Is Required?
You need to give the chemical formula of the acids listed and then name and write
the formula of its oxoanion.
What Is Given?
The names of the acids are given. Chapter 3, Table 3.5, Names and Valences of Some
Common Polyatomic Ions, is given as a reference guide.
Plan Your Strategy
To determine the chemical formula, we use the naming rules for acids and binary
compounds with polyatomic ions. We can apply the same steps as used for Practice
Problems 16–18 in Chapter 3.
The oxoanion of the acid is essentially its conjugate base in an acid-base reaction.
That is, it has the acid formula minus one proton, and acquires a negative charge. If
the conjugate base is able to dissociate further, it becomes an acid in the next reaction; then its own conjugate base is its same formula minus one proton, plus an extra
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negative charge. The charge of a conjugate base becomes one negative charge for
every proton that is lost from its parent acid. Table 3.5 in Chapter 3 can then be used
to identify the name of the oxoanion created.
Act on Your Strategy
(a) HNO3(aq) ; nitrate NO3−
(b) HNO2(aq); nitrite NO2−
(c) HNO(aq) ; hyponitrite NO−
(d) H3PO4(aq) ; dihydrogen phosphate H2PO4− ; hydrogen phosphate HPO42− ;
phosphate PO43−
(e) H3PO3(aq) ; dihydrogen phosphite H2PO3− ; hydrogen phosphite HPO32− ;
phosphite PO33−
(f) HIO4(aq) periodate IO4−
Check Your Solution
The formulas of the oxoanions (conjugate bases) differ by only one proton and one
negative charge to their parent acid formula, as expected. You can search other reference sources or the Internet to reaffirm the names and formulas of the oxoanions.
Solutions for Practice Problems
Student Textbook page 389
6. Problem
Calculate the pH of each solution, given the hydronium ion concentration.
[H3O+] = 0.0027 mol/L
[H3O+] = 7.28 × 10−8 mol/L
[H3O+] = 9.7 × 10−5 mol/L
[H3O+] = 8.27 × 10−12 mol/L
(a)
(b)
(c)
(d)
What Is Required?
You need to calculate the solution pH for the hydronium ion concentrations listed.
What Is Given?
The [H3O+] is given for each solution.
Plan Your Strategy
Apply the equation: pH = −log [H3O+]
Act on Your Strategy
(a) pH = −log 0.0027 = 2.57
(b) pH = −log 7.28 × 10−8 = 7.14
(c) pH = −log 9.75 × 10−5 = 4.01
(d) pH = −log 8.27 × 10−12 = 11.08
Check Your Solution
(a) [H3O+] > 1.0 × 10−7 mol/L, therefore the pH should be less than 7 and the
solution should be acidic, which it is at 2.57.
(b) [H3O+] is slightly lower than 1.0 × 10−7 mol/L, therefore the pH of the solution
should be just above neutral, which it is at 7.14.
(c) [H3O+] > 1.0 × 10−7 mol/L, therefore the pH should be less than 7 and the
solution should be acidic, which it is at 4.01.
(d) [H3O+] < 1.0 × 10−7 mol/L, therefore pH should be more than 7 and the
solution should be basic, which it is at 11.08.
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7. Problem
[H3O+] in a cola drink is about 5.0 × 10−3 mol/L. Calculate the pH of the drink.
State whether the drink is acidic or basic.
What Is Required?
You have to calculate the pH of a cola drink.
What Is Given?
[H3O+] = 5.0 × 10−3 mol/L
Plan Your Strategy
Use the equation: pH = −log [H3O+]
Act on Your Strategy
pH = −log 5.0 × 10−3 = 2.30. Therefore, the cola drink is acidic.
Check Your Solution
[H3O+] > 1.0 × 10−7 mol/L, therefore the pH should be less than 7 and the cola
drink should be acidic, which it is at 2.3.
8. Problem
A glass of orange juice has [H3O+] of 2.9 × 10−4 mol/L. Calculate the pH of the
juice. State whether the result is acidic or basic.
What Is Required?
You have to calculate the pH of a sample of orange juice.
What Is Given?
[H3O+] = 2.9 × 10−4 mol/L
Plan Your Strategy
Use the equation: pH = −log [H3O+]
Act on Your Strategy
pH = −log 2.9 × 10−4 = 3.54. Therefore, the orange juice is acidic.
Check on Your Solution
[H3O+] > 1.0 × 10−7 mol/L, therefore the pH should be less than 7 and the juice
should be acidic, which it is at 3.54.
9. Problem
(a) [H3O+] in a dilute solution of nitric acid, HNO3 , is 6.3 × 10−3 mol/L. Calculate
the pH of the solution.
(b) [H3O+] of a solution of sodium hydroxide is 6.59 × 10−10 mol/L. Calculate the
pH of the solution.
What Is Required?
You have to calculate the pH of the nitric acid and sodium hydroxide solutions.
What Is Given?
[H3O+] = 6.3 × 10−3 mol/L for HNO3
[H3O+] = 6.59 × 10−10 mol/L for NaOH
Plan Your Strategy
Use the equation: pH = −log [H3O+]
Act on Your Strategy
(a) pH = −log 6.3 × 10−3 = 2.2, therefore, the solution is acidic.
(b) pH = −log 6.59 × 10−10 = 9.18 , therefore, the solution is basic.
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Check Your Solution
(a) [H3O+] > 1.0 × 10−7 mol/L, therefore the pH should be less than 7 and the
solution should be acidic, as would be expected for nitric acid.
(b) [H3O+] < 1.0 × 10−7 mol/L, therefore the pH should be more than 7 and the
solution should be basic, as would be expected for NaOH.
Solutions for Practice Problems
Student Textbook page 398
10. 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.
11. 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
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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 mo l
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.
12. 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
(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.
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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.
13. 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.
(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
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(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
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.
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