Chapter 9 Acids and Bases - websites.rcc.eduwebsites.rcc.edu/grey/files/2012/02/Chapter-9-Solutions-Smith.pdf · Chapter 9–1 Chapter 9 Acids and Bases ... 9.4 Use Example 9.4 to

Chapter 9–1

Chapter 9 Acids and Bases Solutions to In-Chapter Problems 9.1 A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a net positive

or negative charge. Use Example 9.1 to help determine which of the compounds are Brønsted–Lowry acids.

a. HI: contains a H atom, a Brønsted–Lowry acid

c. H2PO4–: contains a H atom,

a Brønsted–Lowry acid b. SO4

2–: no H atom d. Cl–: no H atom 9.2 A Brønsted–Lowry base must contain a lone pair of electrons, but it may be neutral or have a net

negative charge. Use Example 9.2 to help determine which of the compounds are Brønsted–Lowry bases.

a. Al(OH)3: lone pairs on OH, a Brønsted–Lowry base

c. NH4+: no lone pair of electrons

b. Br–: lone pairs on Br, a Brønsted–Lowry base

d. CN–: lone pairs on C and N, a Brønsted–Lowry base

9.3 In each equation, the Brønsted–Lowry acid is the species that loses a proton and the Brønsted–Lowry

base is the species that gains a proton. Use Example 9.3 to help determine which reactant is an acid and which is a base.

HCl(g) NH3(g) Cl–(aq) NH4+(aq)+ +a.acid base

b. CH3COOH(l) H2O(l)+ CH3COO–(aq) H3O+(aq)+

acid base

SO42–(aq)OH–(aq) + HSO4

–(aq)c. H2O(l) +acidbase

9.4 Use Example 9.4 to draw the conjugate acid of each species. Conjugate acid–base pairs differ by the

presence of a proton. To draw a conjugate acid from a base, add a proton, H+. Then add +1 to the charge of the base to give the charge on the conjugate acid.

a. H2O: add one H+ to make H3O+. b. I–: add one H+ to make HI. c. HCO3

–: add one H+ to make H2CO3. 9.5 Use Example 9.5 to draw the conjugate base of each species. Conjugate acid–base pairs differ by the

presence of a proton. To draw a conjugate base from an acid, remove a proton, H+. Then, add –1 to the charge of the acid to give the charge on the conjugate base.

a. H2S: remove one H+ to make HS–. b. HCN: remove one H+ to make CN–. c. HSO4

–: remove one H+ to make SO42–.

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Acids and Bases 9–2

9.6 Use Sample Problem 9.6 to label the acid and base in each reaction. The Brønsted–Lowry acid loses a proton to form its conjugate base. The Brønsted–Lowry base gains a proton to form its conjugate acid.

+ H3O+(aq)a. H2O(l) +I–(aq)HI(g)base acid conjugate acidconjugate base

CH3COOH(l) NH3(g)+ CH3COO–(aq) NH4+(aq)b. +

acid base conjugate acidconjugate base

c. HNO3(aq)Br–(aq) + HBr(aq) NO3–(aq)+

base acid conjugate acid conjugate base

9.7 Draw the conjugate acid and base of ammonia as in Examples 9.4 and 9.5.

a. NH3: add one H+ to make NH4+. b. NH3: remove one H+ to make NH2

–. 9.8 The stronger the acid, the more readily it dissociates to form its conjugate base. In molecular art, the

strongest acid has the most A– and H3O+ ions, and the fewest molecules of undissociated HA.

D (smallest amount of A– and H3O+) < F < E (largest amount of A– and H3O+) 9.9 Use Table 9.1 to determine which acid is stronger. The stronger the acid, the weaker the conjugate

base.

a. H2SO4 is the stronger acid; H3PO4 has the stronger conjugate base. b. HCl is the stronger acid; HF has the stronger conjugate base. c. H2CO3 is the stronger acid; NH4

+ has the stronger conjugate base.

d. HF is the stronger acid; HCN has the stronger conjugate base. 9.10 Draw the conjugate acid of each species as in Example 9.4. Then compare the acids.

a. NO2–: The conjugate acid is HNO2.

NO3–: The conjugate acid is HNO3.

b. Since NO2– is the stronger base, it has a weaker conjugate acid. Therefore, HNO3 is the

stronger acid. 9.11 To determine if the reactants or products are favored at equilibrium:

• Identify the acid in the reactants and the conjugate acid in the products. • Determine the relative strength of the acid and the conjugate acid. • Equilibrium favors the formation of the weaker acid.

acid conjugate acidweaker acid

Products are favored.

H2O(l)OH–(aq)+HF(g) F–(aq) +a.

NH3(g)+ Cl–(aq) HCl(aq)NH4

+(aq) +b.conjugate acidacid

weaker acidReactants are favored.


Chapter 9–3

c. HCO3–(aq) H3O+(aq)+ H2CO3(aq) H2O(l)+


Products are favored. 9.12 Compare the acid and conjugate acid to determine if the reactants or products are favored.

conjugate acidacidweaker acid

C3H6O3(aq) + H3O+(aq)H2O(l) +C3H5O3–(aq)a.

Reactants are favored.


Products are favored.

C3H6O3(aq) HCO3–(aq) H2CO3(aq)+ C3H5O3

–(aq) +b.

9.13 Use Table 9.2 to find the Ka for each acid as in Example 9.7. The acid with the larger Ka is the

stronger acid.

a. increasing acid strength: HPO42–, H2PO4

–, H3PO4 b. increasing acid strength: HCN, CH3COOH, HF

9.14 The stronger acid has the larger Ka. The weaker acid has the stronger conjugate base.

a. H3PO4 is a stronger acid than CH3COOH. b. Remove one H+ to draw the conjugate base: H2PO4

– < CH3COO– (the stronger base is formed from the weaker acid).

9.15 To determine the direction of equilibrium, identify the acid in the reactants and the conjugate acid

in the products as in Sample Problem 9.10. Then compare their Ka values. Equilibrium favors the formation of the acid with the smaller Ka value.

HCO3–(aq) NH3(aq)+ CO3

2–(aq) NH4+(aq)+

acidKa = 5.6 x 10–11

conjugate acidKa = 5.6 x 10–10

weaker acid stronger acid

reactants favored

larger Kasmaller Ka

9.16 Compare the acids by comparing their Ka values from Table 9.2. a. H2CO3 has the larger Ka. b. H2CO3 is stronger. c. HCN has the stronger conjugate base. d. H2CO3 has the weaker conjugate base. e. When H2CO3 is dissolved in water, the equilibrium lies further to the right.



9.17 Use the equation [OH–] = Kw/[H3O+] to calculate the hydroxide ion concentration as in Sample Problem 9.11. When [OH–] > [H3O+], the solution is basic. When [OH–] < [H3O+], the solution is acidic.

acidic

1.0 x 10–14

[H3O+][OH–] =

Kw =10–3

= 10–11 Ma.

basic

1.0 x 10–14

[H3O+][OH–] =

Kw =10–11

= 10–3 Mb.

1.0 x 10–14

[H3O+][OH–] =

Kw =2.8 x 10–10

= 3.6 x 10–5 Mc.

basic

1.0 x 10–14

[H3O+][OH–] =

Kw =5.6 x 10–4

= 1.8 x 10–11 Md.

acidic 9.18 Use the equation [H3O+] = Kw/[OH–] to calculate the hydronium ion concentration as in Sample

Problem 9.11.

1.0 x 10–14[H3O+]

[OH–]=

Kw =10–6

= 10–8 Mbasic

a.

1.0 x 10–14[H3O+]

[OH–]=

Kw =10–9

= 10–5 Macidic

b.

1.0 x 10–14[H3O+]

[OH–]=

Kw =5.2 x 10–11

= 1.9 x 10–4 Macidic

c.

1.0 x 10–14[H3O+]

[OH–]=

Kw =7.3 x 10–4

= 1.4 x 10–11 Mbasic

d.

9.19 Since NaOH is a strong base that completely dissociates to form Na+ and OH–, the concentration

of NaOH gives the concentration of OH– ions. The [OH–] can then be used to calculate [H3O+] from the expression for Kw. Similarly, since HCl is a strong acid and completely dissociates, the [H3O+] can then be used to calculate [OH–] from the expression for Kw. See Example 9.8.

a. 1 x 10–14[H3O+]

[OH–]=

Kw =10–3

= 10–11 M

concentration of OH–

concentration of H3O+

b. 1 x 10–14

[H3O+][OH–] =

Kw =10–3

= 10–11 Mconcentration of OH–



Chapter 9–5

c. 1 x 10–14

[H3O+][OH–] =

Kw =1.5

= 6.7 x 10–15 Mconcentration of OH–


d. 1 x 10–14

[H3O+][OH–]

=Kw =

3.0 x 10–1= 3.3 x 10–14 M

concentration of OH–


9.20 When the coefficient of a number written in scientific notation is one, the pH equals the value of

x in 10–x.

a. 1 × 10–6 M: pH = 6 c. 0.000 01 M = 10–5: pH = 5 b. 1 × 10–12 M: pH = 12 d. 0.000 000 000 01 M = 10–11: pH = 11

9.21 The pH equals the value of x in 10–x, and this gives the H3O+ concentration. An acidic solution

has a pH < 7. A basic solution has a pH > 7. A neutral solution has a pH = 7.

a. pH = 13, [H3O+] = 1 × 10–13 M: basic c. pH = 3, [H3O+] = 1 × 10–3 M: acidic b. pH = 7, [H3O+] = 1 × 10–7 M: neutral

9.22 Use a calculator to determine the antilogarithm of (– pH); [H3O+] = antilog(–pH).

[H3O+] = antilog(–pH) = antilog(–4.3)

5 x 10–5 M[H3O+] =

c.


2 x 10–8 M[H3O+] =

b.


6 x 10–11 M[H3O+] =

a.

9.23 Use a calculator to determine the logarithm of a number that contains a coefficient other than one

in scientific notation; pH = –log [H3O+].

pH = –log [H3O+] –log(7.62 x 10–11)=

–(–10.118) = 10.118=

d.

pH = –log [H3O+] –log(8.8 x 10–5)=

–(–4.06) = 4.06=

c.

pH = –log [H3O+] –log(9.21 x 10–12)=

–(–11.036) = 11.036=

b.

pH = –log [H3O+] –log(1.8 x 10–6)=

–(–5.74) = 5.74=

a.

9.24 Label the organs based on the definitions listed in Answer 9.21.

saliva—acidic to slightly basic blood—basic pancreas—basic stomach—acidic small intestines—basic large intestines—acidic to neutral urine—acidic to basic



9.25 Write the balanced equations as in Section 9.7A.

HNO3(aq) NaOH(aq)+a. H2O(l) + NaNO3(aq)water salt

b.

H2SO4(aq) 2 KOH(aq)+ 2 H2O(l) + K2SO4(aq)

Place a 2 to balance K. Place a 2 to balance H and O.

H2SO4(aq) KOH(aq)+ H2O(l) + K2SO4(aq)water salt

9.26 A net ionic equation contains only the species involved in a reaction.

H+(aq) + OH–(aq) H2O(l) for Problem 9.25a, b 9.27 The acid and base react to form a salt and carbonic acid (H2CO3), which decomposes to CO2 and

H2O as in Sample Problem 9.17.

+ CaSO4(aq)CaCO3(s)H2SO4(aq) + H2O(l) + CO2(g)from H2CO3

9.28 The acid and base react to form a salt and carbonic acid (H2CO3), which decomposes to CO2 and

H2O as in Sample Problem 9.17.

+ H2O(l) + CO2(g)

from H2CO3

+a. NaHCO3(aq) NaNO3(aq)HNO3(aq)

+ MgCO3(s) Mg(NO3)2(aq)2 HNO3(aq)b. + H2O(l) + CO2(g)

from H2CO3


Chapter 9–7

9.29 Follow Example 9.10 to determine the acidity when a salt is dissolved in water. Determine what types of acid and base (strong or weak) are used to form the salt. When the ions in the salt come from a strong acid and strong base, the solution is neutral. When the ions come from acids and bases of different strength, the ion derived from the stronger reactant determines the acidity.

KI Ca(NO3)2 BaCl2

K+ I– NO3– Ba2+

from KOH from HIstrong base

Ca2+

from Ca(OH)2strong base

from HNO3strong acid

Cl–

from Ba(OH)2 from HClstrong acid

neutral solution neutral solution

a. c. e.

K2CO3 NH4I Na3PO4

K+ CO32– I– Na+

from KOH from HCO3–

strong base weak acidfrom HI

strong acid strong base

PO43–

from NaOH from HPO42–

weak acid

basic solution basic solution

b. d. f.

strong acid

neutral solution

NH4+

weak basefrom NH3

acidic solution

strong base

9.30 A basic solution has a pH > 7.

LiCl NH4Br

Li+ Cl– Br–

from LiOH from HClstrong base

from HBrstrong acid

acidic solution

a. c.

K2CO3 MgCO3

K+ CO32– CO3

2–

from KOH from HCO3–

strong base weak acidfrom HCO3

–

weak acidbasic solution

pH > 7

b. d.

strong acid

neutral solution

Mg2+

strong basefrom Mg(OH)2

NH4+

weak basefrom NH3

basic solutionpH > 7



9.31 Calculate the molarity of the solution using a titration as in Example 9.11.

25.5 mL NaOH 0.24 mol NaOH1 L

x1 L

1000 mLx = 0.0061 mol NaOH

0.0061 mol NaOH1 mol NaOH1 mol HClx = 0.0061 mol HCl

0.0061 mol HClx

1 L1000 mL=

15 mL solutionM

molarity

molL

= = 0.41 M HClAnswer

9.32 Calculate the number of milliliters of NaOH solution needed.

5.0 mL H2SO46.0 mol H2SO4

1 Lx

1 L1000 mL

x = 0.030 mol H2SO4

0.030 mol H2SO41 mol H2SO4

2 mol NaOHx = 0.060 mol NaOH

0.060 mol NaOH x1 L

1000 mL= 30. mL NaOH solutionx

2.0 mol NaOH1 L

9.33 A buffer is a solution whose pH changes very little when acid or base is added. Most buffers are solutions composed of approximately equal amounts of a weak acid and the salt of its conjugate base.

a. A solution containing HBr and NaBr is not a buffer, because it contains a strong acid, HBr. b. A solution containing HF and KF is a buffer since HF is a weak acid and F– is its conjugate

base. c. A solution containing CH3COOH alone is not a buffer since it contains a weak acid only.

9.34

a. HCO3– and CO3

2– are both needed because the H3O+ concentration depends on two terms—Ka, which is a constant, and the ratio of the concentrations of the weak acid and its conjugate base. If these concentrations do not change much, the concentration of H3O+ and therefore the pH do not change much.

b. When a small amount of acid is added, [HCO3–] increases and [CO3

2–] decreases. c. When a small amount of base is added, [HCO3

–] decreases and [CO32–] increases.


Chapter 9–9

9.35 Calculate the pH of the buffer as in Example 9.12. The pH is the same if equal concentrations of the weak acid and conjugate base are present.

[H3O+] [H2PO4–]

[HPO42–]

Ka= x (6.2 x 10–8) x[0.10 M][0.10 M]

=

[H3O+] = 6.2 x 10–8 M

pH –log [H3O+]= –log(6.2 x 10–8)=

=pH 7.21

a.

[H3O+][H2PO4

–]

[HPO42–]

Ka= x (6.2 x 10–8) x[1.0 M][1.0 M]

=

[H3O+] = 6.2 x 10–8 M

pH –log [H3O+]= –log(6.2 x 10–8)=

=pH 7.21

b.

[HPO42–]

[H3O+][H2PO4

–]Ka= x (6.2 x 10–8) x

[0.50 M][0.50 M]

=

[H3O+] = 6.2 x 10–8 M

pH –log [H3O+]= –log(6.2 x 10–8)=

=pH 7.21

c.

9.36 Calculate the pH of the buffer as in Example 9.12.

pH –log [H3O+]= –log(2.4 x 10–5)=

=pH 4.62

[H3O+][CH3COOH][CH3COO–]

Ka= x (1.8 x 10–5) x[0.20 M][0.15 M]

=

[H3O+] = 2.4 x 10–5 M



Solutions to End-of-Chapter Problems 9.37 A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a net

positive or negative charge. Use Example 9.1 to help determine which of the compounds are Brønsted–Lowry acids.

a. HBr: contains a H atom, a Brønsted–Lowry acid

d. HCOOH: contains a H atom, a Brønsted–Lowry acid

b. Br2: no H atom e. NO2–: no H atom

c. AlCl3: no H atom f. HNO2: contains a H atom, a Brønsted–Lowry acid

9.38 A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a net

positive or negative charge. Use Example 9.1 to help determine which of the compounds are Brønsted–Lowry acids.

a. H2O: contains a H atom, a Brønsted–Lowry acid

d. FeBr3: no H atom

b. I–: no H atom e. CH3CH2COOH: contains a H atom, a Brønsted–Lowry acid

c. HOCl: contains a H atom, a Brønsted–Lowry acid

f. CO2: no H atom

9.39 A Brønsted–Lowry base must contain a lone pair of electrons, but it may be neutral or have a net


a. OH–: lone pairs on OH, a Brønsted–Lowry base

d. PO43–: lone pairs on O,

a Brønsted–Lowry base b. Ca2+: no lone pair of electrons e. OCl–: lone pairs on O and Cl,

a Brønsted–Lowry base c. C2H6: no lone pair of electrons f. MgCO3: lone pairs on O,

a Brønsted–Lowry base 9.40 A Brønsted–Lowry base must contain a lone pair of electrons, but it may be neutral or have a net


a. Cl–: lone pairs on Cl, a Brønsted–Lowry base

d. Na+: no lone pair of electrons

b. BH3: no lone pair of electrons e. Ca(OH)2: lone pairs on O, a Brønsted–Lowry base

c. H2O: lone pairs on O, a Brønsted–Lowry base

f. HCOO–: lone pairs on O, a Brønsted–Lowry base


Chapter 9–11

9.41 Draw the conjugate acid of each species as in Example 9.4.

a. HS–: add one H+ to make H2S. b. CO32–: add one H+ to make HCO3

–.

c. NO2

–: add one H+ to make HNO2. H C

H

H

N H

H

d. H C

H

H

N H

H

H +

add one H+

to make

9.42 Draw the conjugate acid of each species as in Example 9.4.

a. Br–: add one H+ to make HBr

O P

O

O

Ob. O P

O

O

O H

–

add one H+

to makeH

2–

H

H C

H

H

C Oc. add one H+

to make

O –

H C

H

H

C O

O

H

H C

H

H

O Hd. add one H+

to makeH C

H

H

O H

H +

9.43 Draw the conjugate base of each species as in Example 9.5.

a. HNO2: remove one H+ to make NO2–. c. H2O2: remove one H+ to make HO2

–. b. NH4

+: remove one H+ to make NH3.

9.44 Draw the conjugate base of each species as in Example 9.5. a. H3O+: remove one H+ to make H2O c. HSO4

–: remove one H+ to make SO42–

c. H2Se: remove one H+ to make HSe–

9.45 The acid loses a proton (gray sphere) to form its conjugate base, while the base gains a proton to form its conjugate acid.

+ +

+ +

a.

b.


acidbase conjugate acid conjugate base



9.46 The acid loses a proton to form its conjugate base, while the base gains a proton to form its conjugate acid.

+ +

a.

b.



+ +

9.47 Label the conjugate acid–base pairs in each reaction as in Answer 9.6.

HI(g)

NH3(g) NH4+(aq)

I–(aq)a. acidbase conjugate acid

conjugate base

HCOOH(l)H2O(l) H3O+(aq)

HCOO–(aq)b. acidbase

conjugate base

conjugate acid

c. HSO4–(aq) H2SO4(aq)

OH–(aq)H2O(l) acidbase conjugate acid

conjugate base 9.48 Label the conjugate acid–base pairs in each reaction as in Answer 9.6.

Cl—(aq)HSO4

—(aq) SO42—(aq)

HCl(aq)a. baseacid conjugate base

conjugate acid

b. HPO42–(aq) PO4

3–(aq)

H2O(l)–OH(aq) baseacid conjugate base

conjugate acid

HF(aq)

NH3(aq) NH4+(aq)

F–(aq)

c.

acid

base

conjugate base

conjugate acid

9.49 Draw the conjugate acid and base of HCO3

–.

a. conjugate acid: H2CO3 b. conjugate base: CO32–

9.50 Draw the conjugate acid and base of H2PO4

–. a. conjugate acid: H3PO4

b. conjugate base: HPO42–

9.51 Draw the acid–base reaction.

HNO3(aq) + H2O H3O+(aq) + NO3–(aq)(l)


Chapter 9–13

9.52 Draw the acid–base reaction.

HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq) 9.53 A represents HCl because it shows a fully dissociated acid. B represents HF, only partially

dissociated. 9.54 A represents HCN since HCN is a weak acid and will only partially dissociate in water. 9.55 a. B represents a strong acid because HZ is completely dissociated, forming H3O+ and Z–.

b. A represents a weak acid because most of the acid HZ remains, and few ions (H3O+ and Z–) are formed.

9.56

b.

H2Z HZ–+ +

H2O H3O+

a.

H2Z Z2–+ +

H2O H3O+

2 2

9.57 Use Tables 9.1 and 9.2 to determine the stronger acid as in Example 9.7. The acid with the larger Ka is the stronger acid.

a. CH3COOH is stronger than H2O. c. H2SO4 is stronger than HSO4

–. b. H3PO4 is stronger than HCO3

–. 9.58 Use Tables 9.1 and 9.2 to determine the stronger acid as in Example 9.7. The acid with the larger

Ka is the stronger acid.

a. HCN is stronger than HPO42–. c. HF is stronger than H2O.

b. HSO4–

is stronger than NH4+.

9.59 The weaker acid has the stronger conjugate base.

a. H2O b. HCO3– c. HSO4

– 9.60 The weaker acid has the stronger conjugate base. a. HPO4

2– b. NH4+ c. H2O

9.61 a. An acid that dissociates to a greater extent in water is a stronger acid: A is the stronger acid.

b. An acid with a smaller Ka is weaker (A is weaker): B is the stronger acid. c. An acid with a stronger conjugate base is a weaker acid (A is weaker): B is the stronger acid.

9.62 a. An acid that dissociates to a greater extent in water is the stronger acid: B is the stronger acid. b. An acid with a larger Ka is stronger (A is stronger): A is the stronger acid. c. An acid with a stronger conjugate base is a weaker acid (B is weaker): A is the stronger acid.



9.63 The acid with the larger Ka is the stronger acid. The stronger acid has a weaker conjugate base.

a. HSO4– H2PO4

–

stronger acid conjugate base

HPO42–

conjugate basestronger base

SO42–

Ka = 6.2 x 10–8

Ka = 1.2 x 10–2

stronger acidlarger Ka

Ka = 1.3 x 10–5CH3COOHb. CH3CH2COOH

stronger acidCH3COO–

conjugate baseCH3CH2COO–

conjugate basestronger baseKa = 1.8 x 10–5


9.64 The acid with the larger Ka is the stronger acid. The stronger acid has a weaker conjugate base.

a. H3PO4 HCOOH


HCOO–


H2PO4–

Ka = 1.8 x 10–4

Ka = 7.5 x 10–3


b. HCOOH C6H5COOH


C6H5COO–


HCOO–

Ka = 6.5 x 10–5

Ka = 1.8 x 10–4


9.65 The stronger acid has more dissociated ions. The stronger acid has the larger Ka.

a. B has more dissociated ions (A– and H3O+) and is therefore the stronger acid. b. B has the larger Ka since it is the stronger acid.

9.66 The stronger acid has the weaker conjugate base. Since A is the weaker acid it will have the

stronger conjugate base. 9.67 To determine if the reactants or products are favored at equilibrium:


a. H3PO4(aq) CN–(aq)+ H2PO4

–(aq) + HCN(aq)acid conjugate acid

weaker acidProducts are favored.

b. Br–(aq) HSO4

–(aq) SO42–(aq) HBr(aq)+ +




Chapter 9–15

c. H2CO3(aq)CH3COO–(aq) + CH3COOH(aq) + HCO3–(aq)


Reactants are favored. 9.68 To determine if the reactants or products are favored at equilibrium:


a. HF(aq) NH3(aq)+ F–(aq)+NH4

+(aq)acid conjugate acid

weaker acidProducts are favored.

b. Br–(aq) H2O(l)+ –OH(aq)+HBr(aq)



c. HCN(aq) HCO3–(aq)+ –CN(aq)+H2CO3(aq)

acidweaker acid


conjugate acid

9.69 Use the equation [OH–] = Kw/[H3O+] to calculate the hydroxide ion concentration as in Answer

9.17. When [OH–] > [H3O+], the solution is basic. When [OH–] < [H3O+], the solution is acidic.

basic

1.0 x 10–14

[H3O+][OH–] =

Kw =10–8

= 10–6 Ma.

basic

1.0 x 10–14

[H3O+][OH–] =

Kw =10–10

= 10–4 Mb.

acidic

1.0 x 10–14

[H3O+][OH–] =

Kw =3.0 x 10–4

= 3.3 x 10–11 Mc.

basic

1.0 x 10–14

[H3O+][OH–] =

Kw =2.5 x 10–11

= 4.0 x 10–4 Md.

9.70 Use the equation [–OH] = Kw/[H3O+] to calculate the hydroxide ion concentration as in Answer 9.17. When [–OH] > [H3O+], the solution is basic. When [–OH] < [H3O+], the solution is acidic.

acidic

1.0 x 10–14

[H3O+][–OH] =

Kw =10–1

= 10–13 Ma.



basic

1.0 x 10–14

[H3O+][–OH] =

Kw =10–13

= 10–1 Mb.

acidic

1.0 x 10–14

[H3O+][–OH] =

Kw =2.6 x10–7

= 3.8 x 10–8 Mc.

basic

1.0 x 10–14

[H3O+][–OH] =

Kw =1.2 x10–12

= 8.3 x 10–3 Md.

9.71 Use the equation [H3O+] = Kw/[OH–] to calculate the hydronium ion concentration as in Sample Problem 9.11.

1.0 x 10–14

[H3O+][OH–]

=Kw =

10–2= 10–12 M

basica.

1.0 x 10–14[H3O+]

[OH–]=

Kw =4.0 x 10–8


b.

1.0 x 10–14[H3O+]

[OH–]=

Kw =6.2 x 10–7

= 1.6 x 10–8 Mbasic

c.

1.0 x 10–14[H3O+]

[OH–]=

Kw =8.5 x 10–13


d.

9.72 Use the equation [H3O+] = Kw/[–OH] to calculate the hydroxide ion concentration as in Sample

Problem 9.11. When [H3O+] > [–OH], the solution is acidic. When [H3O+] < [–OH], the solution is basic.

1.0 x 10–14

[H3O+][–OH]

=Kw =

10–12= 10–2 M

acidica.

1.0 x 10–14[H3O+]

[–OH]=

Kw =5.0 x 10–10


b.

1.0 x 10–14[H3O+]

[–OH]=

Kw =6.0 x 10–4

= 1.7 x 10–11 Mbasic

c.

1.0 x 10–14[H3O+]

[–OH]=

Kw =8.9 x 10–11


d.


Chapter 9–17

9.73 Use a calculator to determine the logarithm of a number that contains a coefficient other than one in scientific notation; pH = –log [H3O+] as in Answer 9.23.

pH = –log [H3O+] –log(1.2 x 10–2)=

–(–1.92) = 1.92=

d.

pH = –log [H3O+] –log(1.6 x 10–8)=

–(–7.80) = 7.80=

c.

pH = –log [H3O+] –log(2.5 x 10–7)=

–(–6.60) = 6.60=

b.

pH = –log [H3O+] –log(10–12)=

–(–12) = 12=

a.


in scientific notation; pH = –log [H3O+] as in Answer 9.25.

pH = –log [H3O+] –log(1.1 x 10–4)=

–(–3.96) = 3.96=

d.

pH = –log [H3O+] –log(1.7 x 10–11)=

–(–10.77) = 10.77=

c.

pH = –log [H3O+] –log(2.0 x 10–5)=

–(–4.70) = 4.70=

b.

pH = –log [H3O+] –log(10–2)=

–(–2) = 2=

a.

9.75

[H3O+] [OH–] pH Classification 5.3 × 10–3 1.9 × 10–12 2.28 acidic 5.0 × 10–7 2.0 × 10–8 6.30 acidic 4 × 10–5 2.5 × 10–10 4.4 acidic

1.5 × 10–5 6.8 × 10–10 4.82 acidic 9.76

[H3O+] [OH–] pH Classification 6.3 × 10–4 1.6 × 10–11 3.20 acidic 4.0 × 10–6 2.5 × 10–9 5.40 acidic 2.5 × 10–5 4.0 × 10–10 9.4 basic 1.5 × 10–11 6.8 × 10–4 10.83 basic

9.77 Use a calculator to determine the antilogarithm of (–pH); [H3O+] = antilog(–pH).


1.6 x 10–2 M[H3O+] =

c.

[H3O+] = antilog(–pH) = antilog(–1)

1 x 10–1 M[H3O+] =

b.


1 x 10–12 M[H3O+] =

a.


1.3 x 10–9 M[H3O+] =

d.



9.78 Use a calculator to determine the antilogarithm of (–pH); [H3O+] = antilog(–pH).


2.5 x 10–3 M[H3O+] =

c.


1 x 10–8 M[H3O+] =

b.


1 x 10–4 M[H3O+] =

a.


5.0 x 10–12 M[H3O+] =

d.

9.79 Use the equations in Answers 9.77 and 9.69 to calculate the concentrations of H3O+ and OH– in

the sample.


1.3 x 10–6 M[H3O+] =

1.0 x 10–14

[H3O+][OH–] =

Kw =1.3 x 10–6

= 7.7 x 10–9 M

9.80 Use the equations in Problems 9.77 and 9.69 to calculate the concentrations of H3O+ and –OH in

the sample.


6 x 10–9 M[H3O+] =

1.0 x 10–14

[H3O+][–OH] =

Kw =6 x 10–9

= 2 x 10–6 M

9.81 Use the equations in Answers 9.77 and 9.69 to calculate the concentrations of H3O+ and OH– in

the sample.


7.9 x 10–5 M[H3O+] =

1.0 x 10–14

[H3O+][OH–] =

Kw =7.9 x 10–5

= 1.3 x 10–10 M

9.82 Use the equations in Problems 9.77 and 9.69 to calculate the concentrations of H3O+ and –OH in

the sample. [H3O+] = antilog(–pH) = antilog(–3.15)

7.1 x 10–4 M[H3O+] =

1.0 x 10–14

[H3O+][–OH] =

Kw =7.1 x 10–4

= 1.4 x 10–11 M


Chapter 9–19


in scientific notation; pH = –log [H3O+].

1.0 x 10–14[H3O+][OH–]

=Kw =

1.5 x 10–2= 6.7 x 10–13 M pH = –log [H3O+] –log(6.7 x 10–13)=

–(–12.17) = 12.17=

b.

pH = –log [H3O+] –log(2.5 x 10–3)=

–(–2.60) = 2.60=

a.

9.84 Use a calculator to determine the logarithm of a number that contains a coefficient other than one in scientific notation; pH = –log [H3O+].

1.0 x 10–14[H3O+][–OH]

=Kw =

0.0025= 4.0 x 10–12 M pH = –log [H3O+] –log(4.0 x 10–12)=

–(–11.40) = 11.40=

b.

pH = –log [H3O+] –log(0.015)=

–(–1.82) = 1.82=

a.

9.85 The pH of 0.10 M HCl is lower than the pH of 0.1 M CH3COOH solution (1.0 vs. 2.9) because

HCl is fully dissociated to H3O+ and Cl–, whereas CH3COOH is not fully dissociated. 9.86 The pH of 0.0050 M CH3COOH solution is higher than the pH of 0.0050 M HCl (3.5 vs. 2.3)

because HCl is fully dissociated to H3O+ and Cl–, whereas CH3COOH is not fully dissociated. 9.87 Write the balanced equations as in Section 9.7.

HBr(aq)a. KOH(aq)+ KBr(aq) + H2O(l) b. 2 HNO3(aq) Ca(OH)2(aq)+ 2 H2O(l) + Ca(NO3)2(aq)

NaHCO3(aq)c. HCl(aq) + NaCl(aq) + H2O(l) + CO2(g)

Mg(OH)2(aq)H2SO4(aq) +d. 2 H2O(l) + MgSO4(aq)

9.88 Write the balanced equations as in Section 9.7.

HNO3(aq)a. LiOH(aq)+ LiNO3(aq) + H2O(l)

2 NaOH(aq)H2SO4(aq) +b. 2 H2O(l) + Na2SO4(aq)

2 HCl(aq)K2CO3(aq) +c. H2O(l) + CO2(g) + 2 KCl(aq)



NaHCO3(aq)HI(aq) +d. H2O(l) + CO2(g) + NaI(aq) 9.89 Write the balanced equation.

2 HNO3(aq) + CaCO3(s) Ca(NO3)2(aq) + CO2(g) + H2O(l)

9.90 Write the balanced equation. Al(OH)3(s) + 3 HCl(aq) AlCl3(aq) + 3 H2O(l) 9.91 Follow Example 9.10 to determine the acidity when a salt is dissolved in water. Determine what

types of acid and base (strong or weak) are used to form the salt. When the ions in the salt come from a strong acid and strong base, the solution is neutral. When the ions come from acids and bases of different strength, the ion derived from the stronger reactant determines the acidity.

NaI NH4NO3 MgBr2

Na+ I– NO3– Mg2+

from NaOH from HIstrong base

NH4+

from NH3weak base

from HNO3strong acid

Br–

from Mg(OH)2 from HBrstrong acid

acidic solution neutral solution

a. c. e.

LiF KHCO3 NaH2PO4

Li+ F– HCO3– Na+

from LiOH from HFstrong base weak acid

from H2CO3weak acid strong base

H2PO4–

from NaOH from H3PO4weak acid


b. d. f.

strong acid

neutral solution

K+

strong basefrom KOH

basic solution

strong base

9.92 Follow Example 9.10 to determine the acidity when a salt is dissolved in water. Determine what

types of acid and base (strong or weak) are used to form the salt. When the ions in the salt come from a strong acid and strong base, the solution is neutral. When the ions come from acids and bases of different strength, the ion derived from the stronger reactant determines the acidity.


Chapter 9–21

NaBr KCH3COO CaBr2

Na+ Br– CH3COO– Ca2+

from NaOH from HBrstrong base

K+

strong basefrom CH3COOHweak acid

Br–

from Ca(OH)2 from HBrstrong acid

basic solution neutral solution

a. c. e.

NaCN CsF K3PO4

Na+ CN– F– K+

from NaOH from HCNstrong base weak acid

from HFweak acid strong base

PO43–

from KOH from H3PO4weak acid


b. d. f.

strong acid

neutral solution

Cs+

strong basefrom CsOH

basic solution

strong basefrom KOH



x1 L

1000 mLx = 0.0036 mol NaOH


0.0036 mol HClx

1 L1000 mL=

25 mL solutionM

molarity

molL




x1 L

1000 mLx = 0.0026 mol NaOH


0.0026 mol HClx

1 L1000 mL=

5.00 mL solutionM

molarity

molL






x1 L

1000 mLx = 0.0031 mol NaOH

0.0031 mol NaOH1 mol NaOH

1 mol CH3COOHx = 0.0031 mol CH3COOH

0.0031 mol CH3COOHx

1 L1000 mL=

25 mL solutionM

molarity

molL

= = 0.12 M CH3COOHAnswer



x1 L

1000 mLx = 0.0033 mol NaOH

0.0033 mol NaOH2 mol NaOH1 mol H2SO4x = 0.0017 mol H2SO4

0.0017 mol H2SO4x

1 L1000 mL=

25.0 mL solutionM

molarity

molL

= = 0.068 M H2SO4Answer

9.97 Calculate the number of milliliters of solution needed.

10.0 mL CH3COOH 2.5 mol CH3COOH1 L

x1 L

1000 mLx = 0.025 mol CH3COOH

0.025 mol CH3COOH1 mol CH3COOH


0.025 mol NaOH x1 L

1000 mL = 25 mL NaOH solutionx1.0 mol NaOH

1 L

9.98 Calculate the number of milliliters of solution needed.

8.0 mL H2SO43.5 mol H2SO4

1 Lx

1 L1000 mL

x = 0.028 mol H2SO4

0.028 mol H2SO4 1 mol H2SO4


0.056 mol NaOH x1 L

1000 mL = 28 mL NaOH solutionx2.0 mol NaOH

1 L


Chapter 9–23

9.99 B is a buffer because it contains equal amounts of the acid and its conjugate base. A is not a buffer since it contains only acid.

9.100 A is a buffer because it contains equal amounts of the acid and its conjugate base. B is not a buffer since it contains only conjugate base.

9.101 A buffer is most effective at minimizing pH changes when the concentrations of the weak acid

and its conjugate base are equal because when acid or base is added to a buffer, a proton donor and a proton acceptor are available to react with either of them. Since the ratio of the concentration of the acid and conjugate base is one to begin with, the ratio stays close to one after acid or base is added.

9.102 A buffer is a solution whose pH changes very little when acid or base is added. A buffer can also

be prepared from a weak base and its conjugate acid because a proton donor and a proton acceptor are available to react with any added acid or base.

9.103 Yes, a buffer can be prepared from equal amounts of NaCN and HCN. HCN is a weak acid and

CN– is its conjugate base, so in equal amounts they form a buffer. 9.104 No, a buffer cannot be prepared from equal amounts of HNO3 and KNO3. A buffer is prepared

from a weak acid (or weak base) and its conjugate base (or conjugate acid). HNO3 is a strong acid and completely dissociates in solution. KNO3 is a soluble salt and also completely dissociates in solution.

9.105 a. Both HNO2 and NO2

– are needed to prepare the buffer since HNO2 is a proton donor that will react with added base and NO2

– is a proton acceptor that will react with added acid. b. When a small amount of acid is added to the buffer, the concentration of HNO2 increases and

the concentration of NO2– decreases.

c. The concentration of HNO2 decreases and the concentration of NO2– increases when a small

amount of base is added to the buffer. 9.106 a. Both HF and F– are needed to prepare the buffer since HF is a proton donor that will react with

added base and F– is a proton acceptor that will react with added acid. b. When a small amount of acid is added to the buffer, the concentration of HF increases and the

concentration of F– decreases. c. The concentration of HF decreases and the concentration of F– increases when a small amount

of base is added to the buffer. 9.107 Calculate the pH of the buffer as in Example 9.12.

[H3O+] [Na2HPO4][Na3PO4]

Ka= x x[0.10 M][0.10 M]=

[H3O+] = 2.2 x 10–13 M

pH –log [H3O+]= –log(2.2 x 10–13)=

=pH 12.66

a. 2.2 x 10–13



[H3O+] [NaHCO3][Na2CO3]

Ka= x x[0.22 M][0.22 M]

=

[H3O+] = 5.6 x 10–11 M

pH –log [H3O+]= –log(5.6 x 10–11)=

=pH 10.25

b. 5.6 x 10–11


[H3O+] [CH3COOH][NaCH3COO]

Ka= x x[0.55 M][0.55 M]=

[H3O+] = 1.8 x 10–5 M

pH –log [H3O+]= –log(1.8 x 10–5)=

=pH 4.74

a. 1.8 x 10–5

[H3O+] [NaH2PO4][Na2HPO4]

Ka= x x[0.15 M][0.15 M]

=

[H3O+] = 6.2 x 10–8 M

pH –log [H3O+]= –log(6.2 x 10–8)=

=pH 7.21

b. 6.2 x 10–8


[H3O+][CH3COOH][NaCH3COO]

Ka= x x [0.20 M][0.20 M]

=

[H3O+] = 1.8 x 10–5 M

pH –log [H3O+]= –log(1.8 x 10–5)=

=pH 4.74

a. 1.8 x 10–5


Chapter 9–25

[H3O+][CH3COOH]

[NaCH3COO]Ka= x x

[0.20 M][0.40 M]

=

[H3O+] = 9.0 x 10–6 M

pH –log [H3O+]= –log(9.0 x 10–6)=

=pH 5.05

b. 1.8 x 10–5

[H3O+]

[CH3COOH][NaCH3COO]

Ka= x x[0.20 M][0.10 M]

=

[H3O+] = 3.6 x 10–5 M

pH –log [H3O+]= –log(3.6 x 10–5)=

=pH 4.44

c. 1.8 x 10–5


[H3O+] [HCO3–]

[CO32–]

Ka= x x [0.20 M][0.20 M]

=

[H3O+] = 5.6 x 10–11 M

pH –log [H3O+]= –log(5.6 x 10–11)=

=pH 10.25

a. 5.6 x 10–11

[H3O+] [HCO3

–][CO3

2–]Ka= x x [0.20 M]

[0.40 M]=

[H3O+] = 2.8 x 10–11 M

pH –log [H3O+]= –log(2.8 x 10–11)=

=pH 10.55

b. 5.6 x 10–11

[H3O+] [HCO3

–][CO3

2–]Ka= x x [0.20 M]

[0.10 M]=

[H3O+] = 1.1 x 10–10 M

pH –log [H3O+]= –log(1.1 x 10–10)=

=pH 9.55

c. 5.6 x 10–11

9.111 The pH of unpolluted rainwater is lower than that of pure water because CO2 combines with rainwater to form H2CO3, which is acidic, lowering the pH.



9.112 The pH of acid rain is lower than the pH of rainwater because acid rain contains dissolved H2SO4 or HNO3 due to the burning of fossil fuels. Both acids are strong acids and the additional [H3O+] lowers the pH.

9.113 Calculate the concentrations.


3.2 x 10–8 M[H3O+] =

1.0 x 10–14

[H3O+][OH–] =

Kw =3.2 x 10–8

= 3.1 x 10–7 M

9.114


6.6 x 10–5 M[H3O+] =

a.

b. The sample of rainwater most likely contains dissolved sulfuric acid and/or nitric acid as a result of burning fossil fuels, thus leading to a lower pH than expected for unpolluted rainwater.

9.115 By breathing into a bag, the individual breathes in air with a higher CO2 concentration. Thus, the

CO2 concentration in the lungs and the blood increases, thereby lowering the pH. 9.116 Respiratory acidosis results when the body fails to eliminate adequate amounts of CO2 through

the lungs. The increased CO2 concentration leads to an increase in the H3O+ concentration and a lower pH. Respiratory alkalosis is caused by a lower than normal CO2 concentration, which leads to a decrease in the H3O+ concentration and a higher pH.

9.117 A rise in CO2 concentration leads to an increase in H+(aq) concentration by the following

equilibria:

CO2(g) + H2O(l) H2CO3(aq) HCO3–(aq) + H+(aq)

9.118 A lake on a bed of limestone is naturally buffered against the effects of acid rain because the

presence of CaCO3 leads to a natural carbonate/bicarbonate buffer that is able to resist pH changes due to the lower pH of the acid rain.

9.119 Write the acid–base reaction that occurs when OCl– dissolves in water.

OCl–(aq) + H2O(l) HOCl(aq) + OH–(aq)

This reaction forms OH–, which makes the pool water basic.

9.120 The acid–base reaction that occurs when H3PO4 and NaOH react is: 3 NaOH(aq)H3PO4(aq) + 3 H2O(l) + Na3PO4(aq)

Assuming equal volumes of acid and base, NaOH is the limiting reagent, leaving phosphoric acid in solution. Phosphoric acid can then dissociate to form H2PO4

–. This allows the formation of a buffer solution. H3PO4(aq) H2O(l)+ H2PO4

–(aq) + H3O+(aq)


Documents

Chapter 9 Acids and Bases - websites.rcc.eduwebsites.rcc.edu/grey/files/2012/02/Chapter-9-Solutions-Smith.pdf · Chapter 9–1 Chapter 9 Acids and Bases ... 9.4 Use Example 9.4 to