Homework Problems

Homework Problems

Chapter 15 Homework Problems: 4, 6, 8, 24, 34, 36, 38, 39, 42, 52, 56a, 62, 68, 70, 74, 80a, 89, 93, 100, 103a, 116, 117, 124

CHAPTER 15

Acids and Bases

Acids and Bases (Arrhenius)

There are several definitions we can use for acids and bases:

Arrhenius definition

acid - A substance which, when added to water, forms H+ ion

base - A substance which, when added to water, forms OH- ion

Examples:

HCl(aq) H+(aq) + Cl-(aq)

HF(aq) H+(aq) + F-(aq)

NaOH(s) Na+(aq) + OH-(aq)

NH3(aq) + H2O() NH4+(aq) + OH-(aq)

The advantages of the Arrhenius definition are that it is simple and easy to implement. The disadvantages are that it is tied in to a particular solvent (water) and is not a general definition.

Acids and Bases (Bronsted-Lowry)

Bronsted-Lowry definition

acid - a proton (H+) donor; forms a conjugate base

base - a proton (H+) acceptor; forms a conjugate acid

H+

HF(aq) + H2O() H3O+(aq) + F-(aq)

acid base conjugate acid conjugate base

In the Bronsted theory, in an acid-base reaction an acid donates a proton to form a conjugate base, while a base accepts a proton to form a conjugate acid.

In addition, in Bronsted theory acids form hydronium ion (H3O+ ion) instead of hydrogen ion (H+ ion) when added to water.

Examples of Acids

HCl(aq) + H2O() H3O+(aq) + Cl-(aq)

acid: HCl conjugate base: Cl-

base: H2O conjugate acid: H3O+

CH3COOH(aq) + H2O() H3O+(aq) + CH3COO-(aq)

acid: CH3COOH conjugate base: CH3COO-

base: H2O conjugate acid: H3O+

The second reaction goes in both directions, so has an equili-brium constant, the acid ionization constant

KC = Ka = [H3O+] [CH3COO-]

[CH3COOH]

Examples of Bases

NaOH(s) Na+(aq) + OH-(aq)

Considered an ionization reaction in Bronsted theory, not an acid-base reaction.


base: NH3 conjugate acid: NH4+

acid: H2O conjugate base: OH-

The reaction goes in both directions, so has an equilibrium con-stant, the base ionization constant

KC = Kb = [NH4+] [OH-]

[NH3]

Amphoteric Substances

Some substances can act as either Bronsted acids or Bronsted bases. Substances that can act as either acids or bases depending on what they are reacting with are called amphoteric. For example:

H2O (water)

(acid) NH3(aq) + H2O() NH4+(aq) + OH-(aq)

(base) CH3COOH(aq) + H2O() H3O+(aq) + CH3COO-(aq)

HCO3- (hydrogen carbonate ion)

(acid) NH3(aq) + HCO3-(aq) NH4

+(aq) + CO32-(aq)

(base) CH3COOH(aq) + HCO3-(aq) H2CO3(aq) + CH3COO-(aq)

We can picture the reaction of a Bronsted acid with a Bronsted base in several ways.

Given any substance, the conjugate base for the substance is formed by removing a proton (H+), and the conjugate acid is formed by adding a proton (H+).

Example: H2O

The conjugate base of H2O is OH-

The conjugate acid of H2O is H3O+

Example: HSO4-

The conjugate base of HSO4- is SO4

2-

The conjugate acid of HSO4- is H2SO4

Example

Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory.

Example

Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory.

HClO2(aq) + H2O() H3O+(aq) + ClO2-(aq)

C5H5N(aq) + H2O() C5H5NH+(aq) + OH-(aq)

Strong and Weak Acids

There are seven common strong acids:

Binary strong acids Ternary strong acids

HCl HClO3 HNO3

HBr HClO4 H2SO4 (1st proton)

HI

Sulfuric acid is special in that it is a strong acid with respect to the first proton and a weak acid with respect to the second proton.

H2SO4(aq) + H2O() HSO4-(aq) + H3O+(aq)

HSO4-(aq) + H2O() SO4

2-(aq) + H3O+(aq)

Ka2 = [SO42-] [H3O+] = 1.2 x 10-2

[HSO4-]

All other acids are weak acids.

HNO2(aq) + H2O() H3O+(aq) + NO2-(aq)

Ka = [H3O+] [NO2-] = 4.5 x 10-4

[HNO2]

CH3COOH(aq) + H2O() H3O+(aq) + CH3COO-(aq)

Ka = [H3O+] [CH3COO-] = 1.8 x 10-5

[CH3COOH]

The larger the value of Ka, the stronger the acid. Therefore, HNO2 is a stronger acid than CH3COOH.

Strong and Weak Bases

There are seven common strong soluble bases:

Group 1A strong bases Group 2A strong bases

LiOH Sr(OH)2

NaOH Ba(OH)2

KOH

RbOH

CsOH

All other metal hydroxides are insoluble bases. They do not dissolve in water to an appreciable extent, but react as bases in acid-base reactions. Examples: Cu(OH)2, Al(OH)3, Pb(OH)2.

Cu(OH)2(s) + 2 HCl(aq) Cu2+(aq) + 2 Cl-(aq) + 2 H2O()

Weak bases establish an equilibrium in water.


Kb = [NH4+] [OH-] = 1.8 x 10-5

[NH3]

N2H4(aq) + H2O() N2H5+(aq) + OH-(aq)

Kb = [N2H5+] [OH-] = 8.9 x 10-7

[N2H4]

Since Kb is larger for NH3 than for N2H4, NH3 is a stronger base than N2H4.

Autoionization of Water

In pure water there will be a small number of H3O+ and OH- ions. This is due to the autoionization reaction:

H2O() + H2O() H3O+(aq) + OH-(aq)

Kw = [H3O+] [OH-] = 1.0 x 10-14 (at T = 25 C)

Because the above reaction is endothermic (Hrxn = + 55.8 kJ/mol) the value for Kw increases as T increases.

Equilibrium in Pure Water

Since H2O() + H2O() H3O+(aq) + OH-(aq)

Kw = [H3O+] [OH-] = 1.0 x 10-14 (at T = 25 C)

Substance Initial Change Equilibrium

H3O+ 0.0 x x

OH- 0.0 x x

So (x) (x) = x2 = 1.0 x 10-14

x = 1.0 x 10-7

So in pure water at equilibrium at T = 25 C, [H3O+] = [OH-] = 1.0 x 10-7 M.

Acidic, Basic, and Neutral Solutions

We can use the above results to define what we mean by an acidic, basic, and neutral solution, using [H3O+] [OH-] = 1.0 x 10-14.

acidic solution [H3O+] > [OH-]

[H3O+] > 1.0 x 10-7 M

[OH-] < 1.0 x 10-7 M

neutral solution [H3O+] = [OH-]

[H3O+] = 1.0 x 10-7 M

[OH-] = 1.0 x 10-7 M

basic solution [H3O+] < [OH-]

[H3O+] < 1.0 x 10-7 M

[OH-] > 1.0 x 10-7 M

pH

pH represents a convenient way of representing the concentration of hydronium ion in an aqueous solution. pH is defined as follows:

pH = - log10[H+] = - log10[H3O+]

For a neutral solution at T = 25 C

pH = - log10[H3O+] = - log10(1.0 x 10-7) = 7.00

Note that the number of digits to the right of the decimal point is equal to the number of significant figures in the H3O+ concentration.

For acidic and basic solutions

acidic solution [H3O+] > 1.0 x 10-7 M means pH < 7.00

basic solution [H3O+] < 1.0 x 10-7 M means pH > 7.00

The further away the pH is from 7.00 the more acidic (if less than 7.00) or basic (if greater than 7.00) the solution.

pOH and pK

We can use the “p” notation as a general symbol to indicate that we are taking - log10 of something. In particular

pH = - log10[H3O+]

pOH = - log10[OH-]

pK = - log10 K

By reversing the above definitions we get the following relationships

[H3O+] = 10-pH

[OH-] = 10-pOH

K = 10-pK

Relationship Between pH and pOH

There is a simple relationship between pH and pOH.

[H3O+] [OH-] = 1.0 x 10-14

-log10[H3O+] + ( - log10[OH-]) = - log10(1.0 x 10-14)

pH + pOH = 14.00 (at 25 C)

The more general relationship, true at all temperatures, is

pH + pOH = pKw

Relationship Between pH and Concentration

If we know the concentration of hydronium ions in solution we can find the pH of the solution (and vice versa). We can also find pOH and the concentration of hydroxide ions in solution.

Example: A particular solution has pH = 4.62 at T = 25. C. What are the concentrations of H3O+ and OH- in the solution?

Relationship Between pH and Concentration

Example: A particular solution has pH = 4.62 at T = 25. C. What are the concentrations of H3O+ and OH- in the solution?

[H3O+] = 10-pH = 10-4.62 = 2.4 x 10-5 M

[H3O+] [OH-] = 1.0 x 10-14, so

[OH-] = 1.0 x 10-14 = 1.0 x 10-14 = 4.2 x 10-10 M

[H3O+] 2.4 x 10-5

We could also find the OH- concentration as follows:

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 4.62 = 9.38

[OH-] = 10-pOH = 10-9.38 = 4.2 x 10-10 M

pH for Solutions of Strong Acids or Strong Bases

Because strong acids and strong soluble bases are strong electrolytes, and so completely dissociate in solution, finding the value for pH is relatively simple. We may use the following procedure:

1) Use the information in the problem to find the concentration of H3O+ (strong acid) or OH- (strong base).

2) Find the pH

a) For strong acids, find the pH directly

b) For strong bases, find the pOH, then use (at T = 25 C)

pH + pOH = 14.00 ; pH = 14.00 - pOH

to find the pH

Example: Find the pH for the following solutions, at T = 25 C

a) A 3.5 x 10-3 M solution of HBr, a strong acid

b) A solution formed by dissolving 0.200 moles of Sr(OH)2, a strong soluble base, in water, to form a solution with a final volume of 500.0 mL.

a) A 3.5 x 10-3 M solution of HBr, a strong acid

Reaction is

HBr(aq) + H2O() H3O+(aq) + Br-(aq)

[H3O+] = 3.5 x 10-3 mol HBr 1 mol H3O+ = 3.5 x 10-3 M

L soln 1 mol HBr

pH = - log10(3.5 x 10-3) = 2.46

b) A solution formed by dissolving 0.200 moles of Sr(OH)2, a strong soluble base, in water, to form a solution with a final volume of 500.0 mL.

Reaction is

Sr(OH)2(s) Sr2+(aq) + 2 OH-(aq)

[OH-] = 0.200 mol Sr(OH)2 2 mol OH- = 0.800 M

0.5000 L soln 1 mol Sr(OH)2

pOH = - log10(0.800) = 0.097

pH = 14.00 - pOH = 14.00 - 0.097 = 13.90

Weak Acids or Weak Bases

For problems involving solutions containing a single weak acid or weak base we proceed as we do other equilibrium problems.

1) Start with the following information

Balanced chemical reaction

Initial conditions

Value for Ka (weak acid) or Kb (weak base)

2) Set up the problem using the “ICE” method

3) Find [H3O+] (weak acid) or [OH-] (weak base)

4) Find pH (for weak base, first find pOH, then use pH + pOH = 14.00 to find pH)

Example: Find the pH of a 0.100 M solution of HNO2, a weak acid (Ka = 4.5 x 10-4), at T = 25 C

Find the pH of a 0.100 M solution of HNO2, a weak acid (Ka = 4.5 x 10-4), at T = 25 C

HNO2(aq) + H2O() H3O+(aq) + NO2-(aq)

Ka = [H3O+] [NO2-] = 4.5 x 10-4

[HNO2]


H3O+ 0.00 x x

NO2- 0.00 x x

HNO2 0.100 - x 0.100 - x

(x) (x) = 4.5 x 10-4

(0.100 - x)

There are two ways to proceed...

1) Assume x is small (in this case, x << 0.100)

(x) (x) x2 = 4.5 x 10-4

(0.100 - x) 0.100

so x2 = (0.100)(4.5 x 10-4) = 4.5 x 10-5

x = (4.5 x 10-5)1/2 = 6.7 x 10-3

[H3O+] = 6.7 x 10-3 M ; pH = - log10(6.7 x 10-3) = 2.17

Is 6.7 x 10-3 << 0.100? Yes (at least 10 times smaller).

2) Solve the quadratic

x2 = (0.100 - x)(4.5 x 10-4) = (4.5 x 10-5) - (4.5 x 10-4) x

x2 + (4.5 x 10-4) x - (4.5 x 10-5) = 0

x = 6.5 x 10-3 ; - 6.9 x 10-3

[H3O+] = 6.5 x 10-3 M ; pH = - log10(6.5 x 10-3) = 2.19

Percent Dissociation

The percent dissociation of a weak acid is defined as

% dissociation = amount dissociated x 100%

initial amount

For strong acids the percent dissociation is ~ 100%.

For weak acids we can use the equilibrium concentrations to find the percent dissociation.

For the above example, initial HNO2 = 0.100 M

amount dissociated = 6.7 x 10-3 M

% dissociation = 6.7 x 10-3 x 100% = 6.7 %

0.100

pH Calculations involving Weak Bases

Just as we can find concentrations and pH values for solutions of weak acids, we can do the same for solutions of weak bases.

Example: Find the pH of a 0.100 M solution of NH3, a weak base (Kb = 1.8 x 10-5), at T = 25 C

Find the pH of a 0.100 M solution of NH3, a weak base (Kb = 1.8 x 10-5), at T = 25 C


Kb = [NH4+] [OH-] = 1.8 x 10-5

[NH3]


NH4+ 0.00 x x

OH- 0.00 x x

NH3 0.100 - x 0.100 - x

(x) (x) = 1.8 x 10-5

(0.100 - x)

Assume x is small (in this case, x << 0.100)

(x) (x) x2 = 1.8 x 10-5

(0.100 - x) 0.100

so x2 = (0.100)(1.8 x 10-5) = 1.8 x 10-6

x = (1.8 x 10-6)1/2 = 1.3 x 10-3

[OH-] = 1.3 x 10-3 M ; pOH = - log10(1.3 x 10-3) = 2.87

pH = 14.00 - 2.87 = 11.13

Is 1.3 x 10-3 << 0.100? Yes (at least ten times smaller)

If we solve using the quadratic formula, we get pH = 11.12

Polyprotic Acid

A polyprotic acid has two or more ionizable protons that can be donated in an acid-base reaction.

Monoprotic (one ionizable proton) HCl, HNO2, CH3COOH

Diprotic (two ionizable protons) H2CO3, H2SO4

Triprotic (three ionizable protons) H3PO3

For polyprotic acids one can talk about the acid dissociation constant for each proton.

Example: H2CO3

1st proton H2CO3(aq) + H2O() H3O+(aq) + HCO3-(aq)

Ka1 = [H3O+][HCO3-] = 4.3 x 10-7

[H2CO3]

2nd proton HCO3-(aq) + H2O() H3O+(aq) + CO3

2-(aq)

Ka2 = [H3O+][CO32-] = 5.6 x 10-11

[HCO3-]

For a polyprotic acid Ka1 > Ka2 > Ka3...

Calculations Involving Polyprotic Acids

It would seem like calculations with polyprotic acids should be complicated, since there are several sources of H3O+ ions.

However, there is usually a big enough difference in the values of the Kas (acid dissociation constants) that only the first dissociation needs to be considered for polyprotic acid solutions.

Procedure (diprotic acid):

1) Calculate equilibrium concentrations using the first ionization constant.

2) Calculate equilibrium concentrations using the second ionization constant, using the results from the first calculation for the initial conditions.

3) If there are any significant changes in concentrations involving the first ionization, go back and recalculate concentrations using the previous results for the initial conditions.

Relationship Between Ka and Kb

We may find a general relationship between the values for Ka and Kb for a Bronsted acid/conjugate base pair of substances. We proceed as follows:

Let HA be a weak monoprotic acid. A- is the conjugate base.

HA(aq) + H2O() H3O+(aq) + A-(aq) Ka = [H3O+] [A-]

[HA]

A-(aq) + H2O() HA(aq) + OH-(aq) Kb = [HA] [OH-]

[A-]

Ka Kb = [H3O+] [A-] [HA] [OH-] = [H3O+] [OH-] = Kw

[HA] [A-]

Ka Kb = Kw pKa + pKb = pKw

Ka Kb = 1.0 x 10-14 pKa + pKb = 14.00, at T = 25. C

Example: The value for the acid ionization constant for acetic acid (CH3COOH) is Ka = 1.8 x 10-5 at T = 25. C. What is the value for Kb for the acetate ion (CH3COO-)?

Example: The value for the acid ionization constant for acetic acid (CH3COOH) is Ka = 1.8 x 10-5 at T = 25. C. What is the value for Kb for the acetate ion (CH3COO-)?

Ka Kb = Kw = 1.0 x 10-14 for an acid/conjugate base pair

Kb = Kw = (1.0 x 10-14) = 5.6 x 10-10

Ka (1.8 x 10-5)

Note that we could prepare a solution that initially only contained the conjugate base if we added a salt such as NaCH3COO, KCH3COO, etc.

NaCH3COO(aq) Na+(aq) + CH3COO-(aq)

General Statements About Acid and Base Strength

We may use the above results to make the following general statement concerning the relative strengths of acids and their conjugate bases. This is based on the relationship Ka Kb = Kw = 1.0 x 10-14.

The stronger the acid the weaker the conjugate base.

Example: Which is a stronger base, F- or CN-?

Ka(HF) = 3.5 x 10-4

Ka(HCN) = 4.9 x 10-10

Since HF is a stronger acid than HCN, F- is a weaker base than CN-.

Kb(F-) = (1.0 x 10-14)/(3.5 x 10-4) = 2.9 x 10-11

Kb(CN-) = (1.0 x 10-14)/(4.9 x 10-10) = 2.0 x 10-5

Acid-Base Properties of Salts

Recall that a salt is an ionic compound formed from the reaction of an acid with a base. We have the following four possibilities:

1) Salt of a strong acid and a strong base.

Example: HCl + NaOH NaCl + H2O (Na+ and Cl-)

No acid-base properties for the salt. Solutions will be neutral.

2) Salt of a strong acid and a weak base.

Example: HCl + NH3 NH4Cl (NH4+ and Cl-)

NH4+ will act as a weak acid. Solutions will be acidic.

3) Salt of a weak acid and a strong base.

Example: HF + KOH KF + H2O (K+ and F-)

F- will act as a weak base. Solutions will be basic.

4) Salt of a weak acid and a weak base.

Example: HF + NH3 NH4F (NH4+ and F-)

NH4+ will act as a weak acid. F- will act as a weak base.

Solutions will be approximately neutral.

Ka(HF) = 3.5 x 10-4 so Kb(F-) = 2.9 x 10-11

Kb(NH3) = 1.8 x 10-5 so Ka(NH4+) = 5.6 x 10-10

Since NH4+ is a stronger acid than F- is a base, the solution will

be slightly acidic.

Acid-Base Calculations for Salts

We do acid-base problems for salts the same way as we do other weak acid or weak base problems.

Example: What is the pH of a 0.100 M solution of ammonium chloride (NH4Cl), the salt of a strong acid and a weak base. Kb(NH3) = 1.8 x 10-5. Assume T = 25 C

Example: What is the pH of a 0.100 M solution of ammonium chloride (NH4Cl), the salt of a strong acid and a weak base. Kb(NH3) = 1.8 x 10-5. Assume T = 25 C

NH4Cl(s) NH4+(aq) + Cl-(aq)

NH4+(aq) + H2O() H3O+(aq) + NH3(aq)

Ka Kb = 1.0 x 10-14, so Ka = 1.0 x 10-14 = (1.0 x 10-14) = 5.6 x 10-10

Kb (1.8 x 10-5)

Ka = [H3O+] [NH3] = 5.6 x 10-10

[NH4+]


H3O+ 0.00 x x

NH3 0.00 x x

NH4+ 0.100 - x 0.100 - x

Ka = [H3O+] [NH3] = (x) (x) = 5.6 x 10-10

[NH4+] (0.100 - x)

Assume x << 0.100. Then

x2 = 5.6 x 10-10

(0.100)

x2 = (0.100) (5.6 x 10-10) = 5.6 x 10-11

x = (5.6 x 10-11)1/2 = 7.5 x 10-6

pH = - log10(7.5 x 10-6) = 5.13

Note that 7.5 x 10-6 << 0.100, so x is in fact small.

Acid-Base Properties of Cations

Metal cations can act as weak acids in solution. There are several ways in which these reactions can be written

Example: Al3+ ion

Al3+(aq) + H2O() H+(aq) + Al(OH)2+(aq)

Al(H2O)63+(aq) + H2O() H3O+(aq) + Al(H2O)5(OH)2+

Generally speaking, metal cations act as weak acids when they are small and have multiple positive charges. Common examples of metal cations acting as weak acids are Co2+, Ni2+, Zn2+, Fe2+, Fe3+, and Al3+ ions.

Factors Affecting Acid Strength

We can often predict the relative strengths of weak acids by focussing on the factors that control acid strength.

Binary acids

1) In the same column (group). Acid strength increases from top to bottom.

Reason: The H - X bond strength decreases from top to bottom, so it is easier to break the bond and form H+ (or H3O+).

Example: Group 7 binary acids.

HF 567 kJ/mol Ka = 3.5 x 10-4

HCl 431 kJ/mol “strong acid” (Ka ~ 107)

HBr 366 kJ/mol “strong acid” (Ka ~ 109)

HI 299 kJ/mol “strong acid” (Ka ~ 1011)

2) In the same row. Acid strength increases from left to right.

Reason: The electronegativity of the nonmetal increases from left to right, making the conjugate base more stable.

Example: Second row.

acid conj. base nonmetal EN

CH4 CH3- EN(C) = 2.5 insoluble in water (Ka ~ 10-49)

NH3 NH2- EN(N) = 3.0 “weak base” (Ka ~ 10-33)

H2O OH- EN(O) = 3.5 “amphoteric” (Ka = 1.0 x 10-14)

HF F- EN(F) = 4.0 “weak acid” (Ka = 3.5 x 10-4)

Ternary acids

1) Same nonmetal, different number of oxygens. The more oxy-gens, the stronger the acid.

Reason: The more oxygens in the oxyacid the more stable the conjugate base, and so the easier it will form.

acid conj. base

HClO ClO- Ka = 3.5 x 10-8

HClO2 ClO2- Ka = 1.1 x 10-2

HClO3 ClO3- “strong acid” (Ka ~ 5 x 102)

HClO4 ClO4- “strong acid” (Ka = 1 x 103)

2) Same number of oxygens, different nonmetal in the same column (group). Acid strength increases from bottom to top.

Reason: The electronegativity of the nonmetal increases from bottom to top, making the conjugate base more stable, and so easier to form.

acid conj. base nonmetal EN

HClO ClO- EN(Cl) = 3.0 Ka = 3.5 x 10-8

HBrO BrO- EN(Br) = 2.8 Ka = 2.0 x 10-9

HIO IO- EN(I) = 2.5 Ka = 2.3 x 10-11

Lewis Acids and Bases

There is a third way in which acids and bases are defined, de-veloped by G. N. Lewis.

Lewis definition

acid - an electron pair acceptor

base - an electron pair donor

Example:

BF3(g) + NH3(g) BF3NH3(s)

In the above reaction BF3 is an electron pair acceptor (Lewis acid) and NH3 is an electron pair donor (Lewis base). Note that the above reaction would not be an acid-base reaction under the Bronsted definition.

Also note that in the above reaction a coordinate covalent bond is formed (a bond where both electrons in the covalent bond come from the same atom).

Identifying Lewis Acids and Bases

To identify the Lewis acid and Lewis base in a reaction one needs only to identify the species that is accepting an electron pair (Lewis acid) and donating an electron pair (Lewis base).

Example:

CO2 + OH- HCO3-

CO2 accepts an electron pair and so is a Lewis acid; OH- donates an electron pair and so is a Lewis base.

End of Chapter 15

“Things should be made as simple as possible, but not any simpler.” – Albert Einstein

“Never express yourself more clearly than you think.” – Neils Bohr

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