276

ChemActivity

47

Acid/Base Strength of Conjugate Pairs (How Are Ka and Kb Related?)

Model 1: The Mathematical Relationship between Ka and Kb of a Conjugate Pair. Acid Ka HF [H3O+] [F–] hydrofluoric acid [HF] HONO nitrous acid

Conjugate Base F– fluoride ion

Kb – [OH ] [HF]

Ka  Kb

[F–]

NH3 ammonia

Critical Thinking Questions 1. Fill in the missing entries in Model 1. 2. For each acid and each conjugate base in Model 1, write the balanced chemical reaction that has the Ka or Kb as its equilibrium constant. The conjugate base F– is worked as an example: F–(aq) + H2O(l)

HF(aq) + OH–(aq)

Kb =

[OH–] [HF] [F–]

3. For each of the following, describe the common features: a)

all Ka expressions

b)

all Kb expressions

c)

all Ka  Kb products

ChemActivity 47

Acid/Base Strength of Conjugate Pairs

277

4. Provide an expression relating Kw to Ka and Kb of a conjugate acid-base pair.

5. Describe how to determine the value of Kb for a base—given the value of Ka for its conjugate acid.

6. Consider two acids, HA and HX, with HA being a stronger acid than HX. a)

Which acid has a larger value of Ka?

b)

Which conjugate base, A– or X–, has a larger value of Kb?

c)

Provide a qualitative description of the relationship between the relative strength of an acid and the relative strength of its conjugate base.

Exercises 1. Give the conjugate base and the Kb of the conjugate base for each of the acids in the table below: Acid CH3COOH H2CO3 H2S HNO2 NH4+

Ka 1.8  10–5 4.5  10–7 1.0  10–7 5.1  10–4 5.6  10–10

Base

Kb

278

ChemActivity 47

Acid/Base Strength of Conjugate Pairs

2. Give the conjugate acid and the Ka of the conjugate acid for each of the bases in the table below: Base NH3 CH3COO– C6H5NH2 NO2– H2NNH2

Kb 1.8  10–5 5.6  10–10 4.0  10–10 2.0  10–11 1.2  10–6

Acid

Ka

3. J. N. Spencer, G. M. Bodner, and L. H. Rickard, Chemistry: Structure & Dynamics, Fourth Edition, John Wiley & Sons, 2008. Chapter 11: Problems: 95-98.

Model 2: Ions are Potential Acids or Bases. All anions are potential bases: Cl–(aq) + H2O(l)

HCl(aq) + OH–(aq)

NO2–(aq) + H2O(l)

HNO2(aq) + OH–(aq)

All cations are potential acids: NH4+(aq) + H2O(l)

NH3(aq) + H3O+(aq)

C5H5NH+(aq) + H2O(l)

C5H5N(aq) + H3O+(aq)

Critical Thinking Questions 7. Which of the following ions are potential acids: Al3+ ; CH3NH3+ ; HPO42– ; F– ; CH3– ? 8. Which of the following ions are potential bases: Fe3+ ; NH4+ ; F– ?

ChemActivity 47

Acid/Base Strength of Conjugate Pairs

279

Model 3: Some Potential Acids and Some Potential Bases Are So Weak That They Do Not Affect the pH of the Solution. Any acid with a Ka less than 10 if Ka = 0 (in water).

–15

can be treated as

–15

can be treated as

Any base with a Kb less than 10 if Kb = 0 (in water).

The cations of alkali metals and alkaline earth metals act as neither acids nor bases in solution.

The values of Kb for conjugate bases of some strong acids. Conjugate Base Kb Acid Ka 9 HBr 1  10 HCl 1  106 HNO3 28

Table 1.

Critical Thinking Questions 9. Fill in the missing entries in Table 1. 10. The chloride ion is the conjugate base of hydrochloric acid. Explain why Cl– does not produce a basic solution when dissolved in water.

280

11.

ChemActivity 47

Acid/Base Strength of Conjugate Pairs

To 2 significant figures, predict the pH of a 1.00 molar solution of NaBr. your reasoning carefully.

Explain

12. Explain why the conjugate bases of the strong acids do not produce basic solutions when dissolved in water.

Exercises 4. For each of the following, indicate whether the resulting solution would be acidic, basic, or neutral, if 1.0 mole of each were dissolved in 1.0 liter of water. Also provide the predominant acid-base reaction that would occur, and evaluate the equilibrium constant for that process. NaF is worked out as an example: i) ii) iii) iv)

v)

NaF is an ionic compound. Na+ and F– ions exist in solution. Na+ acts neither as an acid nor a base; it has no effect on the pH of the solution. The F– ion is a weak base (the conjugate base of a weak acid). The solution will be basic. The predominant reaction will be F–(aq) + H2O HF(aq) + OH–(aq) . This is the chemical reaction that makes the solution basic. The equilibrium constant is Kb =

1.0 1014 Kw = Ka 1.0 104

= 1.4  10–11

Recall that the strong acids are: HCl; HBr; HI; HClO4; H2SO4; HNO3 a)

NH4NO3

b)

CsI

c)

CH3COONa

d)

KClO4

e)

magnesium(II) acetate

ChemActivity 47

Acid/Base Strength of Conjugate Pairs

281

5. Determine the pH of each of the 1.0 M solutions in Ex. 4.

6. For each of the following, indicate whether the resulting solution would be acidic, basic, or neutral, if 1.0 mole of each were dissolved in 1.0 liter of water. a)

NaCl

b)

KCl

c)

KNO3

d)

NaCH3CO2 (sodium acetate)

e)

NH4Cl

f)

NH4NO3

g)

NaNO2

h)

CaCl2

i)

KCN

j)

KF

k)

NaBr

7. Given that Ka for HCN is 6.2  10–10, calculate the pH of a 0.15 M KCN solution. 8. J. N. Spencer, G. M. Bodner, and L. H. Rickard, Chemistry: Structure & Dynamics, Fourth Edition, John Wiley & Sons, 2008. Chapter 11: Problems: 135, 146, 148.

Problem 1. One mole of CH3CH2NH3+Cl– is added to one liter of water. a) Will the solution be acidic, basic, or neutral? Explain briefly. b) Provide the chemical reaction that occurs to justify your answer in part a). That is, give the chemical reaction that causes the solution to be acidic or basic or neutral in accordance with your answer to part a).

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