Acid-Base Concepts -- Chapter 15

1. Arrhenius Acid-Base Concept (last semester)

Acid: H+ supplier Base: OH– supplier

2. Brønsted-Lowry Acid-Base Concept (more general)

(a) Definition (H+ transfer)

Acid: H+ donor Base: H+ Acceptor

Conjugate Acid-Base Pairs: Base Acid+ H+

– H+

Conjugate Acid-Base Pairse.g.,

NH3(g) + H2O(l) NH4+

(aq) + OH–(aq)

base acid acid base

conjugate pair

conjugate pair

more examples:

NH2– NH3 NH4

+

OH– H2O H3O+

O2– OH– H2O

HSO4– H2SO4 H3SO4

+

CH3– CH4 “CH5

+”

conjugate acids

conjugate bases

Amphoteric Substances

Molecules or ions that can function as both acids and bases (e.g. H2O itself!)

e.g. the bicarbonate ion, HCO3–

HCO3– + OH– --> H2O + CO3

2–

acid base

HCO3– + HCl --> H2CO3 + Cl–

base acid

Autoionization of Water

Water undergoes auto-ionization to a slight extent:H2O(l) + H2O(l) H3O+

(aq) + OH–(aq)

H3O+ = hydronium ion OH– = hydroxide ion

or, simply,

H2O(l) H+(aq) + OH–

(aq)

equilibrium constant:

Kc = [H+][OH–]/[H2O]

But, [H2O] ~ constant ~ 55.6 mol/L at 25 °C

so, instead, use the “ion product” for water = Kw

Kw = [H+][OH–] = 1.0 x 10-14 (at 25 °C)

In pure water: [H+] = [OH–] = 1.0 x 10-7 M

The pH ScaleKw = [H+][OH–] = 1.0 x 10-14

The pH scale: pH = – log [H+]In general: pX = – log X e.g. pOH = – log [OH–]

and, in reverse: [H+] = 10–pH mole/L [OH–] = 10–pOH mole/L

Since Kw = [H+][OH–] = 1.0 x 10–14

pKw = pH + pOH = 14.00

Notice the sig figs!

two sig figs

two sig figs!

Relative Acidity of Solutionsneutral solution [H+] = [OH–] = 1.0 x 10–7 M

pH = pOH = 7.00

acidic solution [H+] > 10–7

(i.e. more H+ than in pure water)pH < 7.00 [OH–] < 10–7 and pOH > 7.00

e.g. if [H+] = 1.00 x 10–3 Mthen pH = 3.000 and pOH = 11.000

Basic solution [H+] < 10–7

(i.e. less H+ than in pure water)pH > 7.00 [OH–] > 10–7 and pOH < 7.00

e.g. if [OH–] = 1.00 x 10–3 Mthen pOH = 3.000 and pH = 11.000

Example Problem

The water in a soil sample was found to have [OH–] equal to 1.47 x 10–9 mole/L. Determine [H+], pH, and pOH.

Answer:[H+] = Kw/[OH–] = (1.00 x 10–14)/(1.47 x 10–9) = 6.80 x 10–6

pH = – log [H+] = – log (6.80 x 10–6) = 5.167 (acidic!)

pOH = 14.00 – pH = 14.00 – 5.167 = 8.833

{or, pOH = – log [OH–] = – log (1.47 x 10–9) = 8.833}

Strong Acids and BasesStrong Acids (e.g. HCl, HNO3, etc) ~ 100% ionized

HNO3(aq) + H2O H3O+(aq) + NO3

–(aq)

or, in simplified form

HNO3(aq) H+(aq) + NO3

–(aq)

[H+] = initial M of HNO3

e.g. in a 0.050 M HNO3 solution:

[H+] = 0.050 and pH = – log (0.050) = 1.30

Strong Bases (metal hydroxides) -- 100% ionized

NaOH(aq) Na+(aq) + OH–

(aq)

[OH–] = initial M of NaOH

100%

100%

100%

Memorize!Table 15.3 p665Table 15.7 p682

Example ProblemWhat mass of Ba(OH)2 (171.34 g/mole) is required to prepare 250 mL of a solution with a pH of 12.50?

First: Write the reaction.Ba(OH)2(aq) Ba2+

(aq) + 2 OH–(aq)

so, [OH–] = 2 x M of Ba(OH)2 solution (2:1 ratio)

100%

Second: Determine [OH–].pOH = 14.00 – pH = 14.00 – 12.50 = 1.50[OH–] = 10–1.50 = 0.032 M

Third: How much Ba(OH)2 is needed for that much OH–?

250 mL soln x0.032 mol OH–

1000 mL soln= 0.0079 mol OH–

0.0079 mol OH– x1 mole Ba(OH)2

2 mol OH–x

171.34 g Ba(OH)2

1 mol Ba(OH)2= 0.68 g Ba(OH)2

Sample Problem(a) Determine the pH of a 1.75 M solution of HNO3.

(b) Now suppose that 500.0 mL of a 1.50 M Ba(OH)2 solution is added to 1.00 L of the above HNO3 solution. Determine the pH of the resulting solution.

(c) What additional volume (in mL) of 1.50 M Ba(OH)2 must be added to the mixture in part (b) to bring the pH to 7.00?

Sample Problem(a) Determine the pH of a 1.75 M solution of HNO3.

a) pH = –0.243

(b) Now suppose that 500.0 mL of a 1.50 M Ba(OH)2 solution is added to 1.00 L of the above HNO3 solution. Determine the pH of the resulting solution.

b) pH = 0.78

(c) What additional volume (in mL) of 1.50 M Ba(OH)2 must be added to the mixture in part (b) to bring the pH to 7.00?

c) 83 mL Ba(OH)2 soln

Weak Acids and Bases

(a) Weak Acids -- Less than 100% ionized (equilibrium !)In general: HA is a weak acid, A– is its conjugate baseHA(aq) + H2O H3O+

(aq) + A–(aq)

or, simply, HA(aq) H+(aq) + A–

(aq)

Acid Dissociation Constant: Ka

Ka =[H+][A–][HA]

Relative acid strength:Weak acid: Ka < ~10–3

Moderate acid: Ka ~ 1 to 10–3

Strong acid: Ka > 1

Example ProblemHypochlorous acid, HOCl, has a pKa of 7.52. What is the pH of an0.25 M solution of HOCl? What is the percent ionization?

First: Write equation and ICE table.HOCl(aq) ⇌ H+

(aq) + OCl–(aq)

0.25 M 0 0

–x +x +x

0.25 – x x x

I

C

E

Second: Write equilibrium expression.

pKa = – log Ka, so Ka = 10–pKa = 10–7.52 = 3.0 x 10–8

3.0 x 10–8 =[H+][Cl–]

[HOCl]=

(x)(x)

(0.25 –x)=

x2

(0.25 –x)

Third: Solve for [H+] and pH.Since Ka is very small, assume x << 0.25

x2/(0.25) ≈ 3.0 x 10–8 x ≈ 8.7 x 10–5 (OK!)

pH = –log (8.7 x 10–5) = 4.06 (solution is acidic!)

Example Problem, cont.Fourth: Determine percent ionization.

• % ionization = (amount HA ionized)/(initial) x 100%

= (8.7 x 10–5)/(0.25) x 100% = 0.035%

Weak Bases(b) Weak Bases• In general: B is a weak base, HB+ is its conjugate acid B(aq) + H2O(l) HB+

(aq) + OH–(aq)

Kb =[HB+][OH–]

[B]

e.g. NH3(aq) + H2O(l) NH4+

(aq) + OH–(aq)

Kb = [NH4

+][OH–]

[NH3]= 1.8 x 10–5

pKb = -log Kb = 4.74

Note: Since OH– rather than H+ appears here, first find [OH–] or pOH, and then convert to pH

Example problem: pH of 0.25 M solution of NH3?

Set up conc. table as usual, solve for x = [OH–][OH–] = 2.1 x 10–3 pOH = 2.68 pH = 11.32 (basic!)

Base Dissociation Constant:

Sample ProblemThe nitrite ion (NO2

–) is a weak base with a pKb value of 10.85.(a) Write a balanced net ionic equation for the major equilibrium reaction that is occurring in an aqueous solution of sodium nitrite (NaNO2).

(b) Calculate the pH of a 0.25 M solution of sodium nitrite (NaNO2). Clearly state and justify any assumptions that you make.

Sample ProblemThe nitrite ion (NO2

–) is a weak base with a pKb value of 10.85.(a) Write a balanced net ionic equation for the major equilibrium reaction that is occurring in an aqueous solution of sodium nitrite (NaNO2).

(b) Calculate the pH of a 0.25 M solution of sodium nitrite (NaNO2). Clearly state and justify any assumptions that you make.

(a) NO2–(aq) + H2O(l) HNO2(aq) + OH–

(aq)

(b) 8.28

Salts of Weak Acids and Bases• A) conjugate acid-base pairs (HA and A–)

– Ka: HA H+ + A–

– Kb: A- + H2O HA + OH–

For any conjugate acid-base pair:

KaKb = Kw

pKa + pKb = 14.00

Salt of a Weak Acid(e.g. NaCN) -- basic solution• Anion acts as a weak base:

Kb: CN–(aq) + H2O(l) HCN(aq) + OH–

(aq)

Kb = Kw/Ka =[OH–][HCN]

[CN–]

e.g. Ka for HCN is 6.2 x 10–10

What is pH of a 0.50 M NaCN solution?Kb = Kw/Ka = (1.0 x 10–14)/(6.2 x 10–10) = 1.6 x 10–5

Use a concentration table based on Kb reaction above:

x = [OH–] = [HCN][CN–] = 0.50 – x ≈ 0.50 (since Kb is small)

Kb = [OH–][HCN]/[CN–] ≈ x2/0.50 ≈ 1.6 x 10–5

x = [OH–] ≈ 2.8 x 10–3

pOH = 2.55 and pH = 11.45 (basic!)

Salt of a Weak Base(e.g. NH4Cl) -- Acidic Solution

• Cation acts as a weak acid:Ka: NH4

+ H+ + NH3

Ka = Kw/Kb =[H+][NH3]

[NH4+]

e.g. Kb for NH3 is 1.8 x 10–5

What is pH of a 0.50 M NH4Cl solution?

Ka = Kw/Kb = (1.0 x 10–14)/(1.8 x 10–5) = 5.6 x 10–10

Use a concentration table based on Ka reaction above:

x = [H+] = [NH3]

[NH4+] = 0.50 – x ≈ 0.50 (since Ka is small)

Ka = [H+][NH3]/[NH4+] ≈ x2/0.50 ≈ 5.6 x 10–10

x = [H+] ≈ 1.7 x 10–5 pH = 4.77 (acidic!)

Sample ProblemThe pKa value for HCN is 9.21. What molar concentration of

NaCN is required to make a solution with a pH of 11.75?

Sample ProblemThe pKa value for HCN is 9.21. What molar concentration of

NaCN is required to make a solution with a pH of 11.75?

Answer: 2.0 M NaCN

Polyprotic Acidse.g. diprotic acids, H2A, undergo stepwise dissociation:

H2A HA– + H+

HA– A2– + H+

Ka1 =[HA–][H+]

[H2A]

Ka2 =[A2–][H+]

[HA–]

Usually, Ka1 >> Ka2 so that:

The 1st equilibrium produces most of the H+ and [HA–] but the 2nd equilibrium determines [A2–]

Example ProblemAscorbic acid (vitamin C), H2C6H2O6, is an example of a diprotic

acid with Ka1 = 7.9 x 10–5 and Ka2 = 1.6 x 10–12. For a 0.10 M solution of ascorbic acid, determine the pH and the concentrations of the mono anion, HC6H2O6

–, and the dianion, C6H2O6

2–.

Based on the first equilibrium:x = [H+] ≅ [HA–] and [H2A] = 0.10 – x ≅ 0.10

Ka1 = 7.9 x 10–5 ≅ x2 / (0.10)

∴ x ≅ 2.8 x 10–3 so pH = 2.55

Must use the 2nd equilibrium to find [A2–]:Ka2 = [A2–][H+] / [HA–] but, from above [H+] ≅ [HA–]

∴ Ka2 ≅ [A2–] (a general result for H2A!)

[A2–] ≅ 1.6 x 10–12

How to tell if it’s acidic or basic?Anions

Anion that is conjugate base of a weak acid is itself a weak base. Exception: H2O!!

An anion that is a conjugate base of a strong acid is pH-neutral.Anion of a polyprotic acid is amphoteric, e.g. H2PO4

–.

CationsCation that is conjugate acid of a weak base is itself a weak acid. A cation that is a conjugate acid of a strong base is pH-neutral.Small, highly charged metal cations form weakly acidic solutions (not

Group I or II), e.g.Al3+

(aq) + 6 H2O(l) Al(H2O)63+

(aq)

NeutralsWeak bases: Table 15.8, p683 and Fig. 15.12, p687Weak acids: Tables 15.4, p666, and 15.12, p687, and Fig. 15.12.

Mixed SaltsSalts where the cation is acidic and the anion basic form solutions

where the pH depends on the relative strengths of the acid and base.

weak acid

More Sample Problems1. Write balanced chemical equations for the important

equilibrium that is occurring in an aqueous solution of each of the following.(a) HClO (b) (NH4)2SO4 (c) KCl

(e) NaCHO2

2. Write the appropriate equilibrium constant expressions (Ka, etc) for the above reactions.

3. Determine the pH of a 0.100 M solution of each of the above compounds. (Use equilibrium constants from the textbook as needed).

Answers1. (a) HClO(aq) H+

(aq) + ClO–(aq)

(b) NH4+

(aq) NH3(aq) + H+(aq)

(c) H2O(l) H+(aq) + OH–

(aq)

(d) CHO2–(aq) + H2O(aq) HCHO2(aq) + OH–

(aq)

2. (a) Ka = [H+][ClO–]/[HClO]

(b) Ka = [NH3][H+]/[NH4+]

(c) Kw = [H+][OH–]

(d) Kb = [HCHO2][OH–]/[CHO2–]

3.(a) 4.26 (b) 4.98 (c) 7.00 (d) 8.38

Relative Strengths of Brønsted Acids

• Binary Acids e.g. HCl, HBr, H2S, etc.

e.g., relative acidity: HCl > H2S (across a period)

HI > HBr > HCl > HF (up in a group)

Periodic Table

Acid Strength Increases

Oxo Acidse.g. HNO3, H2SO4, H3PO4, etc.

1. for same central element, acid strength increases with # of oxygens

acid strength increases

HClO < HClO2 < HClO3 < HClO4

2. for different central element, but same # of oxygensacid strength increases with electronegativity

e.g. H2SO4 > H2SeO4 > H2TeO4

Acid Strength Increases

Periodic Table

Relative Strengths of Conjugate Acid-Base Pairs

For example, HF + H2O H3O+ + F–

acid base acid base

• In this case, the equilibrium lies mainly on the reactant side. Therefore, “HF is a weaker acid than H3O+”

• In general, weaker Brønsted acids have stronger conjugate bases. (and vice versa)

Lewis Acid-Base Concept (most general)

Definition (electron pair transfer)

Acid: e– pair acceptor

Base: e– pair donor

Lewis acids -- electron deficient molecules or cations

Lewis bases -- electron rich molecules or anions. (have one or more unshared e–

pairs)

Lewis Acid-Base Reactions(i.e. all non-redox reactions!)

OH–(aq) + NH4

+(aq) --> H2O(l) + NH3(aq)

N

H

H

H

H

+

H O

Lewis acid

Lewis base

O C O

Lewis acid

H O

H

+ H N

H

H

O C O

OH

O C O

OH

OH–(aq) + CO2(g) --> HCO3

–(aq)

Review Ch. 9, slides 10-15 from CHEM 10113!

Yet More Sample Problems1) Among the following, which is the strongest acid?

HBrO3 HBrO4 H2SeO3 H2SeO4 H2TeO3 H2TeO4

2) Among the following, which is the weakest acid?H2Se NH3 PH3 H2S H2O AsH3

3) The conjugate base of CH3OH is ______________.

4) The conjugate acid of HF is _________.

5) Write the equation for the reaction of H2Se with AsH3, then answer the questions below.

a) Label the Bronsted acid, base, conjugate acid, and conjugate base.

b) If Kc = 0.078, label which is the strongest acid and strongest base in your equation.

Yet More Sample Problems1) Among the following, which is the strongest acid?

HBrO3 HBrO4 H2SeO3 H2SeO4 H2TeO3 H2TeO4

HBrO4

2) Among the following, which is the weakest acid?H2Se NH3 PH3 H2S H2O AsH3

NH3

3) The conjugate base of CH3OH is ______________.

CH3O–

4) The conjugate acid of HF is _________.H2F+

5) Write the equation for the reaction of H2Se with AsH3, then answer the questions below.

a) Label the Bronsted acid, base, conjugate acid, and conjugate base.b) If Kc = 0.078, label which is the strongest acid and strongest base in

your equation.

H2Se(aq) + AsH3(aq) AsH4+

(aq) + HSe–(aq)

1) acid base conj. acid conj. base2) strongest acid is AsH4

+, strongest base is HSe–

And Finally…Consider the reaction of hydroxide ion with the nitrosyl cation, NO2

+, to form nitric acid.

OH–(aq) + NO2

+(aq) --> HNO3(aq)

Write Lewis electron dot formulas (including formal charges and/or resonance forms if needed) for all three species in this reaction. Clearly indicate which reactant is the Lewis acid and which is the Lewis base. Use arrow(s) to illustrate the formation of any new chemical bond(s) during the reaction.

And Finally…Consider the reaction of hydroxide ion with the nitrosyl cation, NO2

+, to form nitric acid.

OH–(aq) + NO2

+(aq) --> HNO3(aq)

Write Lewis electron dot formulas (including formal charges and/or resonance forms if needed) for all three species in this reaction. Clearly indicate which reactant is the Lewis acid and which is the Lewis base. Use arrow(s) to illustrate the formation of any new chemical bond(s) during the reaction.

H O + O N O

Lewis acidLewis base

O N O

OH

O N O

OH