16.1 Acids and Bases: The Brønsted–Lowry Model 16.2 pH and the Autoionization of Water 16.3 Calculations Involving pH, Ka and Kb 16.4 Polyprotic Acids

Chapter 16 - Acids and Bases

16.1

Acids and Bases: Basic Definitions

Sour Taste React with “active” metals (Al, Zn, Fe) to yield H2 gas:

Corrosive React with carbonates to produce CO2:

Change color of vegetable dyes Turn blue litmus red

React with bases to form salts

Properties of Acids

Properties of Bases

Bitter Taste

Also Known as Alkalies

Solutions feel slippery

Change color of vegetable dyes

Turn red litmus blue

React with acids to form salts

Acids and Bases in Solution

Acids ionize in water to form H+ ions. (More precisely, the H+ from the acid molecule is donated to a

water molecule to form hydronium ion, H3O+)

Bases dissociate in water to form OH- ions. (Bases, such as NH3, that do not contain OH- ions, produce OH-

by pulling H+ off water molecules.)

AcidsAcids are molecular compounds that ionize when they dissolve in

water. The molecules are pulled apart by water.

The percentage of molecules that ionize varies.

Acids that ionize virtually 100% are called strong acids.

HCl(aq) ➜ H+(aq) + Cl−(aq)

Acids that only ionize a small percentage are called weak acids.

HF(aq) ⇔ H+(aq) + F−(aq)

Strong Acid

Weak Acid

Molecular Models of Selected Acids

Binary Acid

“Oxy" Acids

Common AcidsName Formula Uses Strength

Perchloric HClO4 explosives, catalysts Strong

Nitric HNO3 explosives, fertilizers, dyes, glues Strong

Sulfuric H2SO4 explosives, fertilizers, dyes, glue, batteries Strong

Hydrochloric HCl metal cleaning, food prep, ore refining, stomach acid Strong

Phosphoric H3PO4 fertilizers, plastics, food preservation Moderate

Chloric HClO3 explosives Moderate

Acetic HC2H3O2 plastics, food preservation, vinegar Weak

Hydrofluoric HF metal cleaning, glass etching Weak

Carbonic H2CO3 soda water, blood buffer Weak

Hypochlorous HClO sanitizer Weak

Boric H3BO3 eye wash Weak

Strong Acids and Their Ionization in Water

Common Weak Acids and Their Ionization in Water

4.74

3.75

3.17

7.54

3.15

pKa = -log KaWhy do we not define Ka for a strong acid ??

Structure of Bases

Most contain OH- ions

Some contain CO32- ions

“Molecular” bases contain structures which react with H+

Name Formula Common Name Uses Strength

Sodium Hydroxide NaOH Lye, Caustic Soda soap, plastic production, petroleum refining

Strong

Potassium Hydroxide KOH Caustic Potash soap, cotton processing,

electroplatingStrong

Calcium Hydroxide

Ca(OH)2 Slaked Lime cement Strong

Sodium Bicarbonate

NaHCO3 Baking Soda food preparation, antacids Weak

Magnesium Hydroxide Mg(OH)2 Milk of Magnesia antacids Weak

Ammonium Hydroxide

NH4OH Ammonia Water fertilizers, detergents, explosives Weak

Common Inorganic Bases

Strong Bases and Their Ionization in Water

Weak Bases and Their Ionization in Water

pKb = -log Kb

acetic acid

Arrhenius Theory

Bases dissociate in water to produce OH- and cations:

Acids ionize in water to produce H+ and anions:

acetate anion

Arrhenius Theory

An Arrhenius Acid-Base Reaction

Problems with Arrhenius Theory

Does not explain why some molecular substance dissolve in water to give basic solutions, but do

not contain OH-. (NH3)

Does not explain why some ionic substances dissolve in water to give basic solutions, but do

not contain OH-. (NaCO3)

Does not explain why some molecular substance dissolve in water to give acidic solutions, but do

not contain H+. (CO2)

Only applicable to aqueous solutions.

Brønsted-Lowry Theory

Acid are proton donors

Bases are proton acceptors

Acid-Base Reactions involve proton transfer

Base structure includes a pair of unshared electrons

Brønsted-Lowry Theory

In a Brønsted-Lowry reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a

base in the reverse process.

Each reactant and product are a conjugate pair.

The original base becomes a conjugate acid, and the original acid becomes a conjugate base.

Conjugate Pairs

H-A + :B ⇔ :A- + H-B+

acid base conjugate conjugate base acid

HCHO2 + H2O ⇔ CHO2- + H3O+

acid base conjugate conjugate base acid

H2O + NH3: ⇔ HO- + NH4+

acid base conjugate conjugate base acid

Conjugate Pairs

Conjugate Pairs

H2O and OH- are an acid/base conjugate

pair.

NH3 and NH4+ are an base/acid conjugate

pair.

Lewis Acid-Base Theory

A Lewis base is an electron donor or nucleophile.

A Lewis acid is an electron acceptor or electrophile.

When a nucleophile donates a pair of electrons to an electrophile, a covalent bond forms.

16.2

pH and the Autoionization of Water

Autoionization of WaterWater is an extremely weak electrolyte.

One out of every 10 million water molecules forms ions.

All aqueous solutions, therefore, contain some H+ and OH-.

Autoionization of Water

The Ion Product of Water is Defined:

at 25ºC in the absence of any other acids or

bases.

A change in [H3O+] causes and inverse change in [OH-].

Autoionization of Water

Basic SolutionNeutral SolutionAcidic Solution

Relationship between [H3O+] and [OH-]

[H3O+] = [OH-][H3O+] > [OH-] [H3O+] < [OH-]

H+ H+ H+ H+ H+

OH-OH-OH-OH-OH-

[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

[OH-] 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100

Relationship between [H+] & [OH-]

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

pH as a Measure of Acidity/Basicity

pH = -log [H3O+]

pH < 7 , acidic

pH > 7, basic

pH = 7, neutral

pH of Common Substances

H+ H+ H+ H+ H+

OH-OH-OH-OH-OH-

[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

[OH-] 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100

pH vs pOH

pH + pOH =14

pOH 14 13 11 9 7 5 3 1 0

pH 0 1 3 5 7 9 11 13 14

1) What are the [OH–] and pH of household ammonia, an aqueous solution that has an [H3O+] of 1.99 x 10-12 M?

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

(1.99 x10-12)[OH-]=1.0 x 10-14

[OH-]=(1.0 x 10-14)/(1.99 x10-12)

[OH-]=5.02 x 10-3

pH = -log[H3O+] =11.70

pOH = -log[OH-] = 2.30

pKa = -logKa

pKa = -logKa

16.3

Calculations Involving pH, Ka and Kb

2) Calculate and compare [H+] and pH for a 0.100 M solution of HClO4 and a 0.100 M solution of HClO (Ka = 2.9 × 10–8).

HClO4 → ClO- + H+

HClO ⇄ ClO- + H+

For a strong acid, [H+] = 0.100 M, pH = -log(0.10) = 1

For the weak acid, a little more work is needed.

x2 0.100-x = 2.9 x 10-8

[H+] = x = 5.4 x 10-5

pH = 4.3

3) The pH of a 0.050 M solution of a weak organic acid is 2.23 . Calculate [H+], Ka and % ionization for the acid.

[H+] = 10-pH = 10-2.23 = 5.89 x 10-3

[A-][H+] [HA]Ka = = [5.89 x 10-3]2

[0.044]= 7.9 x 10-4

% ionization = (5.9 x 10-3/0.050) x 100 = 12%

4) Calculate the percent ionization for 1.28 M HNO2 and 0.0150 M HNO2 solutions (Ka = 4.0 × 10–4).

[A-][H+] [HA]Ka = = [x][x]

[1.28-x] = 4.0 × 10–4

[x]2 [1.28] = 4.0 × 10–4

x = 2.3 × 10–2

% ionization = (2.3 x 10-2/1.28) x 100 = 1.8%

4) Calculate the percent ionization for 1.28 M HNO2 and 0.0150 M HNO2 solutions (Ka = 4.0 × 10–4).

[A-][H+] [HA]Ka = = [x][x]

[0.0150-x] = 4.0 × 10–4

[x]2 [0.0150] = 4.0 × 10–4

x = 2.3 × 10–3

% ionization = (2.4 x 10-3/0.0150) x 100 = 16%

(2.6 x 10–3)

(17%)

5) Calculate and compare [OH-] and pH for 0.200 M LiOH and a 0.200 M solution of methylamine, CH3-NH2 (Kb = 4.4 × 10–4).

LiOH → Li+ + OH-

CH3-NH2 + H2O ⇄ CH3-[NH3]+ + OH-

For a strong base, [OH-] = 0.200 M, pOH = -log(0.20) = 0.7, pH=13.3

For the weak base, once again, a little more work is needed.

[BH+][OH-] [B]Kb = =

[x][x] [0.200-x] = 4.4 × 10–4

[x][x] [0.200] = 4.4 × 10–4

x = 9.4 × 10–3 = [OH-]

pOH = -log(9.4 × 10–3) = 2.03, pH = 11.97

6) Calculate [H+] and pH for 1.0 × 10–8 M HCl. HCl → H+ + Cl-

For a strong acid, [H+] = 1.0 × 10–8 M, pH = 8.0, BUT

THIS DOES NOT MAKE SENSE !!!

H2O + H2O ⇄ H3O+ + OH-

THERE IS ANOTHER EQUILIBRIUM TO CONSIDER

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

(x + 1.00 x 10-8) (x) = 1.0 x 10-14

x2 + (1.00 x 10-8)(x)-1.0 x 10-14= 0solve quadratic x = 9.5 x 10-8 = [OH-]

[H+] = 9.5 x 10-8 + 1.0 x 10-8 = 10.5 x 10-8 = 1.05 x 10-7

IF “x” REPRESENTS [H+] AND [OH-] FROM AUTOIONIZATION

pH = 6.98

16.4 Polyprotic Acids

Polyprotic Acids

7) Calculate the pH of 0.034 M carbonic acid solution (Ka1 = 4.3 × 10–7 and Ka2 = 4.7 × 10–11).

[HCO3-][H+] [HA]Ka1 =

[x][x] [0.034]4.3 x 10-7 =

1.2 x 10-4 = x = [H+]

1.2 x 10-4 = x = [HCO3-]

0.034 = [H2CO3]

7) Calculate the pH of 0.034 M carbonic acid solution (Ka1 = 4.3 × 10–7 and Ka2 = 4.7 × 10–11).

[CO32-][H+] [HCO3-]Ka2 =

[x][x +1.2 x 10-4] [1.2 x 10-4]4.7 x 10-11 =

4.7 x 10-11 = x = [CO32-]

1.2 x 10-4 = [HCO3-]

✔

pH = -log [H+] = -log (1.2 x 10-4 + 4.7 x 10-11) = 3.92

Polyprotic Acids

Polyprotic Acids