Acid and Base Equilibria Electrolytes Strong Conduct electricity Weak Poor conductors of electricity Nonelectrolytes Do not conduct electricity.

Acid and Base Equilibria

Electrolytes StrongConduct electricity WeakPoor conductors of

electricity NonelectrolytesDo not conduct

electricity

Strong Electrolytes Completely ionize or dissociate in

dilute aqueous solutions. Strong acids Strong soluble bases Most soluble salts

HCl + H2O H3O+ + Cl-

KOH K+ + OH-

Concentration of Ions Calculated

directly from molarity of strong electrolyte

What is the concentration of hydrogen ion in a 0.25 M HCl solution?

HCl H+ + Cl-

initial 0.25 Mchange -0.25 M 0.25M 0.25Mfinal 0 M 0.25M 0.25M

Concentration of Ions Calculated

directly from molarity of strong electrolyte

What is the concentration of hydroxide ion in a 0.25 M Ba(OH)2 solution?

Ba(OH)2 Ba2+ + 2OH-

initial 0.25 Mchange -0.25 M 0.25M 2(0.25M)final 0 M 0.25M 0.50M

Weak Electrolyte Slightly ionized in dilute aqueous

solutions. Weak acids Weak bases A few covalent salts

Auto-Ionization of Water

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

Kw = [H3O+][OH-]OH-

Auto-Ionization of Water

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25oC

In a neutral solution [H3O+] = [OH-]

and so [H3O+] = [OH-] = 1.00 x 10-7 M

H2O + H2O H3O+ + OH-

Kw = [H3O+] [OH-]

1000.1

1000.11000.1

H2O + H2O H3O+ + OH-

The Auto-Ionization of Water Calculate the concentrations of H3O+

and OH- in 0.050 M HCl.

0.050OH thus

0.050 0.050 050.0

Cl OH OH + HCl

The Auto-Ionization of Water Calculate the concentrations of H3O+

and OH- in 0.050 M HCl.

100.2OH

100.5100.1

OH100.1

100.1OHOH

0.050OH thus

0.050 0.050 050.0

Cl OH OH+HCl

[H3O+], [OH-] and pH A common way to express acidity

and basicity is with pH

pH = - log [H3O+] In a neutral solution, [H3O+] = [OH-] = 1.00 x 10-7 at 25oCpH = -log (1.00 x 10-7) = - (-7) pH = 7.00

[H3O+], [OH-] and pH

What is the pH of the 0.0010 M NaOH solution?

[H3O+] = 1.0 x 10-11 M

pH = - log (1.0 x 10-11) = 11.00General conclusion — Basic solution pH > 7 Neutral pH = 7 Acidic solution pH < 7

1000.1

001.01000.1

If the pH of Coke is 3.12, What is [H3O+]?

Because pH = - log [H3O+] then

log [H3O+] = - pHTake antilog and get

[H3O+] = 10-pH

[H3O+] = 10-3.12

[H3O+] = 7.6 x 10-4 M

[H3O+], [OH-] and pH

Other pX Scales

In general pX = -log Xand so pOH = - log

[OH-] Kw = [H3O+] [OH-] = 1.00 x 10-14

Take the -log of both sides-log (10-14) = - log [H3O+] + (-log

[OH-])14 = pH + pOH

pKa Values The Ka of acetic acid is 1.8 x 10-5,

what is the pKa?

pKa = -log Ka = -log (1.8 x 10-5) = 4.74

[H+][H+][OH-][OH-]

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

The pH and pOH scales Develop familiarity with pH scale by

looking at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.[H3O

+] [OH-] pH pOH 1.0 M 1.0 x 10-14 M 0.00 14.00

1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00 1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00 2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30 1.0 x 10-14 M 1.0 M 14.00 0.00

The pH and pOH scales Calculate [H3O+], pH, [OH-], and pOH for

0.020 M HNO3 solution.

70.1100.2-logpH 100.2OH

0.020 0.020 020.0

NOOHOHHNO

100%23

The pH and pOH scales Calculate [H3O+], pH, [OH-], and pOH for

0.020 M HNO3 solution.

30.12100.5logpOH

100.5100.2100.1

OH100.1

100.1OHOH

70.1100.2-logpH 100.2OH

0.020 0.020 020.0

NOOHOHHNO

100%23

pH Examples

Ionization Constants for Weak Monoprotic Acids Let’s look at the dissolution of acetic acid, a

weak acid, in water as an example. The equation for the ionization of acetic

acid is:

COOCHOHOH COOHCH -3323

COOHCH

COOCHOHK

Ionization Constants for Weak Monoprotic Acids and Bases Write the equation for the ionization of

the weak acid HCN and the expression for its ionization constant.

a 100.4HCN

HCN H+ + CN-

Equilibria Involving Weak Acids and Bases

Consider acetic acid

HOAc + H2O H3O+ + OAc-

Acid Conj. base

Equilibria Involving Weak Acids and Bases

Consider acetic acid

HOAc + H2O H3O+ + OAc-

Acid Conj. base

(K is designated Ka for ACID)Because [H3O+] and [OAc-] are SMALL, Ka << 1.

Ka [H3O+][OAc- ]

[HOAc] 1.8 x 10-5

Ka From Equilibrium Concentrations A solution of a weak acid has the

following concentrations at equilibrium. What is the Ka? [HA] = 0.049 M; [H+] = [A-] = 0.00084 MHA + H2O H3O+ + A-

Ka = [H3O+][A-]/[HA]Ka = (0.00084)2/0.049Ka = 1.4 x 10-5

Ka From Percent Ionization In a 0.0100 M solution, acetic acid

is 4.2% ionized. What is the Ka? HA + H2O H3O+ + A-

initial 0.0100change -0.00042 0.00042 0.00042equil 0.00958 0.00042 0.00042Ka = [H3O+][A-]/[HA]Ka = (0.00042)2/0.00958Ka = 1.8 x 10-5

Ka From pH The pH of a 0.115 M weak acid is

1.92. What is the Ka? HA + H2O H3O+ + A-

[H3O+] = 10-1.92 = 0.012 Minitial 0.115change -0.012 0.012 0.012equil 0.103 0.012 0.012Ka = [H3O+][A-]/[HA]Ka = (0.012)2/0.103Ka = 1.4 x 10-3

Weak AcidYou have 1.00 M HOAc. Calc. the equilibrium

concs. of HOAc, H3O+, OAc-, and the pH.

Step 1. Define equilibrium concs.

[HOAc] [H3O+] [OAc-]

Initial

Change

Equilib

Weak Acid You have 1.00 M HOAc. Calc. the

equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 1. Define equilibrium concs. [HOAc] [H3O+] [OAc-]Initial 1.00 0 0Change -x +x +xEquilib 1.00-x x x

Weak Acid Step 2. Write Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

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

[HOAc]

1.00 - x

This is a quadratic. Solve using quadratic formula or method of approximations

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

[HOAc]

1.00 - x

Weak Acid Step 3. Solve Ka expression

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

[HOAc]

1.00 - x

First assume x is very small because Ka is so small.

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

[HOAc]

1.00 - x

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

[HOAc]

1.00 - x

Ka 1.8 x 10-5 = x2

And so x = [H3O+] = [OAc-] = [Ka • 1.00]1/2

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

[HOAc]

1.00 - x

Ka 1.8 x 10-5 = x2

Weak Acid Step 3. Solve Ka approximate

expression

x = [H3O+] = [OAc-] = [Ka • 1.00]1/2

Ka 1.8 x 10-5 = x2

expression

x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 x = [H3O+] = [OAc-] = 4.2 x 10-3 M

Ka 1.8 x 10-5 = x2

expression

x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) pH = 2.37

Ka 1.8 x 10-5 = x2

Percent Ionization Calculate the percent ionization of

a 0.10 M solution of acetic acid.HA + H2O H3O+ + A-

initial 0.10change -x x xequil 0.10 - x x xKa = [H3O+][A-]/[HA] = x2/0.10 - x1.8 x 10-5 = x2/0.10 - x = x2/0.10x = 1.3 x 10-3

Calculations Based on Ionization Constants Calculate the concentrations of the

species in 0.15 M hydrocyanic acid, HCN, solution.

Ka = 4.0 x 10-10 for HCNYou do it!

Calculations Based on Ionization Constants Note that the properly applied

simplifying assumption gives the same result as solving the quadratic equation does.

2a4acbb

0100.6100.4

100.415.0

Calculations Based on Ionization Constants

1121010

107.7- and 107.7

12100.614100.4100.4

7.7 x 10-6 is the VALID chemical solution!

Calculations Based on Ionization Constants Let’s look at the percent ionization of two weak acids as a function of their ionization constants.

Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

Solution Ka [H+] pH % ionization0.15 M

CH3COOH1.8 x 10-5 1.6 x 10-3 2.80 1.1

0.15 MHCN

4.0 x 10-10 7.7 x 10-6 5.11 0.0051

Weak BaseYou have 0.010 M NH3. Calc. the pH.

NH3 + H2O NH4+ + OH-

Kb = 1.8 x 10-5

[NH3] [NH4+] [OH-]

initial 0.010 0 0

change -x +x +x

equilib 0.010 - x x x

Kb = 1.8 x 10-5

[NH3] [NH4+] [OH-]

initial 0.010 0 0

change -x +x +x

equilib 0.010 - x x x

Weak Base

You have 0.010 M NH3. Calc. the pH.

Kb = 1.8 x 10-5

Step 2. Solve the equilibrium expressionKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

0.010 - x

Kb = 1.8 x 10-5

Step 2. Solve the equilibrium expression

Assume x is small (100•Kb < Co), so x = [OH-] = [NH4

+] = 4.2 x 10-4 M

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

0.010 - x

Kb = 1.8 x 10-5

Assume x is small (100•Kb < Co), so x = [OH-] = [NH4

+] = 4.2 x 10-4 M

and [NH3] = 0.010 - 4.2 x 10-4 = 0.010 M

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

0.010 - x

Weak Base

You have 0.010 M NH3. Calc. the pH.

Kb = 1.8 x 10-5

Step 3. Calculate pH[OH-] = 4.2 x 10-4 Mso pOH = - log [OH-] = 3.37Because pH + pOH = 14,pH = 10.63

Example 18-15 The pH of household ammonia is

11.50. What is its molarity?NH3 + H2O NH4+ + OH-

pOH = 14 - 11.50 = 2.50[OH-] = 10-2.50 = 0.0032 Minitial xchange - 0.0032 0.0032 0.0032 equil x - 0.0032 0.0032 0.0032 Kb = [NH4

+][OH-]/[NH3 ] = (0.0032 )2/x - 0.0032 1.8 x 10-5 = (0.0032 )2/xx = 0.57 M

Polyprotic Acids Many weak acids contain two or more acidic hydrogens.

polyprotic acids ionize stepwise ionization constant for each step

Consider arsenic acid, H3AsO4, which has three ionization constants1 K1=2.5 x 10-4

2 K2=5.6 x 10-8

3 K3=3.0 x 10-13

Polyprotic Acids The first ionization step is

H AsO H H AsO

KH H AsO

3 4 2 4

2 5 10 4.

Polyprotic Acids The second ionization step is

H AsO H HAsO

KH HAsO

285 6 10.

Polyprotic Acids The third ionization step is

HAsO H AsO

KH AsO

3133 0 10.

Polyprotic Acids Notice that the ionization constants vary in the following

fashion:

This is a general relationship.K K K1 2 3

Polyprotic Acids Calculate the concentration of all species

in 0.100 M arsenic acid, H3AsO4, solution.

1 Write the first ionization ionization step and represent the concentrations.

xMxMMx 100.0

AsOHHAsOH 4243

Polyprotic Acids

2 Substitute into the expression for K1.

applynot does assumption gsimplifyin

0105.2105.2

105.210.0

105.2AsOH

AsOHHK

Polyprotic Acids

3 Use the quadratic equation to solve for x, and obtain two values

095.0100.0AsOH

109.4AsOHH

109.4 and 101.5

105.214105.2105.2

Polyprotic Acids

4 Now we write the equation for the second step ionization and represent the concentrations.

H AsO H + HAsO

from 1st step (4.9 10

algebraically (4.9 10

2 4- +

y M yM yM

Polyprotic Acids

5 Substitute into the second step ionization expression.

K =H O HAsO

K =4.9 10

4.9 10

apply assumption

5 6 10 8.

Polyprotic Acids

K =H O HAsO

K =4.9 10

4.9 10

apply assumption

K =4.9 10

4.9 10

H HAsO

note H H

2nd 42

1st 2nd

5 6 10

Polyprotic Acids

6 Now we repeat the procedure for the third ionization step.

HAsO H AsO

1st and 2nd ionizations 5.6 10 4.9 10 5.6 10

algebraically 5.6 10

-8 -3 -8

z M z M z M

Polyprotic Acids

7 Substitute into the third ionization expression.

K =H O AsO

apply assumption

3 0 10

4 9 10 5 6 10

5 6 10

Polyprotic Acids

K =H O AsO

apply assumption

3rd 43

3 0 10

4 9 10 5 6 10

5 6 10

4 9 10

5 6 103 0 10

3 4 10

Polyprotic Acids Use Kw to calculate the [OH-] in the

0.100 M H3AsO4 solution.

10 10 10 10

4 9 10

2 0 10

Polyprotic Acids A comparison of the various species in

0.100 M H3AsO4 solution follows.Species ConcentrationH3AsO4 0.095 M

H+ 0.0049 MH2AsO4

- 0.0049 MHAsO4

2- 5.6 x 10-8 MAsO4

3- 3.4 x 10-18 MOH- 2.0 x 10-12 M

Solvolysis Solvolysis is the

reaction of a substance with the solvent in which it is dissolved.

Hydrolysis refers to the reaction of a substance with water or its ions.

Hydrolysis …the reaction

of a substance with water or its ions

A- + H2O HA + OH-

Ah ha, BASIC!

BH+ + H2O B + H3O+

Hydrolysis Hydrolysis refers to the reaction of

a substance with water or its ions. Combination of the anion of a

weak acid with H3O+ ions from water to form nonionized weak acid molecules.A- + H2O HA + OH-

H2O H+OH-

NaCNNa+ + H2O NR

CN- + H2O HCN + OH-

BasicHCN + NaOH NaCN + H2O

Hydrolysis

NH4ClCl- + H2O NR

NH4+ + H2O NH3 + H3O+

Acidic

HCl + NH3 NH4Cl

NH4CNNH4

+ + H2O NH3 + H3O+

CN- + H2O HCN + OH-

Acidic or BasicHCN + NH4OH NH4CN + H2O

Hydrolysis

Salts of Weak Bases and Weak Acids Acidic Solution

Ka > KbNH4F NH4

+ + F-

NH4+ + H2O NH3 + H3O+

F- + H2O HF + OH-

Ka = 7.2 x 10-4 and Kb = 1.8 x 10-5

Salts of Weak Bases and Weak Acids Neutral Solution

Ka = KbNH4CH3COO NH4

+ + CH3COO-

NH4+ + H2O NH3 + H3O+

CH3COO- + H2O CH3COOH + OH-

Ka = 1.8 x 10-5 and Kb = 1.8 x 10-5

Salts of Weak Bases and Weak Acids Basic Solution

Ka < KbNH4CN NH4

+ + CN-

NH4+ + H2O NH3 + H3O+

CN- + H2O HCN + OH-

Ka = 4.0 x 10-10 and Kb = 1.8 x 10-5

Salts of Weak Bases and Weak Acids Acidic Solution

Ka > Kb Basic solution

Kb > Ka Neutral solution

Ka = Kb

Hydrolysis The conjugate base of a strong

acid is a very weak base. The conjugate base of a weak acid

is a stronger base. Hydrochloric acid, a typical strong

acid, is essentially completely ionized in dilute aqueous solutions.

HCl H O H O Cl2 3 ~100%

Hydrolysis The conjugate base of HCl, the Cl-

ion, is a very weak base.

True of all strong acids and their anions.

Cl H O No rxn. in dilute aqueous solutions3

Hydrolysis HF, a weak acid, is only slightly ionized in

dilute aqueous solutions. Its conjugate base, the F- ion, is a much

stronger base than the Cl- ion. F- ions combine with H3O+ ions to form

nonionized HF.HF + H O H O F

only slightly

F + H O HF + H O

nearly completely

2 3+ -

Salts of Strong Soluble Bases and Strong Acids Salts made from strong acids and

strong soluble bases form neutral aqueous solutions.

An example is potassium nitrate, KNO3, made from nitric acid and potassium hydroxide.

K NO K NO

H O H O OH + H O

no rxn. to upset H O OH neutral

in H O +3

Salts of Strong Soluble Bases and Weak Acids Salts made from strong soluble bases and

weak acids hydrolyze to form basic solutions. Anions of weak acids (strong conjugate bases) react

with water to form hydroxide ions An example is sodium hypochlorite, NaClO,

made from sodium hydroxide and hypochlorous acid.

Na ClO Na ClO

H O + H O OH + H O

+ - in H O -

Salts of Strong Soluble Bases and Weak Acids

Na ClO Na ClO

H O + H O OH + H O

ClO H O HClO H O

+ - in H O -

Combine these equations into one single equation that represents the reaction:

ClO H O HClO OH-2

Acid-Base Properties of Salts

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

Reaction of NH4+ with H2O

NH4+ + H2O NH3 + H3O+

acid base base acidNH4

+ ion is a moderate acid because its conjugate base is weak.

Therefore, NH4+ is acidic solution

Salts of Weak Bases and Strong Acids Salts made from weak bases and strong

acids form acidic aqueous solutions. An example is ammonium bromide,

NH4Br, made from ammonia and hydrobromic acid.

acidic issolution OH excess generates

OHNHOHNH

OH OH OHOH

Br NH BrNH

100%~OHs

Hydrolysis Constant Relationship of

conjugate acid-base pairs

Kw = KaKb

Example The Ka for acetic acid, CH3COOH, is

1.8 x 10-5, calculate the Kb for the acetate ion, CH3COO-.

106.5108.1

1000.1

Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NH4Cl.

Cl- + H2O is neutral

NH4++ H2O NH3 + H3O+

acid base base acid Ka = 5.6 x 10-10

Step 1. Set up concentration table [NH4

+] [NH3] [H3O+] initial change equilib

0.10 0 0-x +x +x0.10-x x x

OHNHxKa

10.0106.5

Calculate the pH of a 0.10 M solution of NH4Cl.

Assume 0.10-x = 0.10

OHNHxKa

10.0106.5

Assume 0.10-x = 0.10x = [NH3] = [H3O+] = 7.5 x 10-6 M

OHNHxKa

10.0106.5

Cl- + H2O Ž neutral

NH4+ + H2O NH3 + H3O+

Step 3. Calculate the pH

[H3O+] = 7.5 x 10-6 MpH = -log [H3O+] = 5.13

Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NaCH3COO.

Na+ + H2O Ž neutral

base acid acid base Kb = 5.6 x 10-10

Step 1. Set up concentration table [CH3COO-

] [CH3COOH] [OH-] initial change equilib

Step 1. Set up concentration table [CH3COO-

] [CH3COOH] [OH-] initial 0.10 0 0 change -x +x +x equilib 0.10-x x x

Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of

NaCH3COO.

base acid acid base

Kb = 5.6 x 10-10

Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of

NaCH3COO.

base acid acid base

Kb = 5.6 x 10-10

Step 2. Solve the equilibrium expression x

OHCOOHCHxKb

10.0106.5

Assume 0.10-x = 0.10

OHCOOHCHxKb

10.0106.5

Assume 0.10-x = 0.10x = [CH3COO] = [OH-] = 7.5 x 10-6 M

OHCOOCHxKb

10.0106.5

Step 3. Calculate the pH[OH-] = = 7.5 x 10-6 MpOH = -log [OH-] = 5.13pH + pOH = 14.00pH = 8.87

Acid and Base Equilibria Electrolytes Strong Conduct electricity Weak Poor conductors of electricity Nonelectrolytes Do not conduct electricity.

log h3o log oh

h3o ohh2o h2o h3o oh

h3o ohauto

concentrations of h3o

acidbecause h3o

neutral solution h3o

phphh oh

poh scalescalculate

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