ACIDS AND BASES

ACID – A compound that produces hydrogen ions in a water solution

HCl (g) → H+(aq) + Cl-(aq)

BASE – A compound that produces hydroxide ions in a water solution

NaOH (s) → Na+(aq) + OH-(aq)

1884 SVANTE ARRHENIUS

Proposed the first definitions of acids and bases

1E-1 (of 25)

FORMATION OF ACIDS AND BASES

Bases are produced from the reaction of metal oxides with water

Na2O (s) + H2O (l) → 2NaOH (aq)

Acids are produced from the reaction of nonmetal oxides with water

CO2 (g) + H2O (l) → H2CO3 (aq)

1E-2 (of 25)

Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7

NaOH Mg(OH)2 Al(OH)3 Si(OH)4 PO(OH)3 SO2(OH)2 ClO3(OH)

H4SiO4 H3PO4 H2SO4 HClO4

AMPHOTERIC – A substance that can act as an acid or a base

base base base or

acid

acid acid acid acid

strongest base

strongest acid

Products of elemental oxides with water:

1E-3 (of 25)

base

Products of elemental oxides with water:

1+ 2+ 3+ 4+ 5+ 6+ 7+

The higher the element’s oxidation number, the more acidic its hydroxide

Cr(OH)2

Cr(OH)3

Cr(OH)6

amphoteric

acid

1E-4 (of 25)

Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7

NaOH Mg(OH)2 Al(OH)3 Si(OH)4 PO(OH)3 SO2(OH)2 ClO3(OH)

H4SiO4 H3PO4 H2SO4 HClO4

base base base or

acid

acid acid acid acid

1923

Expanded the definitions of acids and bases

THOMAS LOWRY

1E-5 (of 25)

ACID – A hydrogen ion (or proton) donor

BASE – A hydrogen ion (or proton) acceptor

1923

Expanded the definitions of acids and bases

JOHANNES BRØNSTED

1E-6 (of 25)

HCl + H2O →

1E-7 (of 25)

HCl + H2O → Cl- + H3O+

acid base

HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water

+-

1E-8 (of 25)

NH3 + H2O →

1E-9 (of 25)

NH4+ + OH- NH3 + H2O →

base acid

Water is amphoteric

+ -

1E-10 (of 25)

Acids turn into bases, and bases turn into acids

HCl + H2O Cl- + H3O+

acid base

→

1E-11 (of 25)

conjugate base

of HCl

conjugate acid

of H2O

Acids turn into bases, and bases turn into acids

HCl + H2O Cl- + H3O+

acid base

←

NH3 + H2O → NH4+ + OH-

base acid conjugate acid

of NH3

conjugate base of H2O

1E-12 (of 25)

conjugate base

of NH4+

conjugate acid

of NO2-

NH4+ + NO2

- → NH3 + HNO2

acid base

←

The favored reaction direction turns strong acids and bases into weak acids and bases

HNO2 is a stronger acid than NH4+

NH3 is a stronger base than NO2-

1E-13 (of 25)

AUTOIONIZATION OF WATER

Water ionizes itself to a small extent

1E-14 (of 25)

AUTOIONIZATION OF WATER

Water ionizes itself to a small extent

2H2O (l) → H3O+(aq) + OH-(aq)

-+

Makes solutions acidic Makes solutions basic

Equal amounts of H3O+ and OH- make a solution NEUTRAL

pure water is neutral

1E-15 (of 25)

The ionization of water is an endothermic process

2H2O (l) ⇆ H3O+ (aq) + OH-

(aq)

Write the equilibrium constant expression for the autoionization of water

Keq = [H3O+][OH-]

ION-PRODUCT CONSTANT FOR WATER (Kw) – The equilibrium constant for the ionization of water

20 25 30

0.68 x 10-14

1.00 x 10-14

1.47 x 10-14

Temp. (°C) Kw

energy +

1E-16 (of 25)

Find [H3O+] and [OH-] in pure water at 25°C.

x x

2H2O (l) ⇆ H3O+ (aq) + OH- (aq)

Initial M’sChange in M’sEquilibrium M’s

0 0

+ x + x

Kw = [H3O+][OH-]

1.00 x 10-14 M2 = x2

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

1E-17 (of 25)

THE pH SCALE

pH – The negative logarithm of [H3O+] of a solution

The common logarithm of a number is:the exponent to which 10 must be raised to equal the number

100 0.01 0.001

100 = 102 0.01 = 10-2 0.001 = 10-3

0.002

0.002 = 10-2.699

log 100 = 2 log 0.01 = -2 log 0.001 = -3 log 0.002 = -2.699

pOH – The negative logarithm of [OH-] of a solution

1E-18 (of 25)

Calculate the pH of orange juice if its [H3O+] = 2.5 x 10-4 M.

= -log(2.5 x 10-4 M)

pH = -log[H3O+]

= -(0.40 + -4.000000…)

1E-19 (of 25)

= 3.60

For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before

= 3.6 incorrect number of significant figures

= -[log(2.5) + log(10-4)]

Calculate the pH and pOH of pure water at 25ºC.

= -log(1.00 x 10-7 M)

pH = -log[H3O+]

For pure water:

[H3O+] = 1.00 x 10-7 M

= 7.000

= -log(1.00 x 10-7 M)

pOH = -log[OH-]

[OH-] = 1.00 x 10-7 M

= 7.000

1E-20 (of 25)

In any water solution:

Kw = [H3O+][OH-]

Kw = [OH-] ________

[H3O+]

at 25°C

1.00 x 10-14 = [OH-] _____________

[H3O+]

1E-21 (of 25)

For orange juice:

[H3O+] = 2.5 x 10-

4 M

1.00 x 10-14 M2 = [OH-] _________________

2.5 x 10-4 M

= 4.0 x 10-11 M

pH 7 = Neutral< 7 is Acidic > 7 is Basic

7 8 9 10 11 12 13 14650 3 421

[H3O+] = 10-7 M[OH-] = 10-7 M

[H3O+] = 10-3 M[OH-] = 10-11 M

[H3O+] = 10-12 M[OH-] = 10-2 M

-1 15

[H3O+] = 101 M[OH-] = 10-15 M

1E-22 (of 25)

In any water solution:

Kw = [H3O+][OH-]

log Kw = log ([H3O+][OH-])

-log Kw = -log ([H3O+][OH-])

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

pKw = pH + pOH

at 25°C, -log(1.00 x 10-14) = 14.000

∴ 14.000 = pH + pOH

1E-23 (of 25)

Calculate the pH and pOH of soda if its [H3O+] = 1.6 x 10-3 M.

= -log(1.6 x 10-3 M)

pH = -log[H3O+]

[H3O+] = 1.6 x 10-3 M

= 2.80

= 14.000 - 2.80

pOH = pKw - pH

pH + pOH = pKw

= 11.20

1E-24 (of 25)

Calculate the [H3O+] of blood, which has a pH of 7.4.

= 0.0000000398 M

pH = -log[H3O+]

-pH = log[H3O+]

antilog (-pH) = [H3O+]

antilog (-7.4) = [H3O+]

For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before

= 3.98 x 10-8 M = 4 x 10-8 M

1E-25 (of 25)

IONIZATION OF ACIDS

HF (aq) + H2O(l) ⇆ H3O+(aq) + F-(aq)

Write the Keq expression for the ionization of hydrofluoric acid

= [H3O+][F-]

____________

[HF]ACID IONIZATION CONSTANT (Ka) – The equilibrium constant for an acid ionizing

Keq

1F-1 (of 19)

IONIZATION OF ACIDS

HF (aq) + H2O(l) ⇆ H3O+(aq) + F-(aq)

Write the Keq expression for the ionization of hydrofluoric acid

= [H3O+][F-]

____________

[HF]ACID IONIZATION CONSTANT (Ka) – The equilibrium constant for an acid ionizing

Ka

1F-2 (of 19)

Acids are

a) strong if every acid molecule gives up a hydrogen ionHCl, HBr, HI, any acid with 2 or more O’s than H’sKa = large

b) weak if less than every acid molecule gives up a hydrogen ionall other acidsKa = small

1F-3 (of 19)

()

IONIZATION OF BASES

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

Write the Keq expression for the ionization of ammonia

= [NH4+][OH-]

______________

[NH3]BASE DISSOCIATION CONSTANT (Kb) – The equilibrium constant for a base ionizing

Keq

1F-4 (of 19)

IONIZATION OF BASES

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

Write the Keq expression for the ionization of ammonia

= [NH4+][OH-]

______________

[NH3]BASE DISSOCIATION CONSTANT (Kb) – The equilibrium constant for a base ionizing

Kb

1F-5 (of 19)

Bases are

a) strong if every molecule/ion accepts a hydrogen ionalkali metal hydroxides, dilute Ca(OH)2, Sr(OH)2, Ba(OH)2 solutions

Kb = large

b) weak if less than every molecule/ion accepts a hydrogen ionall other bases, including NH3

Kb = small

1F-6 (of 19)

Calculate the pH of 0.010 M nitric acid.

1F-7 (of 19)

CALCULATING THE pH OF STRONG ACID OR BASE SOLUTIONS

Assume complete ionization or dissociation

0.010 0.010

HNO3 (aq) + H2O (l) → H3O+ (aq) + NO3

- (aq)

Initial M’sChange in M’sFinal M’s

0.010 0 0

- x + x + x

0

= -log(0.010 M)

pH = -log[H3O+]

= 2.00

- 0.010 + 0.010 + 0.010

Calculate the pH of 0.0010 M sodium hydroxide.

1F-8 (of 19)

0.0010 0.0010

NaOH (aq) → Na+ (aq) + OH- (aq)

Initial M’sChange in M’sFinal M’s

0.0010 0 0

- x + x + x

0

= -log(0.0010 M)

pOH = -log[OH-]

= 3.00 = 14.000 - 3.00

pH = pKw - pOH

pH + pOH = pKw

= 11.00

- 0.0010 + 0.0010 + 0.0010

the Ka for HNO2 can be looked up

Calculate the pH of 0.010 M nitrous acid.

1F-9 (of 19)

CALCULATING THE pH OF WEAK ACID OR BASE SOLUTIONS

Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium

x x

HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2

- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.010 0 0- x + x + x

0.010 - x

: 4.0 x 10-4

Ka = [H3O+][NO2

-] ________________

[HNO2]

Ka = x2 _____________

(0.010 – x)

4.0 x 10-4 = x2 ______________

(0.010 – x)

If the Ka is < 10-2 the acid does not ionize much, so you may assume that x is very small, and ignore it when it is subtracted from the initial molarity

4.0 x 10-4 = x2 _______

0.010

2.00 x 10-3 = x

For x values that are >1% of the original acid molarity, they should be put back in place of x in the denominator of the Ka expression, and solved again

1F-10 (of 19)

4.0 x 10-4 = x2 _________________________

(0.010 – 2.00 x 10-3)

1.79 x 10-3 = x

Repeat until the answer (to 2 sig fig’s) is the same twice in a row

4.0 x 10-4 = x2 _________________________

(0.010 – 1.79 x 10-3)

1.81 x 10-3 = x

= -log(1.81 x 10-3 M)

pH= -log[H3O+] = 2.74

= [H3O+]

1F-11 (of 19)

Calculate the pH of 0.20 M acetic acid if its Ka = 1.8 x 10-5.

x x

HC2H3O2 (aq) + H2O (l) ⇆ H3O+ (aq) + C2H3O2

- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.20 0 0- x + x + x

0.20 - x

Ka = [H3O+][C2H3O2-]

____________________

[HC2H3O2]

1.8 x 10-5 = x2 ____________

(0.20 – x)

1.8 x 10-5 = x2 ______

0.20

1.90 x 10-3 = x

1F-12 (of 19)

1.8 x 10-5 = x2 ________________________

(0.20 – 1.90 x 10-3)

1.89 x 10-3 = x

= -log(1.89 x 10-3 M)

pH = -log[H3O+] = 2.72

Calculate the pH of 0.20 M acetic acid if its Ka = 1.8 x 10-5.

= [H3O+]

Calculate the percent ionization of acetic acid.

1.89 x 10-3 M _________________

0.20 M

x 100 = 0.95%

1F-13 (of 19)

Calculate the pH of a 0.10 M ammonia solution if its Kb = 1.8 x 10-5.

x x

NH3 (aq) + H2O (l) ⇆ NH4+

(aq) + OH- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.10 0 0- x + x + x

0.10 - x

Kb = [NH4+][OH-]

______________

[NH3]

1.8 x 10-5 = x2 ____________

(0.10 – x)

1.8 x 10-5 = x2 ______

0.10

1.34 x 10-3 = x

1F-14 (of 19)

1.8 x 10-5 = x2 ________________________

(0.10 – 1.34 x 10-3)

1.33 x 10-3 = x

= -log(1.33 x 10-3 M)

pOH= -log[OH-] = 2.88

Calculate the pH of a 0.10 M ammonia solution if its Kb = 1.8 x 10-5.

= [OH-]

= 14.000 - 2.88

pH = pKw - pOH

pH + pOH = pKw

= 11.12

1F-15 (of 19)

A 0.500 M solution of the weak acid HA has a pH 2.010. Calculate its Ka.

x x

HA (aq) + H2O (l) ⇆ H3O+ (aq) + A-

(aq)

Initial M’sChange in M’sEquilibrium M’s

0.500 0 0- x + x + x

0.500 - x

Ka = [H3O+][A-] _____________

[HA]

Ka = x2 ______________

(0.500 – x)

x = [H3O+]

= antilog (-2.010)

= antilog (-pH)

= 0.009772 M

1F-16 (of 19)

CALCULATING THE Ka OF A WEAK ACID FROM pH

A 0.500 M solution of the weak acid HA has a pH 2.010. Calculate its Ka.

Ka = (0.009772 M)2

_____________________________

(0.500 M - 0.009772 M)

= 1.95 x 10-4

What is the pKa of HA?

= -log(1.947 x 10-4)

pKa = -log Ka

= 3.710

The smaller the Ka (or the bigger the pKa) the weaker the acid

1F-17 (of 19)

CALCULATING THE Ka OF A WEAK ACID FROM pH

A 0.500 M solution of the weak acid HX is 3.15% ionized. Calculate its Ka.

x x

HX (aq) + H2O (l) ⇆ H3O+ (aq) + X- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.500 0 0- x + x + x

0.500 - x

Ka = [H3O+][X-] _____________

[HX]

Ka = x2 ______________

(0.500 – x)

3.15 = x (100) _______

0.500

0.01575 = x

1F-18 (of 19)

CALCULATING THE Ka OF A WEAK ACID FROM PERCENT IONIZATION

A 0.500 M solution of the weak acid HX is 3.15% ionized. Calculate its Ka.

Ka = (0.01575 M)2

_____________________________

(0.500 M - 0.01575 M)

= 5.12 x 10-4

What is the pKa of HX?

= -log(5.123 x 10-4)

pKa = -log Ka

= 3.291

1F-19 (of 19)

CALCULATING THE Ka OF A WEAK ACID FROM PERCENT IONIZATION

1G-1 (of 13)

POLYPROTIC ACIDS

Polyprotic acids have Ka values for the ionization of each H+

H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq)

Ka1 = [H3O+][HCO3-]

__________________

[H2CO3]

Ka1 = 4.3 x 10-7

HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO3

2- (aq)

Ka2 = [H3O+][CO32-]

_________________

[HCO3-]

Ka2 = 5.6 x 10-11

Successive H+’s are harder to remove

∴ H2CO3 is a stronger acid than HCO3-

H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq) Ka1 = 4.3 x 10-7

HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO3

2- (aq) Ka2 = 5.6 x 10-11

H2CO3 (aq) + 2H2O (l) ⇆ 2H3O+ (aq) + CO32- (aq) Ka1,2 = ?

Ka1,2 = (4.3 x 10-7)(5.6 x 10-11)

= 2.4 x 10-17

1G-2 (of 13)

[H3O+][HCO3-]

__________________

[H2CO3]

x[H3O+][CO3

2-]_________________

[HCO3-]

[H3O+]2[CO32-]

__________________

[H2CO3]

=

Ka1Ka2 = Ka1,2

Find the concentrations of each species in a 0.10 M H2CO3 solution.

x x

H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.10 0 0- x + x + x

0.10 - x

Ka1 = [H3O+][HCO3-]

_________________

[H2CO3]

4.3 x 10-7 = x2 ____________

(0.10 – x)

x = 2.07 x 10-4

[H2CO3] = 0.10 M – 0.000207 M = 0.10 M

[H3O+] = = 0.00021 M

[HCO3-] = = 0.00021 M

1G-3 (of 13)

Find the concentrations of each species in a 0.10 M H2CO3 solution.

0.000207 + y y

HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO3

2- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.000207 0.000207 0- y + y + y

0.000207 - y

Ka2 = [H3O+][CO32-]

_________________

[HCO3-]

5.6 x 10-11 = (0.000207 + y) y __________________

(0.000207 – y)

y = 5.6 x 10-11

[H2CO3] = = 0.10 M

[H3O+] = 0.000207 + 5.6 x 10-11 = 0.00021 M

[HCO3-] = 0.000207 – 5.6 x 10-11 = 0.00021 M

[CO32-] = = 5.6 x 10-11 M

1G-4 (of 13)

pH OF SALT SOLUTIONS

Acid

HNO3

HNO2

Strong acids have non conjugate basesWeak acids have weak conjugate bases

Strong

Weak

Conjugate Base

NO3-

NO2-

Non

Weak

Base

NH3

NaOH

NaOH(H2O)5

Strong bases have non conjugate acidsWeak bases have weak conjugate acids

Weak

Strong

Conjugate Acid

NH4+

Na+

Na(H2O)6+

Weak

Non

1G-5 (of 13)

from LiOH

which is a strong base

∴ Li+ is a non acid

from H3PO4

which is a weak acid

∴ PO43- is a weak base

Li3PO4

∴ pH > 7

1G-6 (of 13)

Li+ PO43-

PO43- (aq) + H2O (l) ⇆ HPO4

- (aq) + OH- (aq)

HYDROLYSIS – The reaction of a dissolved ion with water

from NH3

which is a weak base

∴ NH4+ is a weak acid

from HCl

which is a strong acid

∴ Cl- is a non base

NH4Cl

∴ pH < 7

1G-7 (of 13)

NH4+ Cl-

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

from KOH

which is a strong base

∴ K+ is a non acid

from HNO3

Which is a strong acid

∴ NO3- is a non base

KNO3

∴ pH = 7

1G-8 (of 13)

K+ NO3-

Find the pH of a 0.10 M potassium acetate solution

x x

C2H3O2- (aq) + H2O (l) ↔ HC2H3O2 (aq) + OH- (aq)

Initial M’sChange in M’sEquilibrium M’s

0.10 0 0- x + x + x

0.10 - x

Kb = [HC2H3O2][OH-] ___________________

[C2H3O2-]

5.6 x 10-10 = x2 ____________

(0.10 – x)

x = 7.48 x 10-6

Potassium ion is a non acid ; acetate ion is a weak base

Need the Kb for acetate : 5.6 x 10-10

1G-9 (of 13)

CALCULATING THE pH OF A SALT SOLUTION

7.48 x 10-6 = x

= -log(7.48 x 10-6 M)

pOH = -log[OH-] = 5.13

= [OH-]

= 14.000 – 5.13

pH = pKw - pOH

pH + pOH = pKw

= 8.87

Find the pH of a 0.10 M potassium acetate solution

1G-10 (of 13)

CALCULATING THE pH OF A SALT SOLUTION

HC2H3O2 (aq) + H2O (l) ⇆ H3O+ (aq) + C2H3O2- (aq) Ka for acetic acid

C2H3O2- (aq) + H2O (l) ⇆ HC2H3O2 (aq) + OH- (aq) Kb for acetate

2H2O (l) ⇆ H3O+ (aq) + OH- (aq) Kw

KaKb = Kw

1G-11 (of 13)

Find the pH of a 0.25 M ammonium chloride solution if the Kb for ammonia is 1.8 x 10-5.

Ammonium ion is a weak acid; chloride ion is a non base

Need the Ka for ammonium

Have the Kb for ammonia : 1.8 x 10-5

Ka = Kw

____

Kb

KaKb = Kw

= 1.00 x 10-14

_______________

1.8 x 10-5

= 5.56 x 10-10

1G-12 (of 13)

Find the pH of a 0.25 M ammonium chloride solution if the Kb for ammonia is 1.8 x 10-5.

x x

NH4+

(aq) + H2O (l) ⇆ H3O+ (aq) + NH3 (aq)

Initial M’sChange in M’sEquilibrium M’s

0.25 0 0- x + x + x

0.25 - x

Ka = [H3O+][NH3] ______________

[NH4+]

5.56 x 10-10 = x2 ____________

(0.25 – x)

= -log(1.18 x 10-5 M)

pH = -log[H3O+] = 4.93

x = 1.18 x 10-5 M

1G-13 (of 13)