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Acids and Bases Part IIDr. Sobers’ Lecture Notes

Acid and Base Ionization Constants Ka and Kb

Acid Ionization Constant

HAn(aq) + H2O(l) ⇄ H3O+ (aq) + An-1

An acid in water undergoes an ionization reaction. The equilibrium constant is the acid ionization constant, Ka.

Ka=[An-1][H3O

+]HAn⎡⎣ ⎤⎦

Acid + Water Hydronium ion

+ ConjugateBase

⇄

Know how to write the chemical equation and the expression

Base Ionization ConstantAn base in water undergoes an ionization reaction. The equilibrium constant is the base ionization constant, Kb.

Base + Water Hydroxide ion

+ ConjugateAcid

⇄

An-1(aq) + H2O(l) ⇄ OH- (aq) + HAn

Kb=[HAn ][OH− ]

An-1⎡⎣ ⎤⎦

Know how to write the chemical equation and the expression

ExamplesWrite the equilibrium reactions and equilibrium expressions for the following acids:

HF and NH4+

HF(aq) + H2O(l) ⇄ H3O+(aq) + F-(aq)Ka=

[F-1][H3O+]

HF[ ]

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

Ka=[NH3][H3O

+]NH4

+⎡⎣ ⎤⎦

ExamplesWrite the equilibrium reactions and equilibrium expressions for the following bases:

F- and CH3NH2

F-(aq) + H2O(l) ⇄ OH-(aq) + HF(aq)

Kb=[CH3NH3

+][OH-]CH3NH2[ ]

CH3NH2(aq) + H2O(l) ⇄ OH-(aq) + CH3NH3+(aq)

Kb=[OH-][HF]

F-⎡⎣ ⎤⎦

Acid and Base Strength


The equilibrium expressions for each equation may be written:


Ka=[An-1][H3O

+]HAn⎡⎣ ⎤⎦

Kb=[HAn ][OH− ]

An-1⎡⎣ ⎤⎦

The larger the Ka, the stronger the acid.

The larger the Kb, the stronger the base.

Sample Ka and Kb Problems

The pH of a 0.10M formic acid (HCOOH) solution is 2.39. Calculate the Ka of formic acid

Ka=[HCOO− ][H3O

+]HCOOH[ ]

[H3O+]=10− pH

What do we know about final equilibrium state?The pH

[H3O+] = 0.0041

HCOOH(aq) + H2O(l) ⇄ H3O+(aq) + HCOO- (aq)


HCOOH(aq) + H2O(l) ⇄ H3O+(aq) + HCOO- (aq)

[ ]initial

Δ[ ]final

0.10M 0M 0M

Ka=[HCOO− ][H3O

+]HCOOH[ ]

[H3O+]=10− pH

What do we know about final equilibrium state? The pH

[H3O+] = 0.0041

0.0041

+0.0041+0.0041

0.0041-0.0041

0.096


HCOOH(aq) + H2O(l) ⇄ H3O+(aq) + HCOO-

[ ]initial

Δ[ ]final

0.10M 0M 0M

Ka=[HCOO− ][H3O

+]HCOOH[ ]

0.0041

+0.0041+0.0041

0.0041-0.0041

0.096

= (0.0041)(0.0041)0.096

=1.8x10-4

The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH and percent ionization of a 0.250M acetic acid solution. Do the same for a 0.100M acetic acid solution.

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO- (aq)

Ka=[CH3COO

− ][H3O+]

CH3COOH[ ]

[ ]initial

Δ[ ]final

0.250M 0M 0M

x

+x+x

x-x

0.250-x

1.8x10−5= (x)(x)(0.250-x)

To avoid the quadratic equation, make the assumption that because the constant is so small that x will be small relative to 0.250.

1.8x10−5= (x)(x)(0.250-x)

1.8x10−5= (x)(x)(0.250-x)

≈ x2

(0.250)

This works for (# + x) or (# - x) when the # is > 100 Keq

x = 2.12x10-3



[ ]initial

Δ[ ]final

0.250M 0M 0M

x

+x+x

x-x

0.250-x

x = 2.12x10-3 pH = -log [H3O+]

pH = -log x = 2.67

Percent Ionization

% Ionization = Amount IonizedTotal amount present

*100

The total amount present is the concentration* regardless of whether it is ionized or not. This is the initial concentration of the weak acid or base in the ICE table.The amount ionized is the concentration* ionized from (products). This is usually defined as x in the equilibrium table. For weak acids it would be the hydronium ion concentration at equilibrium.

Concentrations may be used for amounts because everything is in the same solution volume (molarity instead of moles).



[ ]initial

Δ[ ]final

0.250M 0M 0M

x

+x+x

x-x

0.250-xx = 2.12x10-3

% Ionization = 2.12x10-3 M0.250 M

*100 = 0.848%

The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH and percent ionization of a 0.250M and 0.100M acetic acid solutions. Similar work produces answers for 0.100M

Acetic AcidMolarity

Equilibrium[H3O+] pH % Ionization

0.100M

0.250M

1.34x10-3M

2.12x10-3M

2.87

2.67

1.34%

0.848%

Increasing the overall weak acid concentration increases the acidity of the solution. The percent ionization is less but there is more acid overall.

Relative Acid and Base Strengths

Ka and Kb are like any equilibrium constants. The larger the constant, the more product favored the equilibrium.

The stronger the acid, the larger the Ka.

The stronger the base, the larger the Kb.

The Kb of ammonia is 1.8x10-5 and the Kb of ethyl amine is 5.6x10-4. Which is the stronger base?

Ethyl amine is the stronger base. It has a larger Kb.

Ka, Kb and Kw


The acid in water forms hydronium ion and conjugate base.


Consider a weak acid HAn and its conjugate base of An-1. Each will ionize in water.

The base in water forms hydroxide ion and conjugate acid.


Acid and Base StrengthThe sum of the chemical equations is the auto-ionization of water:

Upon addition of equations, the net equilibrium constant is the product of the original constants.



Ka

Kb

Kw2H2O ⇄ H3O+ + OH-

Kw = Ka • Kb

A strong acid has a weak conjugate base.

A weak acid has a strong conjugate base.

A weak base has a strong conjugate acid.

A strong base has a weak conjugate acid.

Conjugate Acid-Base Pairs

Acid and Base StrengthKw = Ka • Kb

Acid Ka KbConjugate Base

H2O OH-1.8x10-16 55.5

CH3CO2H CH3CO2-1.8x10-5 5.6x10-10

NH4+ NH35.6x10-10 1.8x10-5

HF F-7.4x10-4 1.4x10-11

HCl Cl-large small

acid

stre

ngthbase strength

pKa and pKb


The sum of the pKa and pKb values is 14:

-log(Ka • Kb) = -log(1x10-14)pKa + pKb = 14

Kw = Ka • Kb

Kw = 1x10-14 = Ka • Kb

The pKa and pKb functions are “-log” of each respectively:

pKa = - log Ka pKb = - log Kb

Kw = 1x10-14 = Ka • KbpK

a = -

log

Ka

pKb =

- lo

g K

b

Kb = 10-pKbKa = 10-pKa

Ka Kb

pKa pKbpKa + pKb = 14

Acid and Base StrengthpKa = - log Ka

Acid Ka

CH3CO2H 1.8x10-5

NH4+ 5.6x10-10

HF 7.4x10-4

HCl 1.3x106

acid

stre

ngth

pKa

4.749.25

3.13-6.1

Stronger acids have smaller pKa values. The same for bases.


Acid Conjugate Base

CH3CO2H CH3CO2-

NH4+ NH3

HF F-

HCl Cl-

acid

stre

ngth

base strength

pKa

4.749.25

3.13-6.1

pKb

9.264.75

10.8720.1

pKa = - log Ka pKb = - log Kb

pKa + pKb = 14

The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH of a 0.100M sodium acetate solution.

The pH of a Salt Solution

The sodium acetate dissolves and dissociates into sodium and acetate ions. The acetate ion is a weak base (the conjugate base of a weak acid).

NaCH3COO (aq) → Na+(aq) + CH3COO-

[CH3COO] = 0.100MThe Kb is found from the Ka:

Kw = 1x10-14 = Ka • Kb

Kb = 1x10-14 / (1.8x10-5) = 5.6x10-10

CH3COO-(aq) + H2O(l) ⇄ OH-(aq) + CH3COOH(aq)

[ ]initial

Δ[ ]final

0.100M 0M 0M

x

+x+x

x-x

0.100-x

Kb = 5.6x10-10 =[HCOOH][OH− ]HCOO−⎡⎣ ⎤⎦

= (x)(x)0.100 − x

5.6x10-10 ≈ x2

0.100

x = [OH-] = 7.45x10-6

pOH = -log [OH-]pOH = 5.13

pH = 14 - pOH = 8.87

Common Ion Effect

The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH of a solution that is both 0.250M in acetic acid and 0.350M sodium acetate.

Previously it was determined that the hydronium ion concentration in a 0.250M acetic acid solution is 2.12x10-3M and the pH is 2.67. In this problem there is some conjugate base present too.

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)

0.250M 0.350M

Consider Le Chȃtelier’s principle. If the acetic acid were in equilibrium and the acetate ion were added, it would shift to the left. The hydronium ion concentration should be less.

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)

0.250M 0.350M

Instead of just the acid form, the base form is present too. The pH should be higher.

[ ]initial

Δ[ ]final

0M+x+x-x

0.250 -x 0.300 + xx

Ka=[CH3COO

− ][H3O+]

CH3COOH[ ]1.8x10-5= (0.300+x)(x)

(0.250-x)≈ (0.350)x(0.250)

x = 1.29x10-5

pH = 4.89(higher pH with acetate present)

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