Acids and BasesSection 18.1: Calculations involving Acids and Bases

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

The Ion Product of Water

Kc =[H+][OH-]

[H2O][H2O] = constant

Kc[H2O] = Kw = [H+][OH-]

15.2

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

Kc =[H3O +][OH-]

[H2O]2 [H2O] = constant

Kc[H2O] 2 = Kw = [H+][OH-]

The Ion Product of Water

The ion-product constant (Kw) is

the product of the molar concentrations of H+ and OH- ions

at a particular temperature.

At 250CKw = [H+][OH-] = 1.0 x 10-14

when [H+] = [OH-]

when [H+] > [OH-]

when [H+] < [OH-]

Solution Is

neutral

acidic

basic

15.2

The Ion Product of Water

The ion-product constant (Kw) is the product of the molar concentrations of H+ and

OH- ions at a particular temperature. At 250CKw = [H+][OH-] = 1.0 x 10-14

15.2

Note: The autoionization of water is an endothermic process.

What will happen to the Kw if the temperature increases?

• equil shifts to the right – more H+ and OH- produced…• Kw increases… pH decreases• water is still neutral… but pH is slightly less than 7 at

higher temps • ex: @ 50°C [H+] = [OH-] = 3.05 x 10-7 ;pH is 6.5

heat + H2O (l) H+ (aq) + OH- (aq)

What is the concentration of OH- ions in a solution whose hydrogen ion concentration is 1.00 x 10-4 M?

15.2

pH – A Measure of Acidity

pH = -log [H+]

[H+] = 10-pH

pH [H+]

15.3

pOH = -log [OH-]

[OH-] = 10-pOH

pH

[OH-]

[H+] [OH-]

15.3

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

-log [H+] – log [OH-] = 14.00

pH + pOH = 14.00

What is the pH of a 2 x 10-3 M HNO3 solution?

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

15.4

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

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

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.615.4

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

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

Ka is the acid ionization constant

if Ka

strength of weak acid

15.5

if Ka is small, acid is weak

What is the pH of a 0.5 M HF solution (at 250C)?

Ka =

Eqn:

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.00

+x

15.5

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]

= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50= 7.1 x 10-4

0.50 – x 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M15.5

When can I use the approximation?

0.50 – x 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.0190.019 M0.50 M

x 100% = 3.8%Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka x2

0.05= 7.1 x 10-4 x = 0.006 M

0.006 M0.05 M

x 100% = 12%More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or method of successive approximation. IB WILL NOT REQUIRE THIS!!! 15.5

NOTES: Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• assume you can ignore the autoionization of water – this works in most cases (pH < 6) ,

• ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations.

15.5

Solving weak acid ionization problems cont:

4. Solve for x by the approximation method.

• approximation assumes that the acid is quite weak… so not much dissociates… (<5%) [If approximation is not valid, solve for x exactly… NOT REQD FOR IB.]

5. Use the value for x to calculate concentrations of all species and/or pH of the solution.

15.5

What is the pH of a 0.122 M monoprotic acid whose Ka is 2.5 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

Ka =

Ka

If Ka << 1; do approximation

15.5

Your Turn: Practice Problem # 1

A 0.0100 M solution of a weak acid has a pH of 5.00. What is the dissociation constant of the acid?

Answer: 1.00 x 10-8 M

Ka is often expressed as pKa

pKa = -log Ka

Ka = 10-pKa

15.3

acid strength

pKa

Your Turn: Practice Problem # 2

Benzoic acid has a pKa of 4.2. What is the pH of a 0.100 M solution of this acid?

Answer: 2.6

Your Turn: Practice Problem # 3a

a. What concentration of hydrofluoric acid is required to give a solution of pH 2.00?

Answer: 0.148 M

percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

15.5

Your Turn: Practice Problem # 3b

b. What percentage of the hydrofluoric acid is dissociated at this pH, if the dissociation constant of the acid is 6.76 x 10-4 M?

Answer: 6.76 %

B (aq) + H2O (l) BH+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb =[BH+][OH-]

[B]

Kb is the base ionization constant

Kb

weak basestrength

15.6

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

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

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH3]

Kb is the base ionization constant

Kb

weak basestrength

15.6

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

Kb is often expressed as pKb

pKb = -log Kb

Kb = 10-pKb

15.3

basestrength

pKb

Conjugate acid pKa pKb Conjugate base

Stronger acid

(but still weak)

Weaker base

H3PO4 2.1 11.9 H2PO4–

HF 3.3 10.7 F–

CH3COOH 4.8 9.2 CH3COO–

H2CO3 6.4 7.6 HCO3–

NH4+ 9.3 4.7 NH3

Weaker acid Stronger base

(but still weak)

Your Turn: Practice Problem # 4

What is the pH of a 0.0500 M solution of ethylamine (pKb = 3.40)?

Answer: 11.60

15.7

Ionization Constants of Conjugate Acid-Base Pairs

HA (aq) H+ (aq) + A- (aq)

A- (aq) + H2O (l) OH- (aq) + HA (aq)

Ka

Kb

H2O (l) H+ (aq) + OH- (aq) Kw

KaKb = Kw

Weak Acid and Its Conjugate Base

Ka = Kw

Kb

Kb = Kw

Ka

pKa+ pKb = 14.00

Homework

Read Section 18.1 pp. 217-220

Do Ex 18.1 p 220-221

#1-5, 6a,6c,6e, 7, 8, 9, 10