Acid-BaseAcid-BaseEquilibriaEquilibria

Chapter 16Chapter 16

Conjugate Acid-Base Pairs

Eqn 16.7 p 670

Conjugate Acid-Base Pairs

Eqn 16.8 p 671

Fig 16.4 Relative strengths of some conjugate acid-base pairs

The strongerthe acid…

the weaker its conjugate base.

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

The Ion Product of Water

Kc =[H+][OH-]

[H2O][H2O] = 55.6 M = constant

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

Ion-product constant (Kw) - the product of the molar concentrations of H+ and OH- ions at a particular temperature.

At 25°C:Kw = [H+][OH-] = 1.0 x 10-14

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution is:

neutral

acidic

basic

pH – A Measure of Acidity

pH = −log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution is:

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7

pH < 7

pH > 7

At 25°C

pH [H+]

pH = −log [H3O+]

pOH = -log [OH-]

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

−log [H+] – log [OH-] = 14.00

pH + pOH = 14.00

Fig 16.5 H+ concentrations and pH of common substances

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

AcidsandBases

Other “p” Functions

The “p” in pH tells us to take the negative base - 10 logarithm of the quantity (in this case, hydronium ions)

Some similar examples:

pOH = -log [OH−]

pKw = -log Kw

pCl = -log [Cl−]

AcidsandBases

How Do We Measure pH?

• For less accurate measurements:– Litmus paper

• “Red” paper turns blue above ~pH = 8• “Blue” paper turns red below ~pH = 5

– Indicator:

AcidsandBases

For more accurate measurements: ◦ pH meter, which measures the voltage in the solution

Fig 16.6 Digital pH meter

Table 4.2

Strong Electrolyte – 100% dissociation

Strong Acids and Strong Bases are strong electrolytes (p 130)

HCl (aq) + H2O (l) H3O+ (aq) + Cl− (aq)

Strong Acids and Bases

NaOH (aq) Na+ (aq) + OH− (aq)

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)

Weak Acids

Weak electrolytes - only partially ionized in aqueous solution

• Most acidic substances are weak acids

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 ≡ acid ionization constant

Kaweak acidstrength

16.5

Table 16.2 Some Weak Acids in Water at 25 °C

Calculating Ka from the pHSample Exercise 16.10 p 682

The pH of a 0.10 M solution of formic acid, HCOOH, at 25Cis 2.38. Calculate Ka for formic acid at this temperature.

HCOOH (aq) H+ (aq) + HCOO- (aq)

pH = -log [H+]

2.38 = -log [H+]

-2.38 = log [H+]

10-2.38 = 10log [H+] = [H+]

4.2 10-3 = [H+] = [HCOO-]

Ka =[H+][HCOO-][HCOOH]

[HCOOH], M [H3O+], M [HCOO-], M

Initially 0.10 0 0

Change - 4.2 10-3 + 4.2 10-3 + 4.2 10-3

At Equilibrium 0.10 - 4.2 10-3

= 0.0958 = 0.10

4.2 10-3 4.2 10-3

4.2 10-3 = [H+] = [HCOO-]

Calculating Ka from the pH

HCOOH (aq) H+ (aq) + HCOO- (aq)

Ka =[H+][HCOO-][HCOOH]

=[4.2 10-3] [4.2 10-

3][0.10]

= 1.8 10-4

Percent ionization =

For a monoprotic acid HA:

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

Concentration ionized

Original concentrationx 100%

Fig 16.9The more dilute the acid,the greater the percent

ionization:

What is the pH of a 0.50 M HF solution (at 25°C)?

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

= 6.8 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= 6.8 x 10-4

Ka x2

0.50= 6.8 x 10-4

0.50 – x 0.50If [HF] > 100 Ka

x2 = 3.40 x 10-4 x = 0.018 M

[H+] = [F-] = 0.018 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M

When can I use the approximation?

Then 0.50 – x 0.50

x = 0.0180.018 M0.50 M

x 100% = 3.6%Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 25°C)?

Ka x2

0.05= 6.8 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...

Let’s determine error introduced:

If [HF] > 100 Ka

Solving weak acid ionization problems:

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

• In most cases, the autoionization of water can be ignored.

• Ignore [OH¯] because it is determined by [H+].

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

3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.

4. Calculate concentrations of all species and/or pH of the solution.

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-3?

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

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-3

NO!!

Approximation not ok.

Is [HF] > 100 Ka ?

Ka =x2

0.122 - x= 5.7 x 10-3 x2 + 5.7 X 10-3 x – 6.95 X 10-4 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

x = 0.0237 x = - 0.0294

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

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0237 M pH = -log[H+] = 1.625

Polyprotic AcidsHave more than one ionizable proton

If difference between the Ka1 and subsequent Ka values > 103, the pH generally depends only on the first dissociation.