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Chapter 16Acid–Base Equilibria

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Solutions of a Weak Acid or Base1. Acid-Ionization Equilibria2. Polyprotic Acids3. Base-Ionization Equilibria4. Acid–Base Properties of Salt Solutions

Solutions of a Weak Acid or Base with Another Solute5. Common-Ion Effect6. Buffers7. Acid–Base Titration Curves

Contents and Concepts

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Learning Objectives

Solutions of a Weak Acid or Base Acid-Ionization Equilibria

a. Write the chemical equation for a weak acid undergoing acid ionization in aqueous solution.

b. Define acid-ionization constant and degree of ionization.

c. Determine Ka from the solution pH.d. Calculate concentrations of species in a

weak acid solution using Ka (approximation method).

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1. Acid-Ionization Equilibria (cont)

a. State the assumption that allows for using approximations when solving problems.

b. Calculate concentrations of species in a weak acid solution using Ka (quadratic formula).

2. Polyprotic Acids

a. State the general trend in the ionization constants of a polyprotic acid.

b. Calculate concentrations of species in a solution of a diprotic acid.

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3. Base-Ionization Equilibria

a. Write the chemical equation for a weak base undergoing ionization in aqueous solution.

b. Define base-ionization constant.

c. Calculate concentrations of species in a weak base solution using Kb.

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4. Acid–Base Properties of Salt Solutions

a. Write the hydrolysis reaction of an ion to form an acidic solution.

b. Write the hydrolysis reaction of an ion to form a basic solution.

c. Predict whether a salt solution is acidic, basic, or neutral.

d. Obtain Ka from Kb or Kb from Ka.

e. Calculating concentrations of species in a salt solution.

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Solutions of a Weak Acid or Base with Another Solute

5. Common-Ion Effect

a. Explain the common-ion effect.

b. Calculate the common-ion effect on acid ionization (effect of a strong acid).

c. Calculate the common-ion effect on acid ionization (effect of a conjugate base).

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6. Buffers

a. Define buffer and buffer capacity.

b. Describe the pH change of a buffer solution with the addition of acid or base.

c. Calculate the pH of a buffer from given volumes of solution.

d. Calculate the pH of a buffer when a strong acid or a strong base is added.

e. Learn the Henderson–Hasselbalch equation.

f. State when the Henderson–Hasselbalch equation can be applied

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7. Acid–Base Titration Curves

a. Define equivalence point.

b. Describe the curve for the titration of a strong acid by a strong base.

c. Calculate the pH of a solution of a strong acid and a strong base.

d. Describe the curve for the titration of a weak acid by a strong base.

e. Calculate the pH at the equivalence point in the titration of a weak acid by a strong base.

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7. Acid–Base Titration Curves (cont)

f. Describe the curve for the titration of a weak base by a strong acid.

g. Calculate the pH of a solution at several points of a titration of a weak base by a strong acid.

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The simplest acid–base equilibria are those in which a weak acid or a weak base reacts with water.

We can write an acid equilibrium reaction for the generic acid, HA.

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

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Acetic acid is a weak acid. It reacts with water as follows:

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

Here A from the previous slide = C2H3O2-(aq)

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The equilibrium constant for the reaction of a weak acid with water is called the acid-ionization constant (or acid-dissociation constant), Ka.

Liquid water is not included in the equilibrium constant expression.

HA

A OH3a

K

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16 | 15

16 | 16

HCN

Highest pH

Lowest pH

HF

HNO2

HC2H3O2

HCN

HF

HNO2

HC2H3O2

1.7 × 10-5

4.5 × 10-4

Ka

6.8 × 10-4

4.9 × 10-10

In order of decreasing Ka:

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Calculations with Ka

Given the value of Ka and the initial concentration of HA, you can calculate the equilibrium concentration of all species.

Given the value of Ka and the initial concentration of HA, you can calculate the degree of ionization and the percent ionization.

Given the pH of the final solution and the initial concentration of HA, you can find the value of Ka and the percent ionization.

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We can be given pH, percent or degree ionization, initial concentration, and Ka.

From pH, we can find [H3O+].

From percent or degree ionization, we can find Ka.

Using what is given, we can find the other quantities.

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Sore-throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C.

a. What is the acid-ionization constant, Ka, for phenol at 25°C?

b. What is the degree of ionization?

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We were told that pH = 5.43. That allows us to find [H3O+] = 3.7 × 10-6 M = x = [C6H5O-].

Now we find [HC6H5O] = 0.10 – x = 0.10 M.

HC6H5O(aq) + H2O(l) H3O+(aq) + C6H5O-(aq)

Initial 0.10 0 0

Change –x +x +x

Equilibrium 0.10 – x x x

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Finally, we write the expression for Ka and substitute the concentrations we now know.

HC6H5O + H2O H3O+ + C6H5O-

[H3O+] = [C6H5O-] = 3.7 × 10-6 M

[HC6H5O] = 0.10 M

OHHC

OHC OH

56

563a

K

0.10

)10 (3.7 26

a

K

-10a 10 1.4 K

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5

6

10 3.7ionization of Degree0.10

10 3.7

0.100 ionization of Degree

x

The degree of ionization is the ratio of ionized concentration to original concentration:

Percent ionization is the degree of ionization × 100%:

Percent ionization = 3.7 x 10-3% or 0.0037%

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Simplifying Assumption for Acid and Base IonizationsThe equilibrium concentration of the acid is most often ([HA]0 – x).

If x is much, much less than [HA]0, we can assume that subtracting x makes no difference to [HA]:

([HA]0 – x) = [HA]0

This is a valid assumption when the ratio of [HA]0 to Ka is > 103. If it is not valid, you must use the quadratic equation to solve the problem.

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Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of this acid, of hydronium ion, and of para-hydroxybenzoate ion in a 0.200 M aqueous solution at 25°C? What is the pH of the solution and the degree of ionization of the acid? The Ka of this acid is 2.6 × 10-5.

We will use the generic formula HA for para-hydroxybenzoic acid and the following equilibrium:

HA + H2O H3O+ + A-

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

Initial 0.200 0 0

Change –x +x +x

Equilibrium 0.200 – x x x

HA

A OH3

a

K

x

x

-0.20010 2.6

25

0.200 - 0.200 so 1000, 10 2.6

0.2005

x

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)10 (2.3 log- ]O[H log pH -3

3-

2610 5.2 x

0.200

2

a

xK

]O[H 10 2.3 3

3 x

2.64 pH

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Polyprotic Acids

A polyprotic acid has more than one acidic proton—for example, H2SO4, H2SO3, H2CO3, H3PO4.

These acids have successive ionization reactions with Ka1, Ka2, . . .

The next example illustrates how to do calculations for a polyprotic acid.

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Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. What is the pH of a 0.10 M solution? What is the concentration of the C4H4O6

2 ion in the same solution?

Ka1 = 9.2 104; Ka2 = 4.3 105.

First, we will use the first acid-ionization equilibrium to find [H+] and [HC4H4O6

-]. In these calculations, we will use the generic formula H2A for the acid.

Next, we will use the second acid-ionization equilibrium to find [C4H4O6

2-].

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H2A(aq) + H2O(l) H3O+(aq) + HA-(aq)

Initial 0.10 0 0

Change –x +x +x

Equilibrium 0.10 – x x x

AH

HA OH

2

31a

Kx

x

-0.1010 9.2

24

.assumption gsimplifyin the use cannot We

1000 10 9.2

0.104

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-5-4 10 9.2- 10 9.21 cba

a

acbbx

2

42

10 9.210 9.2 245 xx

24 )(0.1010 9.2 xx

0 10 9.2 - 10 9.2 -5-42 xx

2(1)

)10 4(1)(-9.2)10 (9.2)10 (9.2 -5244

x

3424

10 9.6 10 4.62

)10 (1.92)10 (9.2

x

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2-3 10 1.0and10 9.1 xx

ion.concentrat negative a have to impossible physically

is it because value negative the eliminate We

M 10 9.1]O[H 33

M 10 9.1][HA 3

m.equilibriu ionization

-acid second the for these use Now we

M 10 9.0A][H 42

are ionsconcentrat

the m,equilibriu ionization acid first the of end the At

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HA-(aq) + H2O(l) H3O+(aq) + A2-(aq)

Initial 0.0091 0.0091 0

Change –x +x +x

Equilibrium 0.0091 – x 0.0091 + x x

-

3

a HA

A OH

2

2K

)- (0.0091

) (0.009110 4.3 5

x

xx

0.0091 - 0.0091 and 0.0091 0.0091

that assume can We

xx

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0.0091

0.009110 4.3 5 x

valid. asw assumption urO

Mx 10 4.3 5

M 10 4.3 ]OH[C 5-2644

M 10 9.1 ]OH[HC 3-644

M 10 9.1 ]O[H 3

3

M 10 9.0 ]OHC[H 46442

)10 (9.1 logpH 3 2.04pH

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Base-Ionization Equilibrium

The simplest acid–base equilibria are those in which a weak acid or a weak base reacts with water.

We can write a base equilibrium reaction for the generic base, B.

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

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Ammonia is a weak base. It reacts with water as follows:

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

Here the generic B from the previous slide is NH3(aq).

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The equilibrium constant for the reaction of a weak base with water is called the base-ionization constant, Kb.

Liquid water is not included in the equilibrium constant expression.

B

HO HBb

K

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Writing Kb Reactions

The bases in Table 16.2 are nitrogen bases; that is, the proton they accept adds to a nitrogen atom. Next we’ll practice writing the Kb reactions.

Ammonium becomes ammonium ion:

NH3+ H2O NH4+ + OH-

Ethylamine becomes ethyl ammonium ion:

C2H5NH2 + H2O C2H5NH3+ + OH-

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Dimethylamine becomes dimethylammonium ion:

(CH3)2NH2 + H2O (CH3)2NH3+ + OH-

Pyridine becomes pyridinium ion:

C5H5N+ H2O C5H5NH+ + OH-

Hydrazine becomes hydrazinium ion:

N2H4+ H2O N2H5+ + OH-

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We can be given pH, initial concentration, and Kb.

From the pH, we can find first [H3O+] and then [OH-]. Using what is given, we can find the other quantities.

We can also use a simplifying assumption: When [B]0 / Kb > 103, the expression ([B]0 – x) = [B]0.

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Aniline, C6H5NH, is used in the manufacture of some perfumes. What is the pH of a 0.035 M solution of aniline at 25°C?

Kb= 4.2 × 10-10 at 25°C.

We will construct an ICE chart and solve for x.

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We are told that Kb = 4.2 × 10-10. That allows us to substitute into the Kb expression to solve for x.

C6H5H(aq) + H2O(l) C6H5NH+(aq) + OH-(aq)

Initial 0.035 0 0

Change –x +x +x

Equilibrium 0.035 – x x x

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C6H5N + H2O C6H5NH+ + OH-

[B] = 0.200 – x and [B+] = [OH-] = x

NHCNHHC

56

56

HO

bK

)-(0.20010 4.2

21

x

x 0

xx - 0.200 that assume can weso 1000, 10 4.2

0.20010-

-112 10 8.4 x0.200

10 4.22

10- x

Mx 10 9.2 -6

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The question asks for the pH:

)6-10 (9.2 log - 14.00 pH

][OH log14.00pOH14.00 pH

04.500.41 pH

96.8pH

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Salt Solutions

We will look at the cation and the anion separately, and then combine the result to determine whether the solution is acidic, basic, or neutral.

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The conjugate acid of a strong base is very weak and does not react with water. It is therefore considered to be neutral.

Na+ + H2O No Reaction (NR)

The conjugate base of a strong acid is very weak and does not react with water. It is therefore considered to be neutral.

Cl- + H2O NR

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The conjugate acid of a weak base reacts with water to form an acidic solution:

NH4+ + H2O <==> NH3 + H3O+

The conjugate base of a weak acid reacts with water to form a basic solution:

F- + H2O <==> HF + OH-

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We classify each salt by examining its cation and its anion, and then combining the result.

NaBrNa+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is neutral.

Br- is the conjugate base of HBr, a strong acid. It does not react with water, so is neutral.

The cation is neutral; the anion is neutral.

NaBr is neutral.

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NaC2H3O2

Na+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is neutral.

C2H3O2- is the conjugate base of HC2H3O2

-, a weak acid. It reacts with water to give a basic solution.

The cation is neutral; the anion is basic.

NaC2H3O2 is basic.

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NH4Cl

NH4+ is the conjugate acid of NH3, a weak base. It

reacts with water to give an acidic solution.

Cl- is the conjugate base of HCl, a strong acid. It does not react with water, so it is neutral.

The cation is acidic; the anion is neutral.

NH4Cl is acidic.

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NH4F

NH4+ is the conjugate acid of NH3, a weak base. It

reacts with water to give an acidic solution.

F- is the conjugate base of HF, a weak acid. It does not react with water, so it is basic.

The cation is acidic; the anion is basic. We need more information in this case. We compare Ka for HF to Kb for NH3.

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If Ka > Kb, the solution is acidic. If Ka < Kb, the solution is basic. If Ka = Kb, the solution is neutral.

105

14

b

wa4 10 5.6

10 1.8

10 1.0 :NH For

K

KK

acidic. is FNH4

114

14

a

wb 10 1.5

10 6.8

10 1.0 :F For

K

KK

F for KNH for K b4a

?

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Ammonium nitrate, NH4NO3, is administered as an intravenous solution to patients whose blood pH has deviated from the normal value of 7.40.

Would this substance be used for acidosis (blood pH < 7.40) or alkalosis (blood pH > 7.40)?

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NH4+ is the conjugate acid of a NH3, a weak base.

NH4+ is acidic.

NO3- is the conjugate base of HNO3, a strong acid.

NO3- is neutral.

NH4NO3 is acidic, so it could be used for alkalosis.

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The hydrolysis equilibrium constant can be used in problems to determine the pH of a salt solution. To use the hydrolysis equilibrium, we need to compute the K value for it.

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We can determine K for the reaction of the ion with water (hydrolysis reaction).

NH4+ + H2O <==> NH3 + H3O+

is the sum of

NH4+ + OH- <==> NH3 + H2O K1 = 1/Kb

2H2O <==> H3O+ + OH- K2 = Kw

Overall reaction

NH4+ + H2O <==> NH3 + H3O+

b

w21a K

KK KK

?

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What is Kb for the F- ion, the ion added to the public water supply to protect teeth?

For HF, Ka = 6.8 × 10-4.

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F- + H2O <==> HF + OH-

Is the sum of

H3O+ + F- <==> HF + H2O; K1 = 1/Ka

2H2O <==> H3O+ + OH-; K2 = Kw

Overall reaction:

F- + H2O <==> HF + OH-

10-

14

a

w21b 10 6.8

10 1.0

K

KK KK

11b

- 10 1.5 :F For K

?

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Household bleach is a 5% solution of sodium hypochlorite, NaClO. This corresponds to a molar concentration of about 0.70 M NaClO.

What is the [OH-] and the pH of the solution?

For HClO, Ka = 3.5 × 10-8.

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We are told that Ka = 3.5 × 10-8. That means that Kb= 2.9 × 10-7.

This allows us to substitute into the Kb expression to solve for x.

ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)

Initial 0.70 0 0

Change –x +x +x

Equilibrium 0.70 – x x x

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ClO- + H2O HClO + OH-

[H3O+] = [A-] = x and [HA] = 0.200 – x

NHCNHHC

56

56

HO

bK

)-(0.7010 2.9

2

x

x 7

0.70. - 070 that assume can weso 1000, 10 2.9

0.707-

x

-72 10 2.0 x0.70

10 2.92

7- x

Mx 10 4.5 -4

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The question asks for [OH-] and the pH:

)4-10 (4.5 log - 14.00 pH

][OH log14.00pOH14.00 pH

35.300.41 pH

65.10pH

Mx 10 4.5 ][OH -4

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Common-Ion Effect

The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that takes part in the equilibrium.

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Buffers

A buffer solution is characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

A buffer is made by combining a weak acid with its conjugate base or a weak base with its conjugate acid.

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This problem involves dilution first. Once we know the concentrations of benzoic acid and benzoate ion, we can use the acid equilibrium to solve for x.

We will use HBz to represent benzoic acid and Bz- to represent benzoate ion.

What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC7H5O2, with 3.00 L of 0.060 M sodium benzoate, NaC7H5O2? Ka for benzoic acid is 6.3 × 10-5.

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MM

0.0050L 3.0) (1.0

L) )(1.0 (0.020

][HBz

MM

0.045L 3.0) (1.0

L) )(3.0 (0.060

][Bz

HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)

Initial 0.0050 0 0.045

Change –x +x +x

Equilibrium 0.0050 – x x 0.045 + x

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HBz

B OH3a

z

K

)-(0.0050

) (0.04510 6.3 5

x

xx

0.0050 - 0.0050 and 0.045 0.045

that assume can We

xx

Mx 10 7.0 6

0.0050

0.04510 6.3 5 x

valid. wasassumption urO

16 | 68

Mx 10 7.0 ]O[H 63

5.15 pH

)10 (7.0 log ]O[H log pH 63

?

16 | 69

Calculate the pH of a 0.10 M HF solution to which sufficient sodium fluoride has been added to make the concentration 0.20 M NaF. Ka for HF is 6.8 × 10-4.

We will use the acid equilibrium for HF.

NaF provides the conjugate base, F-, so [F-] = 0.20 M.

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M 0.10]HF[ M 0.20]F[

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

Initial 0.10 0 0.20

Change –x +x +x

Equilibrium 0.10 – x x 0.20 + x

[HF]

][F ]O[H -3

a

K)-(0.10

) (0.2010 6.8 4

x

xx

0.10 - 0.10 and 0.20 0.20

that assume can We

xx

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0.10

0.20x10 6.8 4

Mx 10 3.4 4

valid. wasassumption urO

Mx 10 3.4 ]O[H 43

3.47 pH

)10 (3.4 log ]O[H log pH 43

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Henderson–Hasselbalch Equation

Buffers at a specific pH can be prepared using the Henderson-Hasselbalch Equation.

Buffers are prepared from a conjugate acid-base pair in which the ionization is approximately equal to the desired H3O+ concentration.

Consider a buffer made up of a weak acid HA and its conjugate base A-

The acid-ionization constant is

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The preceding equation can be used to derive an equation for the pH of the buffer.

Take the negative logarithm of both sides of the equation.

The pKa of a weak acid is defined in a manner similar to pH and pOH

pKa = - log Ka

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The equation can then be expressed as:

This equation is generally shown as:

This equation relates buffer pH for different concentrations of conjugate acid and base and is known as the Henderson-Hasselbalch equation.

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16 | 75

Question: What is the [H3O+] for a buffer solution that is 0.250 M in acid and 0.600 M in the corresponding salt if the weak acid Ka = 5.80 x 10-7?

Use as example the following equation of a conjugate acid-base pair

pKa = - log(5.80 x 10-7) = -(-6.24) = 6.24

[H3O+] = pKa + log [A-]/[HA] = 6.24 + log 0.600/0.250

[H3O+] = 10-6.62 = 2.40 x 10-7

[H3O+] = 2.40 x 10-7

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Acid–Base Titration

An acid–base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it.

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An acid–base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid).

Such curves are used to gain insight into the titration process. You can use the titration curve to choose an indicator that will show when the titration is complete.

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This titration plot shows the titration of a strong acid with a strong base. Note that the pH at the equivalence point is 7.0.

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The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added.

The indicator must change color near the pH at the equivalence point.

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When solving problems in which an acid reacts with a base, we first need to determine the limiting reactant. This requires that we use moles in the calculation.

The next step is to review what remains at the end of the reaction. In the case of a strong acid and a strong base, the solution is neutral.

When a weak acid or weak base is involved, the product is a salt. After finding the salt concentration, we use a hydrolysis equilibrium to find the pH.

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Calculate the pOH and the pH of a solution in which 10.0 mL of 0.100 M HCl is added to 25.0 mL of 0.100 M NaOH.

First we calculate the moles of HCl and NaOH.

Then we determine what remains after the reaction.

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The overall reaction is

HCl + NaOH H2O + NaCl

Moles HCl = (0.100 M)(10.0 × 10-3 L) = 1.00 × 10-3 mol

Moles NaOH = (0.100 M)(25.0 × 10-3 L) = 2.50 × 10-3 mol

All of the HCl reacts, leaving 1.50 × 10-3 mol NaOH.

The new volume is 10.0 mL + 25.0 mL = 35.0 mL.

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M 10 4.29L 10 35.0

mol 10 1.50OH 2

3-

-3

][

]) log[OH(14.00pOH14.00pH

368.100.14pH

12.632 pH

?

16 | 84

Calculate the [OH] and the pH at the equivalence point for the titration of 500. mL of 0.10 M propionic acid with 0.050 M calcium hydroxide. (This can be used to prepare a preservative for bread.)

Ka = 1.3 105.

First we need to find the moles of propionic acid to find the volume of calcium hydroxide. Then we will examine the salt that is produced and use the hydrolysis equilibrium to find the pH. We will use HPr to represent propionic acid and Pr- to represent propionate ion.

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Moles propionic acid = (0.10 M)(500. × 10-3 L)

= 0.050 mol HPr

To find the moles of calcium hydroxide, we need the balanced chemical equation:

2HPr + Ca(OH)2 Ca(Pr)2 + 2H2O

Moles calcium hydroxide =

22 Ca(OH) mol 0.025

HPr mol 2

Ca(OH) mol 1 HPr mol 0.050

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Now we find the volume of Ca(OH)2 required to give 0.025 mol:

L 0.50

Lmol

0.050

mol 0.025Ca(OH) Volume 2

MM 0.050 10 5.0L 1.0

mol 0.050][Pr 2-

The new volume is 500. × 10-3 L + 0.50 L = 1.00 L

The concentration of Pr- is 2(0.025 M) = 0.050 M

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The hydrolysis reaction is

Pr- + H2O HPr + OH-

10-5

14

a

wb 10 7.7

10 1.3

10 1.00

K

KK

Pr-(aq) + H2O(l) HPr(aq) + OH-(aq)

Initial 0.050 0 0

Change –x +x +x

Equilibrium 0.050 – x x x

16 | 88

][Pr

][OH [HPr]-

-

a K)-(0.050

10 7.72

10

x

x

0.080. 0.050 that assume can We

1000 10 7.7

0.05010

x

0.05010 7.7

210 x

112 10 3.85 x

Mx 10 6.20 -6

8.79pH

Mx 10 6.20 ][OH -6-

5.21pOH