ap chemistry: acids, bases, and salts...2017/02/24 · 1 ap chemistry: acids, bases, and salts unit...

1

AP Chemistry: Acids, Bases, and Salts

Unit Objectives 1. Understand the acid-base theories of Arrhenius, Brønsted-Lowry, and Lewis.

2. Identify strong acids and bases and calculate their pH’s.

3. Calculate the pH of a weak acid or base.

4. Calculate the concentration of a strong or weak acid or base from its pH.

5. Calculate the pH and ion concentrations in a polyprotic acid.

6. Predict the pH of a salt from its formula and then calculate the pH of the salt.

7. Identify the components of a buffer and perform calculations involving the preparation of a

buffer and the addition of a strong acid or strong base to a buffer.

8. Perform calculations involving strong acid-strong base titrations as well as weak acid-strong base

and weak base-strong acid calculations.

9. Be familiar with titration curves and selection of an acid-base indicator.

Key Terms acid-base indicatior

acid dissociation constant

Acidic oxides

ampohoteric substance

Arrhenius Concept

autoionization

basic oxides

Brønsted-Lowry model

buffered solution

buffering capacity

conjugate base

conjugate acid

conjugate acid-base pair

diprotic acid

equivalence point

Henderson-Hasselbalch equation

hydronium ion

ion-product (dissociation) constant

Lewis acid

Lewis base

major species

monoprotic acids

oxyacids

organic acids

percent dissociation

pH scale

polyprotic acid

salt

strong acids

strong bases

titration curve

triprotic acid

weak acids

weak bases

Equations and Constants

K Ka3

+ -

a

+ -H O ][A

HA] which is often simplified as

H ][A

HA]

[ ]

[

[ ]

[

Kb [ ]

[

HB ][OH

B]

+ -

K

K K

w

a b

[OH ][H ] @ 25 C

=

10 10 14.

pH = - log [H pOH = - log [OH

14 = pH + pOH

+ -], ]

pH = p + log [A

HA]

-

Ka]

[

pOH = p + log [HB

B]

+

Kb]

[ p = - log p = - log K K K Ka a b b,

2

AP Chemistry: Acid-Base Chemistry

Text References

Pages 658-660

Note: H+ does

not really exist

in aqueous

solutions. It is

hydrated in

water to form

H3O+

(hydronium

ion). However

H+ is often used

to make things

easier.

I. Properties of Aqueous Solutions of Acids and Bases

A. Acids

1. Sour taste.

Ex. Vinegar

2. Change colors of indicators.

Ex. Acids turn blue litmus red.

3. Non-oxidizing acids react with active metals to liberate

hydrogen gas.

Ex. Zn + 2HCl → ZnCl2 + H2

4. React with metal oxides, metal hydroxides, yielding a salt and

water.

Ex. CaO + HCl → CaCl2 + H2O

5. React with carbonates, yielding CO2.

Ex. 2HCl + Na2CO3 → 2NaCl + H2O +CO2

6. Electrolytes, because they ionize in water.

B. Bases

1. Bitter taste

Ex. Alka selter and other antacids

2. Slippery (slimy) to the touch.

Ex. Soap

3. Change colors of indicators.

Ex. Bases turn red litmus blue.

4. React with acids to produce a salt and water.

Ex. HCl + NaOH → NaCl + H2O

5. Electrolytes, because they ionize in water.

II. Acid-Base Theories

A. Arrhenius – this is the most simplistic definition of acids and bases

1. Acid – produces H+ in solution. Ex. HCl, HBr, H3PO4

a. monoprotic acid – contains one ionizable hydrogen.

Ex. HCl

b. diprotic acid – contains two ionizable hydrogens.

Ex. H2SO4, H2CO3

c. triprotric acid – contains three ionizable hydrogens.

Ex. H3PO4

Note: Only one hydrogen ionizes at a time

2. Base – produces OH- in solution. Ex. NaOH

Ammonia is not defined as a base according to the Arrhenius

definition of bases.

B. Brønsted-Lowry

1. Acid – donates a proton in water.

2. Base – accepts a proton in water.

Ammonia is defined as a base under this theory.

3. Conjugate Acid (CA) – acid formed when a base accepts a

proton.

4. Conjugate base (CB) – remaining part of an acid after it has

released a proton.

5. Identifying Conjugate Acid-Base Pairs

NH3 + H2O ⇄ NH4+ + OH

-

base acid CA CB

3

a. Label the acid, base, CA, and CB

HNO3 + H2O ⇄ H3O+ + NO3

-

acid base CA CB

C2H3O2-

+

H2O ⇄ HC2H3O2 + OH-

base acid CA CB

b. amphoteric – substances that can act as both an acid

and a base.

Write two equations to show that HS- is amphoteric.

HS- + H2O ⇄ H2S

+ OH

-

HS-

+ H2O ⇄ H3O+

+ S2-

6. Writing Disssociation Reactions and Equilibrium Expressions

a. The general reaction that occurs when an acid is

dissolved in water can best be represented as

HA(aq) + H2O(l) ⇄ H3O+(aq) + A

-(aq)

Or HA(aq) ⇄ H+(aq) + A

-(aq)

The equilibrium expression for the reaction is

Ka

3+ - + -H O A

HA]

H A

HA]

[ ][ ]

[

[ ][ ]

[ b. The general reaction that occurs when a base is

dissolved in water can be represented as

B(aq) + H2O(l) ⇄ HB+(aq) + OH

-(aq)

The equilibrium expression for the reaction is

Kb

[ ]

[

HB ][OH

B]

+ -

c. Write the dissociation reaction and the corresponding

Ka equilibrium expression for each of the following

acids in water.

a. HCN

HCN(aq) + H2O(l) ⇄ H3O+(aq) + CN

-(aq)

Or HCN(aq) ⇄ H+(aq) + CN-(aq)

Ka

3+ - + -H O CN

HCN]

H CN

HCN]

[ ][ ]

[

[ ][ ]

[

b. HOC6H5

HOC6H5(aq) + H2O(l) ⇄ H3O+(aq) + OC6H5

-(aq)

Or HOC6H5(aq) ⇄ H+(aq) + OC6H5-(aq)

Ka3

+6 5

-

6 5

+6 5

-

6 5

H O OC H

HOC H ]

H OC H

HOC H ]

[ ][ ]

[

[ ][ ]

[

c. C6H5NH3+

C6H5NH3+

(aq) + H2O(l) ⇄ H3O+(aq) + C6H5NH2(aq)

Or C6H5NH3+

(aq) ⇄ H+(aq) + C6H5NH2(aq)

Ka3

+6 5 2

6 5 3+

+6 5 2

6 5 3+

H O C H NH

C H NH ]

H C H NH

C H NH ]

[ ][ ]

[

[ ][ ]

[

4

Pages 704-707

Pages 661-665

701-704

Strong Acids

HCl, HBr, HI,

HNO3, H2SO4,

HClO4

Strong Soluble

Bases

Hydroxides or

oxides of IA and

IIA metals

C. Lewis Model

1. Acid – Electron-pair acceptor

A Lewis acid has an empty atomic orbital that it can use to

accept (share) an electron pair from a molecule that has a lone

pair of electrons (Lewis Base). The bond formed is called a

coordinate covalent bond.

2. Base – Electron-pair donor

3. Examples

a. Reaction between boron trifluoride and ammonia.

b. Hydration of a metal ion, such as Al3+

.

4. Tell whether each of the following is a Lewis acid or base.

a. PH3 Base

b. BCl3 Acid

c. H2S Base

5. Zinc hydroxide is an amphoteric substance. Write equations

that describe Zn(OH)2 acting as a Brønsted-Lowry base

toward H+ and as a Lewis acid toward OH

-.

Brønsted-Lowry Base:

Zn(OH)2(s) + 2H+(aq) → Zn

2+(aq) + 2H2O(l)

Lewis Acid:

Zn(OH)2(s) + 2OH-(aq) Zn(OH)4

2-(aq)

III. Strengths of Acids and Bases (Do not confuse concentration with strength)

The strength of an acid or base is determined by the extent to which it ionizes

or dissociates in aqueous solutions.

A. Strong Acids and Bases

1. Strong acids and bases dissociate (ionize) completely in water.

2. They have very large equilibrium constants (K values). The

dissociation (ionization) equilibrium position lies all the way

to the right.

3. The stronger the acid the weaker its conjugate base. The

stronger the base the weaker its conjugate acid.

4. The strength of binary acids (HA) depends upon two factors.

a. The larger the molecule A is the stronger the acid will

be.

b. The more polar the bond between H and A is, the

weaker the acid will be.

Ex. HF << HCl < HBr < HI

5

Pages 664-668

5. The strength of oxyacids (HOX) depends upon two factors.

a. The more electronegative element X is, the stronger

the acid HOX will be. Ex. HClO3 > HBrO3.

b. The more oxygen present in the polyatomic ion, the

stronger the acid is within that group.

Ex. HClO4 > HClO3 > HClO2 > HClO.

6. For polyprotic acids (acids where more than one proton can be

removed), each successive proton becomes more difficult to

remove. Therefore, for polyprotic acids, the anions formed by

the dissociation of the acidic hydrogen are always less acidic

than their parent molecule.

B. Weak Acids and bases

1. Weak acids and bases only partially ionize in water.

2. They have small equilibrium constants (K values). The

dissociation (ionization) equilibrium position lies far to the

left.

3. Weak acids have relatively strong conjugate bases. Weak

bases have relatively strong conjugate acids.

4. The percent dissociation of a weak acid or base can be

calculated using the following formula.

Percent Dissociation =

amount dissociated (mol / L)

initial concentration (mol / L)100

a. The smaller the percent dissociation, the weaker the

acid or base.

b. Dilution of a weak acid increases its percent

dissociation.

IV. The pH Concept

A. Self-Ionization of Water

1. Since the water molecule is amphoteric, it may dissociate with

itself to a slight extent. In the self-ionization of water, two

water molecules collide producing a hydronium ion and a

hydroxide ion.

H2O(l) + H2O(l) ⇄ H3O+(aq) + OH

-(aq)

H2O(l) ⇄ H+(aq) + OH

-(aq)

2. The equilibrium expression used here is referred to as

ionization constant for water, Kw

Kw = [H3O+][OH

-] = 1.0 × 10

-14 (at 25°C)

Kw = Ka × Kb

3. All aqueous solutions have H3O+ and OH

- present.

a. Neutral: [H3O+] = [OH

-] = 1.0 × 10

-7

b. Acidic: [H3O+] > [OH

-]

c. Basic (alkaline): [H3O+] < [OH

-]

4. Calculate the [OH-] if the [H3O

+] is 1.8 × 10

-8. Is the solution

acidic, basic or neutral?

Kw 3+ -H O OH [ ][ ] .10 10 14

[ ][ ]

.

..OH

H O

-

3+

KMw 10 10

18 1056 10

14

8

7

H3O+] < [OH

-] basic

6

Pages 666-669

Reminder: [H

+]

and [H3O+] are

used

interchangeably.

Pages 669-671

Pages 671-675

B. pH Scale

1. The pH scale was developed by Sorenson in 1909. The pH

scale is an easy way to express very small [H3O+]. The pH

scale ranges from 0 to 14.

2. pH<7: acidic solution

pH = 7 neutral solution

pH >7: basic solution

3. pH = -log [H3O+] [H3O

+] = antilog (-pH)

pOH = -log [OH-] [OH

-] = antilog (-pOH)

pH + pOH = 14

4. When expressing the pH or pOH you should use as many

decimal places as there are significant figures in the problem.

5. The pH of a sample of human blood was measured to be 7.41

at 25°C. Calculate pOH, [H+] and [OH

-] for the sample.

pOH = 14 – 7.41 = 6.59

[H+] = 3.9×10

-8 M

[OH-] = 2.6×10

-7 M

C. Calculating the pH of Strong Acid (or Base) Solutions

1. Calculating the pH of a solution of a strong monoprotic acid is

straightforward because [H+] equals the original concentration

of the acid.

2. What is the pH of a 0.040 M solution of HClO4?

pH = -log(0.040) = 1.40

3. An aqueous solution of HNO3 has a pH of 2.34. What is the

concentration of the acid?

[H+] = antilog(-2.34) = 10

-2.34 = 0.0046 M

4. What is the pH of a 0.50 M solution of NaOH?

pOH = -log(0.50) = 0.30

pH = 14 – 0.30 = 13.70

D. Calculating the pH of Weak Acid (or Base) Solutions

1. Calculating the pH of a weak acid (or base) involves setting up

an equilibrium expression. Always start by writing the

equation, setting up the acid (or base) equilibrium expression

(Ka or Kb), defining initial concentrations, changes and final

concentrations in terms of x, substituting values and variable

into the Ka (or Kb) expression and solving for x. We will use

RICE tables.

R = reaction

I = initial concentrations

C = change taking place

E = equilibrium concentrations

7

Often the –x in a

Ka expression

can be treated as

negligible if the

Ka value is

small. When

you assume that

it is negligible,

you must check

the validity of

this assumption.

To be valid, x

must be less

than 5% of the

original

number.

2. Calculate the pH of a 1.00×10-4

M solution of acetic acid. The

Ka of acetic acid is 1.8×10-5

.

R HC2H3O2 ⇄ H

+ C2H3O

2-

I 1.00×10-4

0 0

C -x +x + x

E 1.00×10-4

– x x x

Ka [ ][

[ ].

H C H O ]

HC H O

+2 3 2

-

2 3 2

18 10 5

18 10100 10

5

4.

( )( )

( . )

x x

x

Let’s assume that –x is negligible.

We will assume that 1.00×10-4

–x ≈ 1.00×10-4

18 10

100 104 2 105

4

5.( )( )

( . ).

x xx

We must now check the validity of our assumption.

4 2 10

10 10100 42%

5

4

.

.

Unfortunately, this is greater than 5% so the assumption is

invalid and we must use the quadratic equation to solve for x.

18 10

100 10

5

4.

( )( )

( . )

x x

x 1.8×10

-5(1.00×10

-4 – x) = x

2

1.8×10-9

– 1.8×10-5

x = x2

x2 + 1.8×10

-5x – 1.8×10

-9 =0

x = 3.5×10-5

and -5.2×10-5

(discard)

pH = - log(3.5×10-5

) = 4.56

8

Pages 675-677

Since the Ka for

HNO2 is the

largest it will be

the primary

source of H+.

3. The hypochlorite ion (OCl-) is a strong oxidizing agent often found in

household bleaches and disinfectants. It is also the active ingredient that

forms when swimming pool water is treated with chlorine. In addition to

its oxidizing abilities, the hypochlorite ion has a relatively high affinity for

protons (it is a much stronger base than Cl-, for example) and forms the

weakly acidic hypochlorous acid (HClO, Ka = 3.5×10-8

). Calculate the pH

of a 0.100 M aqueous solution of hypochlorous acid.

R HClO ⇄ H+ ClO

-

I 0.100 0 0

C -x +x + x

E 0.100 – x x x

Ka [ ][

[ ].

H ClO ]

HClO

+ -

35 10 8

35 10

0100

8.( )( )

( . )

x x

x Assume 0.100 – x ≈ 0.100

35 100100

59 108 5.( )( )

( . ). x x

x

Check the validity of the assumption.

59 10

0100100 0 059%

5.

..

Since it is less than 5% the assumption is valid.

pH = - log(5.9×10-5

) = 4.23

E. Determination of the pH of a Mixture of Weak Acids

1. Only the acid with the largest Ka value will contribute an appreciable [H+].

Determine the pH based on this acid and ignore any others.

2. Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2×10-10

)

and 5.00 M HNO2 (Ka = 4.0×10-4

). Also calculate the concentration of

cyanide ion (CN-) in this solution at equilibrium.

R HNO2 ⇄ H+ NO2

-

I 5.00 0 0

C -x +x + x

E 5.00 – x x x

Ka [ ][

[ ].

H NO ]

HNO

+2

-

2

4 0 10 4

4 0 10

500

4.( )( )

( . )

x x

x Assume 5.00 – x ≈ 5.00

4 0 10

5000 454.

( )( )

( . ). x x

x


0 45

500100 0 90%

.

..

Since it is less than 5% the assumption is valid. pH = - log(0.45) = 1.35

Now we need to calculate [CN-].

Ka [ ][

[ ].

H CN ]

HCN

+ -

6 2 10 2

( . )[

( . ).

0 45

1006 2 10 2CN ]-

[ ]

( . )( . )

..CN-

6 2 10 100

0 4514 10

108M

9

Pages 677-681

Pages 688-694

F. Calculating Percent Dissociation (also known as percent ionization)

1. It is often useful to specify the amount of weak acid that

dissociated in achieving equilibrium in an aqueous solution.

The percent dissociation is defined as follows:


amount dissociated (mol / L)

initial concentration (mol / L)100

2. Calculate the percent dissociation of acetic acid in a 1.00 M

HC2H3O2 solution. The Ka of acetic acid is 1.8×10-5

.

R HC2H3O2 ⇄ H

+ C2H3O2

-

I 1.00 M 0 0

C -x +x + x

E 1.00 – x x x

Ka [ ][

[ ].

H C H O ]

HC H O

+2 3 2

-

2 3 2

18 10 5

18 10

100

5.( )( )

( . )

x x

x Let’s assume that 1.00 –x ≈ 1.00

18 10

1004 2 105 3.

( )( )

( . ). x x

x

We must now check the validity of our assumption.

4 2 10

100100 0 42%

3.

..

This is less than 5% so the assumption is valid.


[H

HC H O ]

+

2 3 2

]

[

.

..

1004 2 10

100100 0 42%

3

G. Calculating the pH of a Polyprotic Acid

1. A polyprotic acid always dissociates in a stepwise manner,

one proton at a time.

2. Each step has a characteristic Ka value.

3. Typically for a weak polyprotic acid, Ka1 > Ka2 > Ka3

4. Write out the stepwise Ka reactions for the diprotic acid

H2SO3.

H2SO3 + H2O ⇄ H3O+ + HSO3

-

HSO3- + H2O ⇄ H3O

+ + SO3

2-

5. For a typical polyprotic acid in water, only the first

dissociation step is important in determining the pH.

10

Sulfuric Acid is

unique. It is a

strong acid in the

first dissociation

step (Ka1 is very

large). It is a

weak acid in the

second step.

For relatively

concentrated

solutions of H2SO4

(1.0 M or higher),

the large

concentration of

H+ from the first

dissociation step

represses the

second step, which

can be neglected

as a contributor of

H+ ions. For

dilute solutions of

sulfuric acid, the

second step does

make a significant

contribution, and

the quadratic

equation must be

used to obtain the

total H+

concentration.

6. Calculate the pH of a 5.0 M H3PO4 solution and the

equilibrium concentrations of the species H3PO4, H2PO4-,

HPO42-

, and PO43-

. Ka1 = 7.5×10-3

, Ka2 = 6.2×10-9

, 4.8×10-13

.

The dominant equilibrium is the dissociation of H3PO4.

R H3PO4 ⇄ H

+ H2PO4

-

I 5.0 0 0

C -x +x + x

E 5.0 – x x x

Substituting the equilibrium concentrations into the expression

for Ka1 and making the usual approximation gives

K

x x

x

xxa

+2 4

3 41

H H PO

H PO

7 5 105 0 50

01932

.[ ][ ]

[ ]

( )( )

( . ) ..


019

50100 38%

.

..

The assumption is valid.

Therefore [H+] = [H2PO4

-] = 0.19

pH = -log(0.19) = 0.72

The concentration of HPO42-

can be obtained by using the

expression for Ka2.

K Ma

+4

2 4

442

H HPO

H PO

HPO HPO

6 2 10

019

0196 2 108

2 22 8.

[ ][ ]

[ ]

( . )[ ]

( . )[ ] .

The concentration of PO43- can be obtained by using

the expression for Ka3.

Ka

+4

4

4

3

H PO

HPO

PO

4 8 10

019

6 2 10

133

2

3

8.

[ ][ ]

[ ]

( . )[ ]

( . )

[ ]

( . )( . )

..PO4

313 8

194 8 10 6 2 10

01916 10

M

7. Calculate the pH of a 1.0 M H2SO4 solution.

The initial concentration of H+ is at least 1.0 M. We must

determine if the HSO4- ion dissociates enough to produce a

significant contribution to the concentration of H+.

R HSO4

- ⇄ H

+ SO4

2-

I 1.00 M 1.0 0

C -x +x + x

E 1.00 – x x x


for Ka2 and making the usual approximation gives

Kx x x

x Ma

+4

42

H

HSO

(1.0)( ) = 1.2

12 1010

10 10102

22.

[ ][SO ]

[ ]

( . )( )

( . ) ( . )


0 012

10100 12%

.

..


[H+] = 1.0 + x = 1.0 + 0.012 = 1.0 M (to the correct number of

significant figures)

pH = -log(1.0 M) = 0.00

11

Pages 694-700

V. Acid Base Properties of Salts

A. A salt is made by neutralizing an acid with a base. When a salt

dissolves in water, it releases ions having an equal number of positive

and negative charges. Thus a solution of a salt should be neither

acidic nor basic.

B. Some salts do form neutral solutions, but others react with water

(hydrolye) to form acidic or basic solutions.

C. Types of Salts

1. Neutral Salts – Salts that are formed from the cation of a

strong base and the anion of a strong acid form neutral

solutions when dissolved in water. A salt such as NaNO3

gives a neutral solution.

2. Basic Salts – Salts that are formed from the cation of a strong

base and the anion of a weak acid form basic solutions when

dissolved in water. The anion hydrolyzes the water molecule

to produce hydroxide ions and thus a basic solution. K2S

should be basic since S2-

is the conjugate base of the very

weak acid HS-, while K+ does not hydrolyze appreciably.

S2-

+ H2O ⇄ OH- + HS

-

strong base weak acid

3. Acidic Salts – Salts that are formed from the cation of a weak

base and the anion of a strong acid form acidic solutions when

dissolved in water. The cation hydrolyzes the water molecule

to produce hydronium and thus an acidic solution. NH4Cl

should be weakly acidic, sinc NH4+ hydrolyzes to give an

acidic solution, while Cl- does not hydrolyze.

NH4+ + H2O ⇄ H3O

+ + NH3

Strong acid weak base

4. The salt produced from the anion of a weak acid and the

cation of a weak base may form an acidic, basic, or neutral

solution. The Ka and Kb must be compared to predict the

nature of the salt.

Ka > Kb pH < 7 (acidic)

Kb > Kb pH > 7 (basic)

Ka = Kb pH = 7 (neutral)

D. Use the information at the right to predict whether an aqueous

solution of each of the following salts will be acidic, basic or neutral.

1. NaCl

neutral: anion of a strong acid

and cation of a strong base

2. NH4SO4

acidic: anion of a strong acid and cation of a weak base

3. NH4C2H3O2

neutral: Ka = Kb

4. NH4CN

basic: Kb > Ka

KK

Kb

w

a

(for CN - ).

..

10 10

6 2 1016 10

14

10

5

NH4+ Ka = 5.6×10

-10

C2H3O2- Kb = 5.6×10

-10

HCN Ka = 6.2×10-10

12

NH4

+ is a

conjugate acid

of a weak base

Cl- is a

conjugate base

of a strong acid

NH4Cl is the salt

of a strong acid

and weak base. It

will be acidic.

F

- is a conjugate

base of a weak

acid

Na+ is the

conjugate acid

of a strong base NaF is the salt of

a weak acid and a

strong base. It

will be basic.

E. Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3

is 1.8×10-5

.

The major species in solution are NH4+, Cl

-, and H2O. Both the NH4

+

and the H2O can produce H+.

NH4+ ⇄ NH3 + H

+

Ka [ ][ ]

[ ]

NH H

NH

3+

4+

We must determine the value for the Ka of NH4

+ from the relationship

Ka × Kb = Kw

K

K

Ka

w

b

(for NH (for NH

4+

3

))

.

..

10 10

18 1056 10

14

5

10

This means that NH4

+ is a stronger acid than H2O (Ka = 1.0×10

-14) and

will dominate in the production of H+

R NH4+ ⇄ H

+ NH3

I 0.10 0 0

C -x +x + x

E 0.10 – x x x

Substituting the equilibrium concentrations into the expression for Ka

and making the usual approximation gives

K

x x

x

xa

56 10

010 010

102

.[ ][ ]

[ ]

( )( )

. .

NH H

NH

3+

4+

x ≈7.5×10

-6


7 5 10

010100 0 0075%

6.

..

The assumption is valid

[H+] = 7.5×10-6

M and pH = 5.13

F. Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is

7.2×10-4

.

The major species in solution are Na+, F

- (from the weak acid HF) and

H2O. Both the F- and the H2O can produce OH

-.

F- + H2O ⇄ HF + OH

-

We must determine the value for the Kb of HF from the relationship

Ka × Kb = Kw

K

K

Kb

w

a

(for HF (for HF

))

.

..

10 10

7 2 1014 10

14

4

11

This means that F

- is a stronger base than H2O (Kb = 1.0×10

-14) and

will dominate in the production of OH-

R F- ⇄ HF OH

-

I 0.30 0 0

C -x +x + x

E 0.30 – x x x

Substituting the equilibrium concentrations into the expression for Kb

and making the usual approximation gives

K

x x

x

xb

14 10

0 30 0 30

112

.[ ][ ]

[ ]

( )( )

. .

HF OH

F

-

-

x ≈ 2.0×10

-6 The approximation is valid by the 5% rule.

[OH-] = 2.0×10

-6 M pOH = 5.69; pH = 14 - 5.69 = 8.31

13

Pages 722-736

VI. Buffers

A. A buffer is a solution of a weak acid or base and its salt which resists

changes in pH when either OH- or H

+ ions are added.

B. Buffer capacity – The amount of acid or base that can be absorbed by

a buffer system without a significant change in pH. In order to have a

large buffer capacity, a solution should have large concentrations of

both buffer components.

C. Calculating the pH of a Buffered Solution

1. A buffered solution contains 0.50 M acetic acid (HC2H3O2,

Ka = 1.8×10-5

) and 0.50 M sodium acetate (NaC2H3O2).

Calculate the pH of this solution.

The major species in solution are

HC2H3O2 Weak acid

Na+

Neither acid nor base

C2H3O2- Conjugate Base of HC2H3O2

H2O very weak acid or base

The acetic acid dissociation equilibrium, which involves both

HC2H3O2 and C2H3O2-, will control the pH of the solution.

R HC2H3O2 ⇄ H+ C2H3O2

-

I 0.50 0 0.50

C -x +x + x

E 0.50 – x x 0.50 + x


for Ka and making the usual approximation gives

K

x x

x

xa

18 10

050

050

050

050

5.[ ] ]

[ ]

( )( . )

.

( . )

.

H [C H O

HC H O

+2 3 2

-

2 3 2 x ≈ 1.8×10

-5

Check the validity of the approximation.

18 10

050100 0 0036%

5.

..


[H+] = 1.8×10

-5 pH = 4.74

2. Another way to calculate the pH of a buffer system is with the

Henderson-Hasselbach equation.

pH = p + log

[A

HA]

-

Ka]

[ This equation must be used with caution. This equation is

only valid for solutions that contain weak monoprotic acids

and their salts or weak bases and their salts. The buffered

solution cannot be too dilute and the Ka/Kb cannot be too large.

Let’s calculate the pH of the buffer system in question 1 using

the Henderson-Hasselbach equation.

pKa = -log(1.8×10-5

) = 4.74

pH = p + log

[A

HA]

-

Ka]

[. log

.

.. 4 74

050

0504 74

14

Page 736

Ideally, the end

point and the

equivalence

point in a

titration should

coincide.

D. Calculating pH Changes in Buffered Solutions.

1. Calculate the change in pH that occurs when 0.010 mol solid

NaOH is added to 1.0 L of the buffered solution described in

the example above. Compare this pH change with that which

occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

The major species in solution are HC2H3O2, Na+, C2H3O2

-,

OH-, and H2O.

HC2H3O2 + OH

- → C2H3O2

- + H2O

Before 0.50 M 0.010 M 0.50 M

After 0.50-0.010

=0.49 M

0.010-0.010

= 0

0.50 + 0.010

=0.51 M

pH = p + log

[A

HA]

-

Ka]

[. log

.

.. 4 74

051

0 494 76

ΔpH = 4.76 -4.74 = + 0.02

[ ]

.

..OH

mol

L

0 010

100 010 M

pOH = -log(0.010) = 2 pH = 14-2 = 12

ΔpH = 12.00 – 7.00 = + 5.00

VII. Titrations

A. Titration – the process by which the volume of a standard solution

required to react with a specific amount of a substance is determined.

B. Titration Curve – plot of the pH versus the amount (usually volume)

of acid or base added.

C. End Point – the point at which the color of an indicator changes in a

titration. It is determined by the Ka value for the indicator.

D. Equivalence Point – the point at which chemically equivalent amounts

of acid and base have reacted.

E. Steps in carrying out a titration to determine the concentration of an

unknown acid solution.

1. A measured amount of an acid of unknown concentration is

added to an Erlenmeyer flask.

2. An appropriate indicator (such as phenolphthalein) is added to

the solution.

3. Measured amounts of a base of known concentration are

mixed into the acid. The solution of known concentration is

called the standard solution. The addition of the base is

carried out using a buret. This process is continued until the

indicator indicates that the end point has been reached.

4. The point at which the two solutions used in a titration are

present in chemically equivalent amounts is the equivalence

point.

F. Choosing an Indicator - Choose an indicator that will change color as

close as possible to the end point.

15

Pages 737-740

VIII. Titration Problems

A. Titration of a Strong Acid with a Strong Base

1. A titration is performed by adding 0.600 M KOH to 40.0 mL of

0.800 M HCl. The net ionic equation for this reaction is:

H+ + OH

- → H2O

a. Calculate the pH before the addition of any KOH.

Remember before you add any base all you have is the

strong acid in the flask. Strong acids completely ionize in

water.

HCl + H2O → Cl- + H3O

+

pH = -log [H3O+]

pH = -log[0.800] = 0.097

b. Calculate the pH after the addition of 5.0 mL of the base.

H+ + OH

- → H2O

(.040 L)(0.8 M) = 0.032 moles (0.005 L)(0.6 M) = 0.003 moles

-0.003 moles -0.003 moles

0.029 moles 0 moles

The total volume of solution is 45 mL (40.0 mL + 5.0 mL).

pH = - log[H+]

pH log0.029 moles

0.045 L LNM

OQP 0191.

c. Calculate the pH after the addition of 20.0 mL of the

base.

H+ + OH

- → H2O

(.040 L)(0.8 M) = 0.032 moles (0.020L)(0.6 M) = 0.012 moles


0.020 moles 0 moles


pH = - log[H+]

pH log0.020 moles

0.060 L LNM

OQP 0 477.

16

d. Calculate the pH after the addition of 40.0 mL of the

base.

H+ + OH

- → H2O

(.040 L)(0.8 M) = 0.032 moles (0.040L)(0.6 M) = 0.024 moles


0.008 moles 0 moles


pH = - log[H+]

pH log0.008 moles

0.080 L LNM

OQP 100.

e. Calculate the pH after the addition of 52.0 mL of the

base.

H+ + OH

- → H2O

(.040 L)(0.8 M) = 0.032 moles (0.052L)(0.6 M) = 0.0312 moles

-0.0312 moles -0.0312 moles

0.0008 moles 0 moles


pH = - log[H+]

pH log0.0008 moles

0.092 L LNM

OQP 2 06.

f. Calculate the volume of base needed to reach the

equivalence point.

The equivalence point is where the moles of acid equals

the moles of base.

We know we have 0.032 moles of acid. The volume of

base needed is equal to:

Molarity

mol

L( )M

Lmol

= 0.032 mol

= 0.533 L = 53.3 mL

M M0 6.

g. What is the pH at the equivalence point?

H+ + OH

- → H2O

pH = 7 (Strong acid and a strong base)

17

h. Calculate the pH after adding 5.00 mL of NaOH past the

equivalence point.

H+ + OH

- → H2O

(.040 L)(0.8 M) = 0.032 moles (0.0583 L)(0.6 M) = 0.03498

moles


0 moles 0.00298 moles

The total volume of solution is 98.3 mL (40.0 mL + 58.3 mL).

pOH = - log[OH-]

pOH log0.00298 moles

0.0983 L LNM

OQP 152.

pH = 14.00 – 1.52 = 12.48

2. A sketch of the titration curve from the problem just

completed.

3. What are some characteristics of the curve?

a. Extreme pH at the beginning

b. No Buffering

c. Sharp rise in pH at equivalence point

d. The equivalence point is at 7

18

Pages 740-749

Why is it okay

to use the moles

of the acids and

base instead of

the

concentrations?

The volume was

the same for

both.

B. Titration of a Weak Acid with a Strong Base

1. A titration is performed by adding 0.200 M NaOH to 24.0 mL

of 0.350 M HOCl. The net ionic equation for this reactions

is:

HOCl + OH- OCl

- + H2O

a. Calculate the pH before the addition of any NaOH. The

Ka of HOCl is 3 x 10-8

.

0.350 – x ≈ 0.350 (This assumption is made because the Ka

value for HOCl is so small – less than 10-4

.)

HOCl + H2O OCl- + H3O

+

Ka [H O ][OCl ]

[HOCl]

3

3 100 350

82

= = 1.02 10 -4xx M

. pH = -log[H3O

+] = -log[1.02×10

-4] = 3.99

b. Calculate the pH after the addition of 5.0 mL of the

base.

The Henderson-Hasselbach equation can be used because a weak

acid and a salt are left over.

pH p log [base]

[acid]

FHGIKJKa

pH 0.0010

0.0074 6.65

FHG

IKJ

log ( ) log3 10 8

R HOCl + H2O → OCl- + H3O

+

I 0.350 M

0

0

C - x

+ x

+ x

E 0.350-x

x

x

HOCl + OH- → OCl

- + H2O

(0.350 M)(0.024L)=

0.0084 moles

(0.200M)(0.0050 L)

= 0.0010 moles

0

- 0.0010 - 0.0010

+ 0.0010

0.0074 0

0.0010

19

c. Calculate the pH after the addition of 15.0 mL of the

base.

HOCl + OH- → OCl

- + H2O

(0.350 M)(0.024L)=

0.0084 moles

(0.200M)(0.0150 L)

= 0.0030 moles

0

-0.0030 -0.0030

+0.0030

0.0054 0

0.0030



pH = p + log [base]

[acid]Ka

FHGIKJ

pH = - log (3 10 + log 0.0030

0.0054 = 7.27-8

FHG

IKJ)

d. Calculate the pH after the addition of 25.0 mL of the

base.

HOCl + OH- → OCl

- + H2O

(0.350 M)(0.024L)=

0.0084 moles

(0.200M)(0.0250 L)

= 0.0050 moles

0

-0.0050 -0.0050

+0.0050

0.0034 0

0.0050



pH = p + log [base]

[acid]Ka

FHGIKJ

pH = - log (3 10 + log 0.0050

0.0034 = 7.69-8

FHG

IKJ)

e. Calculate the pH after the addition of 35.0 mL of the

base.

HOCl + OH- → OCl

- + H2O

(0.350 M)(0.024L)=

0.0084 moles

(0.200M)(0.0350 L)

= 0.0070 moles

0

-0.0070 -0.0070

+0.0070

0.0014 0

0.0070



pH = - log (3 10 log 0.0070

0.0014 = 8.22-8

FHG

IKJ)

20

f. Calculate the volume of base needed to reach the

equivalence point.

The equivalence point is where the moles of acid equals

the moles of base.

We know we have 0.0084 moles of acid. The volume

of base needed is equal to:

Molarity

mol

L( )M

Lmol

= 0.0084 mol

= 0.0420 L = 42.00 mL

M M0 2. g. What is the pH at the equivalence point?

HOCl + OH- → OCl

- + H2O

(0.350 M)(0.024L)=

0.0084 moles

(0.200M)(0.0420 L)

= 0.0084 moles

0

-0.0084 -0.0084

+0.0084

0 0

0.0084

Molarity

mol

L =

0.0084 mol

L + 0.042 L ( )

..M M

0 0240127

R OCl- + H2O → HOCl + OH

-

I 0.127 M

0

0

C - x

+ x

+ x

E 0.127-x

x

x

Kw = Ka × Kb = 1 × 10-14

KK

Kb

w

a

=

1 10

3 10

-14

-8333 10 7.

0.127 – x ≈ 0.127 (since the Ka and Kb are so small)

Kb [ ][

[ ]

OH HOCl]

OCl

-

-

333 100127

7.[ ][ ]

. x x

x = [OH

-] = 2.056 × 10

-4 M

pOH = -log[2.056 × 10-4

] = 3.687

pH = 14 – 3.687 = 10.31

21

At the midpoint

(half-

equivalence

point) the pH =

pKa. Why?

This is because

pH = p + log [A ]

[HA]

-

Ka

and since [A-] =

[HA], the

log[A

HA]

-]

[ 0

.

h. Calculate the pH after adding 5.00 mL of NaOH past

the equivalence point.

HOCl + OH- → OCl

- + H2O

(0.350 M)(0.024L)=

0.0084 moles

(0.200M)(0.0470 L)

= 0.0094 moles

-0.0084 -0.0084

0 0.0010

The total volume of solution is 71.0 mL (24.0 mL + 47.0 mL).

pOH = - log[OH-]

pOH log0.0010 moles

0.0710 L LNM

OQP 185.

pH = 14.00 – 1.85 = 12.15

2. A sketch of the titration curve from the problem just

completed.

3. What are some characteristics of the curve?

a. Not as an extreme of start in pH

b. Buffering occurs

c. Not as sharp of a rise in pH at equivalence point

d. The equivalence point is not at 7

ap chemistry: acids, bases, and salts...2017/02/24 · 1 ap chemistry: acids, bases, and salts unit...

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