chapter · chapter 16 aqueous ionic equilibrium ... hbo 2 (aq) + h+(aq) ... 2 = 1.8 x 10−5 tro:...

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Copyright 2011 Pearson Education, Inc.

Chapter 16

Aqueous Ionic

Equilibrium

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA

Chemistry: A Molecular Approach, 2nd Ed.

Nivaldo Tro


ethylene glycol

(aka 1,2–ethandiol)

The Danger of Antifreeze• Each year, thousands of pets and wildlife

species die from consuming antifreeze

• Most brands of antifreeze contain ethylene

glycolsweet taste

initial effect drunkenness

• Metabolized in the liver to glycolic acidHOCH2COOH

2

glycolic acid

(aka a-hydroxyethanoic acid)

Tro: Chemistry: A Molecular Approach, 2/e


Why Is Glycolic Acid Toxic?• If present in high enough concentration in the

bloodstream, glycolic acid overwhelms the

buffering ability of the HCO3− in the blood,

causing the blood pH to drop

• When the blood pH is low, its ability to carry O2

is compromised

acidosis

HbH+(aq) + O2(g) HbO2(aq) + H+(aq)

• One treatment is to give the patient ethyl

alcohol, which has a higher affinity for the

enzyme that catalyzes the metabolism of

ethylene glycol 3Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Buffers

• Buffers are solutions that resist changes in pH

when an acid or base is added

• They act by neutralizing acid or base that is

added to the buffered solution

• But just like everything else, there is a limit to

what they can do, and eventually the pH changes

• Many buffers are made by mixing a solution of a

weak acid with a solution of soluble salt

containing its conjugate base anion

blood has a mixture of H2CO3 and HCO3−

4Tro: Chemistry: A Molecular Approach, 2/e


Making an Acid Buffer

5Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

How Acid Buffers Work:

Addition of Base

HA(aq) + H2O(l) A−(aq) + H3O

+(aq)

• Buffers work by applying Le Châtelier’s Principle

to weak acid equilibrium

• Buffer solutions contain significant amounts of the

weak acid molecules, HA

• These molecules react with added base to

neutralize it

HA(aq) + OH−(aq) → A−(aq) + H2O(l)

you can also think of the H3O+ combining with the OH−

to make H2O; the H3O+ is then replaced by the shifting

equilibrium6Tro: Chemistry: A Molecular Approach, 2/e

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H2O

HA

How Buffers Work

HA + H3O+

A−

Added

HO−

new

A−

A−


How Acid Buffers Work:

Addition of Acid


+(aq)

• The buffer solution also contains significant amounts of the conjugate base anion, A−

• These ions combine with added acid to make more HA

H+(aq) + A−(aq) → HA(aq)

• After the equilibrium shifts, the concentration of H3O

+ is kept constant



H2O

How Buffers Work

HA + H3O+A−

A−

Added

H3O+

new

HA

HA


Common Ion Effect


+(aq)

• Adding a salt containing the anion NaA, which

is the conjugate base of the acid (the common

ion), shifts the position of equilibrium to the left

• This causes the pH to be higher than the pH of

the acid solution

lowering the H3O+ ion concentration



Common Ion Effect


Example 16.1: What is the pH of a buffer that

is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?

12

write the reaction

for the acid with

water

construct an ICE

table for the

reaction

enter the initial

concentrations –

assuming the

[H3O+] from

water is ≈ 0

HC2H3O2 + H2O C2H3O2 + H3O

+

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change

equilibrium


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[HA] [A−] [H3O+]

initial 0.100 0.100 0

change

equilibrium



13

represent the

change in the

concentrations in

terms of x

sum the columns

to find the

equilibrium

concentrations in

terms of x

substitute into the

equilibrium

constant

expression

+x+xx

0.100 x 0.100 + x x


+

Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.



14

determine the

value of Ka

because Ka is very

small, approximate

the [HA]eq = [HA]init

and [A−]eq = [A−]init ,

then solve for x

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 x0.100 x 0.100 +x

Ka for HC2H3O2 = 1.8 x 10−5





15

check if the

approximation

is valid by

seeing if x <

5% of

[HC2H3O2]init

Ka for HC2H3O2 = 1.8 x 10−5

the approximation is valid

x = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 x




16

substitute x

into the

equilibrium

concentration

definitions

and solve x = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 1.8E-50.100 + x x 0.100 x





17

substitute [H3O+]

into the formula

for pH and solve

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 1.8E−5




18

check by

substituting the

equilibrium

concentrations

back into the

equilibrium

constant

expression and

comparing the

calculated Ka to the

given Ka

the values

match

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x


Ka for HC2H3O2 = 1.8 x 10−5


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Practice − What is the pH of a buffer that is

0.14 M HF (pKa = 3.15) and 0.071 M KF?


Practice − What is the pH of a buffer that is

0.14 M HF (pKa = 3.15) and 0.071 M KF?

20

write the

reaction for the

acid with water

construct an

ICE table for

the reaction

enter the initial

concentrations

– assuming the

[H3O+] from

water is ≈ 0

HF + H2O F + H3O+

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change

equilibrium



[HA] [A−] [H3O+]

initial 0.14 0.071 0

change

equilibrium

Practice – What is the pH of a buffer that is

0.14 M HF (pKa = 3.15) and 0.071 M KF?

21

represent the

change in the

concentrations in

terms of x

sum the columns

to find the

equilibrium

concentrations in

terms of x

substitute into

the equilibrium

constant

expression

+x+xx

0.14 x 0.071 + x x

HF + H2O F + H3O+


pKa for HF = 3.15Ka for HF = 7.0 x 10−4


0.14 M and 0.071 M KF?

22

determine the

value of Ka

because Ka is

very small,

approximate the

[HA]eq = [HA]init

and [A−]eq = [A−]init

solve for x

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.012 0.100 x0.14 x 0.071 +x




0.14 M HF (pKa = 3.15) and 0.071 M KF?

23

check if the

approximation

is valid by

seeing if

x < 5% of

[HC2H3O2]init

Ka for HF = 7.0 x 10−4


x = 1.4 x 10−3

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x




0.14 M HF (pKa = 3.15) and 0.071 M KF?

24

substitute x into

the equilibrium

concentration

definitions and

solve

x = 1.4 x 10−3

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x



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0.14 M HF (pKa = 3.15) and 0.071 M KF?

25

substitute

[H3O+] into the

formula for pH

and solve

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x




0.14 M HF (pKa = 3.15) and 0.071 M KF?

26

check by

substituting the

equilibrium

concentrations

back into the

equilibrium

constant

expression and

comparing the

calculated Ka to

the given Ka

the values are

close enough

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x


Ka for HF = 7.0 x 10−4



Henderson-Hasselbalch Equation

• Calculating the pH of a buffer solution can be

simplified by using an equation derived from

the Ka expression called the Henderson-

Hasselbalch Equation

• The equation calculates the pH of a buffer

from the pKa and initial concentrations of the

weak acid and salt of the conjugate base

as long as the “x is small” approximation is valid


Deriving the Henderson-Hasselbalch

Equation





29

assume the [HA] and

[A−] equilibrium

concentrations are

the same as the

initial

substitute into the

Henderson-

Hasselbalch

Equation

check the “x is small”

approximation


+

Ka for HC7H5O2 = 6.5 x 10−5



0.14 M HF (pKa = 3.15) and 0.071 M KF?


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0.14 M HF (pKa = 3.15) and 0.071 M KF?

31

find the pKa from the

given Ka

assume the [HA]

and [A−] equilibrium

concentrations are

the same as the

initial

substitute into the

Henderson-

Hasselbalch

equation

check the “x is

small”

approximation

HF + H2O F + H3O+


Do I Use the Full Equilibrium Analysis or

the Henderson-Hasselbalch Equation?• The Henderson-Hasselbalch equation is

generally good enough when the “x is small” approximation is applicable

• Generally, the “x is small” approximation will work when both of the following are true:

a) the initial concentrations of acid and salt are not very dilute

b) the Ka is fairly small

• For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of Ka



How Much Does the pH of a Buffer

Change When an Acid or Base Is Added?• Though buffers do resist change in pH when acid

or base is added to them, their pH does change

• Calculating the new pH after adding acid or base requires breaking the problem into two parts

1. a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A− to make more HA

added base reacts with the HA to make more A−

2. an equilibrium calculation of [H3O+] using the new

initial values of [HA] and [A−]


Example 16.3: What is the pH of a buffer that has

0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in

1.00 L that has 0.010 mol NaOH added to it?

34

If the added

chemical is a

base, write a

reaction for OH−

with HA. If the

added chemical

is an acid, write a

reaction for H3O+

with A−.

construct a

stoichiometry

table for the

reaction

HC2H3O2 + OH− C2H3O2 + H2O

HA A− OH−

mols before 0.100 0.100 0

mols added ─ ─ 0.010

mols after






35

fill in the table

– tracking the

changes in the

number of

moles for each

component


HA A− OH−

mols before 0.100 0.100 ≈ 0


mols after 0.090 0.110 ≈ 0





36

If the added

chemical is a base,

write a reaction for

OH− with HA. If the

added chemical is

an acid, write a

reaction for it with A−.

construct a

stoichiometry table

for the reaction

enter the initial

number of moles for

each


HA A– OH−

mols before 0.100 0.100 0.010

mols change

mols end

new molarity


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37

using the added

chemical as the

limiting reactant,

determine how the

moles of the other

chemicals change

add the change to

the initial number of

moles to find the

moles after reaction

divide by the liters

of solution to find

the new molarities


HA A− OH−

mols before 0.100 0.100 0.010

mols change

mols end

new molarity

−0.010−0.010 +0.010

00.1100.090

0.090 0.110 0





38

write the reaction

for the acid with

water

construct an ICE

table for the

reaction

enter the initial

concentrations –

assuming the

[H3O+] from water

is ≈ 0, and using

the new

molarities of the

[HA] and [A−]


+

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium



[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium




39

represent the

change in the

concentrations in

terms of x

sum the columns

to find the

equilibrium

concentrations in

terms of x

substitute into

the equilibrium

constant

expression

+x+xx

0.090 x 0.110 + x x


+





40

determine the

value of Ka

because Ka is very

small, approximate

the [HA]eq = [HA]init

and [A−]eq = [A−]init

solve for x

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.090 0.110 x0.090 x 0.110 +x

Ka for HC2H3O2 = 1.8 x 10−5






41

check if the

approximation

is valid by

seeing if

x < 5% of

[HC2H3O2]init

Ka for HC2H3O2 = 1.8 x 10−5


x = 1.47 x 10−5

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x






42

substitute x

into the

equilibrium

concentration

definitions

and solve x = 1.47 x 10−5

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x



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43

substitute

[H3O+] into

the formula

for pH and

solve

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x






44

check by

substituting the

equilibrium

concentrations

back into the

equilibrium

constant

expression and

comparing the

calculated Ka to

the given Ka

the values

match

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x


Ka for HC2H3O2 = 1.8 x 10−5



or, by using the

Henderson-Hasselbalch

Equation





46

write the reaction for

the acid with water

construct an ICE

table for the reaction

enter the initial

concentrations –

assuming the [H3O+]

from water is ≈ 0,

and using the new

molarities of the [HA]

and [A−]

fll in the ICE table


+

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium

+x+xx

0.090 x 0.110 + x x






47

find the pKa from

the given Ka

assume the [HA]

and [A−] equilibrium

concentrations are

the same as the

initial


+

Ka for HC2H3O2 = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x






48

substitute into the

Henderson-

Hasselbalch

equation

check the “x is

small”

approximation


+

pKa for HC2H3O2 = 4.745


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Example 16.3: Compare the effect on pH of adding

0.010 mol NaOH to a buffer that has 0.100 mol

HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to

adding 0.010 mol NaOH to 1.00 L of pure water

49


+

pKa for HC2H3O2 = 4.745


Practice – What is the pH of a buffer that has

0.140 moles HF (pKa = 3.15) and 0.071 moles

KF in 1.00 L of solution when 0.020 moles of

HCl is added?

(The “x is small” approximation is valid)




0.140 moles HF and 0.071 moles KF in 1.00 L of

solution when 0.020 moles of HCl is added?

51

If the added

chemical is a

base, write a

reaction for OH−

with HA. If the

added chemical

is an acid, write a

reaction for H3O+

with A−.

construct a

stoichiometry

table for the

reaction

F− + H3O+ HF + H2O

F− H3O+ HF

mols before 0.071 0 0.140

mols added – 0.020 –

mols after





52

fill in the table

– tracking the

changes in the

number of

moles for each

component

F− + H3O+ HF + H2O

F− H3O+ HF

mols before 0.071 0 0.140

mols added – 0.020 –

mols after 0.051 0 0.160






53

If the added chemical

is a base, write a

reaction for OH− with

HA. If the added

chemical is an acid,

write a reaction for

H3O+ with A−.

construct a

stoichiometry table

for the reaction

enter the initial

number of moles for

each

F− + H3O+ HF + H2O

F− H3O+ HF

mols before 0.071 0.020 0.140

mols change

mols after

new molarity


F− H3O+ HF

mols before 0.071 0.020 0.140

mols change

mols after

new molarity




54

using the added

chemical as the

limiting reactant,

determine how the

moles of the other

chemicals change

add the change to

the initial number of

moles to find the

moles after reaction

divide by the liters

of solution to find

the new molarities

0.16000.051

F− + H3O+ HF + H2O

+0.020−0.020 −0.020

0.16000.051


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55

write the reaction for

the acid with water

construct an ICE

table.

assume the [HA] and

[A−] equilibrium

concentrations are

the same as the

initial

substitute into the

Henderson-

Hasselbalch

equation

HF + H2O F + H3O+

[HF] [F−] [H3O+]

initial 0.160 0.051 ≈ 0

change −x +x +x



Basic Buffers

B:(aq) + H2O(l) H:B+(aq) + OH−

(aq)

• Buffers can also be made by mixing a weak

base, (B:), with a soluble salt of its conjugate

acid, H:B+Cl−

H2O(l) + NH3 (aq) NH4+

(aq) + OH−(aq)



• The Henderson-Hasselbalch equation is written for a

chemical reaction with a weak acid reactant and its

conjugate base as a product

• The chemical equation of a basic buffer is written with a

weak base as a reactant and its conjugate acid as a

product

B: + H2O H:B+ + OH−

• To apply the Henderson-Hasselbalch equation, the

chemical equation of the basic buffer must be looked at like

an acid reaction

H:B+ + H2O B: + H3O+

this does not affect the concentrations, just the way we are looking

at the reaction

Henderson-Hasselbalch Equation for

Basic Buffers


Relationship between pKa and pKb

• Just as there is a relationship between the Ka

of a weak acid and Kb of its conjugate base,

there is also a relationship between the pKa of

a weak acid and the pKb of its conjugate base

Ka Kb = Kw = 1.0 x 10−14

−log(Ka Kb) = −log(Kw) = 14

−log(Ka) + −log(Kb) = 14

pKa + pKb = 14



Example 16.4: What is the pH of a buffer that is

0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?

59

find the pKa of the

conjugate acid (NH4+)

from the given Kb

assume the [B] and

[HB+] equilibrium

concentrations are the

same as the initial

substitute into the

Henderson-

Hasselbalch equation


approximation

NH3 + H2O NH4+ + OH−


• The Henderson-Hasselbalch equation is written for a

chemical reaction with a weak acid reactant and its

conjugate base as a product

• The chemical equation of a basic buffer is written

with a weak base as a reactant and its conjugate

acid as a product

B: + H2O H:B+ + OH−

• We can rewrite the Henderson-Hasselbalch equation

for the chemical equation of the basic buffer in terms

of pOH

Henderson-Hasselbalch Equation for

Basic Buffers


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Example 16.4: What is the pH of a buffer that is

0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?

61

find the pKb if given Kb

assume the [B] and

[HB+] equilibrium

concentrations are the

same as the initial

substitute into the

Henderson-Hasselbalch

equation base form,

find pOH


approximation

calculate pH from pOH

NH3 + H2O NH4+ + OH−


Buffering Effectiveness

• A good buffer should be able to neutralize moderate amounts of added acid or base

• However, there is a limit to how much can be added before the pH changes significantly

• The buffering capacity is the amount of acid or base a buffer can neutralize

• The buffering range is the pH range the buffer can be effective

• The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base



HA A− OH−



mols after 0.17 0.030 ≈ 0

Effect of Relative Amounts of Acid

and Conjugate Base

Buffer 1

0.100 mol HA & 0.100 mol A−

Initial pH = 5.00

Buffer 2

0.18 mol HA & 0.020 mol A−

Initial pH = 4.05pKa (HA) = 5.00

after adding 0.010 mol NaOH

pH = 5.09

HA + OH− A + H2O

HA A− OH−



mols after 0.090 0.110 ≈ 0


pH = 4.25

A buffer is most effective with equal

concentrations of acid and base


HA A− OH−



mols after 0.49 0.51 ≈ 0

HA A− OH−



mols after 0.040 0.060 ≈ 0

Effect of Absolute Concentrations of Acid

and Conjugate Base

Buffer 1

0.50 mol HA & 0.50 mol A−

Initial pH = 5.00

Buffer 2

0.050 mol HA & 0.050 mol A−

Initial pH = 5.00pKa (HA) = 5.00


pH = 5.02

HA + OH− A + H2O


pH = 5.18

A buffer is most effective when the

concentrations of acid and base are largest


Buffering Capacity

a concentrated

buffer can

neutralize

more added

acid or base

than a dilute

buffer


Effectiveness of Buffers

• A buffer will be most effective when the

[base]:[acid] = 1

equal concentrations of acid and base

• A buffer will be effective when

0.1 < [base]:[acid] < 10

• A buffer will be most effective when the [acid]

and the [base] are large


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Buffering Range• We have said that a buffer will be effective when

0.1 < [base]:[acid] < 10

• Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective

Lowest pH Highest pH

Therefore, the effective pH range of a buffer is pKa ± 1

When choosing an acid to make a buffer, choose

one whose is pKa closest to the pH of the buffer


Formic acid, HCHO2 pKa = 3.74Formic acid, HCHO2 pKa = 3.74

Example 16.5a: Which of the following acids

would be the best choice to combine with its

sodium salt to make a buffer with pH 4.25?

Chlorous acid, HClO2 pKa = 1.95

Nitrous acid, HNO2 pKa = 3.34

Hypochlorous acid, HClOpKa = 7.54

The pKa of HCHO2 is closest to the desired

pH of the buffer, so it would give the most

effective buffering range



Example 16.5b: What ratio of NaCHO2 : HCHO2

would be required to make a buffer with pH 4.25?

69

Formic acid, HCHO2, pKa = 3.74

To make a buffer with

pH 4.25, you would use

3.24 times as much

NaCHO2 as HCHO2


Practice – What ratio of NaC7H5O2 : HC7H5O2


Benzoic acid, HC7H5O2, pKa = 4.19



Practice – What ratio of NaC7H5O2 : HC7H5O2


Benzoic acid, HC7H5O2, pKa = 4.19

to make a buffer with

pH 3.75, you would use

0.363 times as much

NaC7H5O2 as HC7H5O2


Buffering Capacity• Buffering capacity is the amount of acid or base

that can be added to a buffer without causing a large change in pH

• The buffering capacity increases with increasing absolute concentration of the buffer components

• As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves

• Buffers that need to work mainly with added acid generally have [base] > [acid]

• Buffers that need to work mainly with added base generally have [acid] > [base]


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Titration• In an acid-base titration, a solution of unknown

concentration (titrant) is slowly added to a solution of known concentration from a buretteuntil the reaction is completewhen the reaction is complete we have reached the

endpoint of the titration

• An indicator may be added to determine the endpointan indicator is a chemical that changes color when the

pH changes

• When the moles of H3O+ = moles of OH−, the

titration has reached its equivalence point


Titration



Titration Curve

• A plot of pH vs. amount of added titrant

• The inflection point of the curve is the equivalence point of the titration

• Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH

• The pH of the equivalence point depends on the pH of the salt solutionequivalence point of neutral salt, pH = 7

equivalence point of acidic salt, pH < 7

equivalence point of basic salt, pH > 7

• Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH


Titration Curve:

Unknown Strong Base Added

to Strong Acid



Before Equivalence

(excess acid)

After Equivalence

(excess base)

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

Equivalence Point

equal moles of

HCl and NaOH

pH = 7.00

Because the

solutions are equal

concentration, and

1:1 stoichiometry, the

equivalence point is

at equal volumes


HCl NaCl NaOH

mols before 2.50E-3 0 5.0E-4

mols change −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new

5.0 x 10−4 mole NaOH added


• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

• Initial pH = −log(0.100) = 1.00

• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence point added 5.0 mL NaOH


4/19/2016

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Titration of 25 mL of 0.100 M HCl with

0.100 M NaOH

• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• To reach equivalence, the added moles NaOH =

initial moles of HCl = 2.50 x 10−3 moles

• At equivalence, we have 0.00 mol HCl and 0.00

mol NaOH left over

• Because the NaCl is a neutral salt, the pH at

equivalence = 7.00


HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0




• Initial pH = −log(0.100) = 1.00


• At equivalence point added 25.0 mL NaOH



HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new



• Initial pH = −log(0.100) = 1.00


• After equivalence point added 30.0 mL NaOH3.0 x 10−3 mole NaOH added

L of added NaOH0.100 mol NaOH

1 L

moles added NaOH


added 5.0 mL NaOH

0.00200 mol HCl

pH = 1.18

added 10.0 mL NaOH

0.00150 mol HCl

pH = 1.37

added 25.0 mL NaOH

equivalence point

pH = 7.00

added 30.0 mL NaOH

0.00050 mol NaOH

pH = 11.96

added 40.0 mL NaOH

0.00150 mol NaOH

pH = 12.36

added 50.0 mL NaOH

0.00250 mol NaOH

pH = 12.52

added 35.0 mL NaOH

0.00100 mol NaOH

pH = 12.22

added 15.0 mL NaOH

0.00100 mol HCl

pH = 1.60

added 20.0 mL NaOH

0.00050 mol HCl

pH = 1.95

Adding 0.100 M NaOH to 0.100 M HCl

82

25.0 mL 0.100 M HCl

0.00250 mol HCl

pH = 1.00



Practice – Calculate the pH of the solution that

results when 10.0 mL of 0.15 M NaOH is added to

50.0 mL of 0.25 M HNO3


HNO3 NaNO3 NaOH


mols change −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new

HNO3 NaNO3 NaOH


mols change −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new 0.018 0.025 0



50.0 mL of 0.25 M HNO3

• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

• Initial pH = −log(0.250) = 0.60

• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• Before equivalence point added 10.0 mL NaOH


4/19/2016

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Practice – Calculate the amount of 0.15 M NaOH

solution that must be added to 50.0 mL of 0.25 M

HNO3 to reach equivalence


Practice – Calculate the amount of 0.15 M NaOH

solution that must be added to 50.0 mL of 0.25 M

HNO3 to reach equivalence


• Initial pH = −log(0.250) = 0.60


• At equivalence point: moles of NaOH = 1.25 x 10−2





50.0 mL of 0.25 M HNO3


HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new



50.0 mL of 0.25 M HNO3


• Initial pH = −log(0.250) = 0.60


• After equivalence point added 100.0 mL NaOH


HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new



50.0 mL of 0.25 M HNO3


• Initial pH = −log(0.250) = 0.60

• initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• After equivalence point added 100.0 mL NaOH


Titration of a Strong Base with a Strong Acid

• If the titration is run

so that the acid is in

the burette and the

base is in the flask,

the titration curve

will be the reflection

of the one just

shown


4/19/2016

16


Titration of a Weak Acid with a Strong Base• Titrating a weak acid with a strong base results in

differences in the titration curve at the equivalence

point and excess acid region

• The initial pH is determined using the Ka of the

weak acid

• The pH in the excess acid region is determined as

you would determine the pH of a buffer

• The pH at the equivalence point is determined

using the Kb of the conjugate base of the weak acid

• The pH after equivalence is dominated by the

excess strong base

the basicity from the conjugate base anion is negligible91Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Titration of 25 mL of 0.100 M HCHO2 with

0.100 M NaOH

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)

• Initial pH

[HCHO2] [CHO2−] [H3O

+]

initial 0.100 0.000 ≈ 0

change −x +x +x

equilibrium 0.100 − x x x

Ka = 1.8 x 10−4




0.100 M NaOH• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalenceadded 5.0 mL NaOH

HA A− OH−

mols before 2.50E-3 0 0

mols added – – 5.0E-4

mols after 2.00E-3 5.0E-4 ≈ 0


HA A− OH−

mols before 2.50E-3 0 0

mols added – – 2.50E-3

mols after 0 2.50E-3 ≈ 0

[HCHO2] [CHO2−] [OH−]

initial 0 0.0500 ≈ 0

change +x −x +x

equilibrium x 5.00E-2-x x


0.100 M NaOH

added 25.0 mL NaOHCHO2

−(aq) + H2O(l) HCHO2(aq) + OH−

(aq)

Kb = 5.6 x 10−11

[OH−] = 1.7 x 10−6 M

• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)


• At equivalence


HA A− NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new

HA A− NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

HA A− NaOH


mols change

mols end

molarity, new


0.100 M NaOH

95

added 30.0 mL NaOH


• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)


• After equivalence


added 35.0 mL NaOH

0.00100 mol NaOH xs

pH = 12.22

initial HCHO2 solution

0.00250 mol HCHO2

pH = 2.37

added 5.0 mL NaOH

0.00200 mol HCHO2

pH = 3.14

added 10.0 mL NaOH

0.00150 mol HCHO2

pH = 3.56

added 25.0 mL NaOH

equivalence point

0.00250 mol CHO2−

[CHO2−]init = 0.0500 M

[OH−]eq = 1.7 x 10−6

pH = 8.23

added 30.0 mL NaOH

0.00050 mol NaOH xs

pH = 11.96

added 20.0 mL NaOH

0.00050 mol HCHO2

pH = 4.34

added 15.0 mL NaOH

0.00100 mol HCHO2

pH = 3.92

added 12.5 mL NaOH

0.00125 mol HCHO2

pH = 3.74 = pKa

half-neutralization

Adding NaOH to HCHO2

added 40.0 mL NaOH

0.00150 mol NaOH xs

pH = 12.36

added 50.0 mL NaOH

0.00250 mol NaOH xs

pH = 12.52


4/19/2016

17


Titrating Weak Acid with a Strong Base

• The initial pH is that of the weak acid solution

calculate like a weak acid equilibrium problem

e.g., 15.5 and 15.6

• Before the equivalence point, the solution becomes a buffer

calculate mol HAinit and mol A−init using reaction

stoichiometry

calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−

init

• Half-neutralization pH = pKa


Titrating Weak Acid with a Strong Base

• At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established

mol A− = original mole HA

calculate the volume of added base as you did in Example 4.8

[A−]init = mol A−/total liters

calculate like a weak base equilibrium problem

e.g., 15.14

• Beyond equivalence point, the OH is in excess

[OH−] = mol MOH xs/total liters

[H3O+][OH−]=1 x 10−14



Example 16.7a: A 40.0 mL sample of 0.100 M HNO2

is titrated with 0.200 M KOH. Calculate the volume

of KOH at the equivalence point.

99

write an equation for

the reaction for B

with HA

use stoichiometry to

determine the volume

of added B

HNO2 + KOH NO2 + H2O


Example 16.7b: A 40.0 mL sample of 0.100 M HNO2

is titrated with 0.200 M KOH. Calculate the pH after

adding 5.00 mL KOH.

100

write an

equation for the

reaction for B

with HA

determine the

moles of HAbefore

& moles of

added B

make a

stoichiometry

table and

determine the

moles of HA in

excess and

moles A made


HNO2 NO2− OH−

mols before 0.00400 0 ≈ 0

mols added – – 0.00100

mols after ≈ 00.00300 0.00100




is titrated with 0.200 M KOH. Calculate the pH after

adding 5.00 mL KOH.

101

write an

equation for the

reaction of HA

with H2O

determine Ka

and pKa for HA

use the

Henderson-

Hasselbalch

equation to

determine the

pH

HNO2 + H2O NO2 + H3O

+

HNO2 NO2− OH−



mols after 0.00300 0.00100 ≈ 0

Table 15.5 Ka = 4.6 x 10−4



is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

102

write an

equation for the

reaction for B

with HA

determine the

moles of HAbefore

& moles of added

B

make a

stoichiometry

table and

determine the

moles of HA in

excess and

moles A made


HNO2 NO2− OH−



mols after ≈ 00.00200 0.00200

at half-equivalence, moles KOH = ½ mole HNO2


4/19/2016

18



is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

103

write an equation

for the reaction of

HA with H2O

determine Ka and

pKa for HA

use the

Henderson-

Hasselbalch

equation to

determine the pH

HNO2 + H2O NO2 + H3O

+

HNO2 NO2− OH−



mols after 0.00200 0.00200 ≈ 0

Table 15.5 Ka = 4.6 x 10-4


Titration Curve of a Weak Base with a

Strong Acid



Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =

4.75) with 0.10 M HCl. Calculate the initial pH of the

NH3 solution.


Practice – Titration of 25.0 mL of 0.10 M NH3 with

0.10 M HCl. Calculate the initial pH of the NH3(aq)

106

• NH3(aq) + HCl(aq) NH4Cl(aq)

• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

change +x +x −x

equilibrium x x 0.10−x

pKb = 4.75

Kb = 10−4.75 = 1.8 x 10−5



Practice – Titration of 25.0 mL of 0.10 M NH3 with

0.10 M HCl. Calculate the initial pH of the NH3(aq)

107


• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

change +x +x −x

equilibrium x x 0.10−x

pKb = 4.75

Kb = 10−4.75 = 1.8 x 10−5



4.75) with 0.10 M HCl. Calculate the pH of the solution

after adding 5.0 mL of HCl.


4/19/2016

19



4.75) with 0.10 M HCl. Calculate the pH of the

solution after adding 5.0 mL of HCl.

109


• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl


mols change −5.0E-4 −5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

NH4+

(aq) + H2O(l) NH4+

(aq) + H2O(l) pKb = 4.75

pKa = 14.00 − 4.75 = 9.25



4.75) with 0.10 M HCl. Calculate the pH of the solution

after adding 5.0 mL of HCl.



• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl


mols change −5.0E-4 −5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0





solution at equivalence.





112



• At equivalence mol NH3 = mol HCl = 2.50 x 10−3

added 25.0 mL HClNH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0






113

NH3(aq) + HCl(aq) NH4Cl(aq)

at equivalence [NH4Cl] = 0.050 M

[NH3] [NH4+] [H3O

+]

initial 0 0.050 ≈ 0

change +x −x +x

equilibrium x 0.050−x x

NH4+

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

(aq)






4/19/2016

20





115



• After equivalence: after adding 30.0 mL HCl

NH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

when you mix a strong acid, HCl,

with a weak acid, NH4+, you only

need to consider the strong acidTro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Titration of a Polyprotic Acid

• If Ka1 >> Ka2, there will be two equivalence

points in the titration

the closer the Ka’s are to each other, the less

distinguishable the equivalence points are

titration of 25.0 mL of 0.100 M

H2SO3 with 0.100 M NaOH



Monitoring pH During a Titration

• The general method for monitoring the pH during

the course of a titration is to measure the

conductivity of the solution due to the [H3O+]

using a probe that specifically measures just H3O+

• The endpoint of the titration is reached at the

equivalence point in the titration – at the inflection

point of the titration curve

• If you just need to know the amount of titrant

added to reach the endpoint, we often monitor the

titration with an indicator


Monitoring pH During a Titration



Indicators• Many dyes change color depending on the pH of the

solution

• These dyes are weak acids, establishing an

equilibrium with the H2O and H3O+ in the solution

HInd(aq) + H2O(l) Ind(aq) + H3O

+(aq)

• The color of the solution depends on the relative

concentrations of Ind:HInd

when Ind:HInd ≈ 1, the color will be mix of the colors of

Ind and HInd

when Ind:HInd > 10, the color will be mix of the colors

of Ind

when Ind:HInd < 0.1, the color will be mix of the colors

of HInd119Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Phenolphthalein


4/19/2016

21


Methyl Red

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

H

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

NaOOC

H3O+ OH-


Monitoring a Titration with an

Indicator

• For most titrations, the titration curve shows a

very large change in pH for very small

additions of titrant near the equivalence point

• An indicator can therefore be used to

determine the endpoint of the titration if it

changes color within the same range as the

rapid change in pH

pKa of HInd ≈ pH at equivalence point



Acid-Base Indicators


Solubility Equilibria

• All ionic compounds dissolve in water to some

degree

however, many compounds have such low solubility

in water that we classify them as insoluble

• We can apply the concepts of equilibrium to

salts dissolving, and use the equilibrium

constant for the process to measure relative

solubilities in water



Solubility Product

• The equilibrium constant for the dissociation of a

solid salt into its aqueous ions is called the

solubility product, Ksp

• For an ionic solid MnXm, the dissociation reaction is:

MnXm(s) nMm+(aq) + mXn−(aq)

• The solubility product would be

Ksp = [Mm+]n[Xn−]m

• For example, the dissociation reaction for PbCl2 is

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

• And its equilibrium constant is

Ksp = [Pb2+][Cl−]2

125Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.126Tro: Chemistry: A Molecular Approach, 2/e

4/19/2016

22


Molar Solubility• Solubility is the amount of solute that will

dissolve in a given amount of solution

at a particular temperature

• The molar solubility is the number of moles of

solute that will dissolve in a liter of solution

the molarity of the dissolved solute in a saturated

solution

for the general reaction MnXm(s) nMm+(aq) + mXn−(aq)


Example 16.8: Calculate the molar solubility of

PbCl2 in pure water at 25 C

128

write the

dissociation

reaction and Ksp

expression

create an ICE

table defining

the change in

terms of the

solubility of the

solid

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S


Ksp = [Pb2+][Cl−]2



Example 16.8: Calculate the molar solubility of

PbCl2 in pure water at 25 C

129

substitute into

the Ksp

expression

find the value of

Ksp from Table

16.2, plug into

the equation,

and solve for S

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

Ksp = [Pb2+][Cl−]2

Ksp = (S)(2S)2


Practice – Determine the Ksp of PbBr2 if its molar

solubility in water at 25 C is 1.05 x 10−2 M





131

write the

dissociation

reaction and Ksp

expression

create an ICE

table defining

the change in

terms of the

solubility of the

solid

[Pb2+] [Br−]

initial 0 0

change +(1.05 x 10−2) +2(1.05 x 10−2)

equilibrium (1.05 x 10−2) (2.10 x 10−2)

PbBr2(s) Pb2+(aq) + 2 Br−(aq)

Ksp = [Pb2+][Br−]2




132

substitute into

the Ksp

expression

plug into the

equation and

solve

Ksp = [Pb2+][Br−]2

Ksp = (1.05 x 10−2)(2.10 x 10−2)2

[Pb2+] [Br−]

initial 0 0

change +(1.05 x 10−2) +2(1.05 x 10−2)

equilibrium (1.05 x 10−2) (2.10 x 10−2)


4/19/2016

23


Ksp and Relative Solubility

• Molar solubility is related to Ksp

• But you cannot always compare solubilities of

compounds by comparing their Ksps

• To compare Ksps, the compounds must have

the same dissociation stoichiometry


The Effect of Common Ion on

Solubility• Addition of a soluble salt that contains one of

the ions of the “insoluble” salt, decreases the

solubility of the “insoluble” salt

• For example, addition of NaCl to the solubility

equilibrium of solid PbCl2 decreases the

solubility of PbCl2


addition of Cl− shifts the

equilibrium to the left



Example 16.10: Calculate the molar solubility

of CaF2 in 0.100 M NaF at 25 C

135

write the

dissociation

reaction and Ksp

expression

create an ICE

table defining

the change in

terms of the

solubility of the

solid

[Ca2+] [F−]

initial 0 0.100

change +S +2S

equilibrium S 0.100 + 2S

CaF2(s) Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2


Example 16.10: Calculate the molar solubility

of CaF2 in 0.100 M NaF at 25 C

136

substitute into

the Ksp

expression,

assume S is

small

find the value of

Ksp from Table

16.2, plug into

the equation,

and solve for S

[Ca2+] [F−]

initial 0 0.100

change +S +2S

equilibrium S 0.100 + 2S

Ksp = [Ca2+][F−]2

Ksp = (S)(0.100 + 2S)2

Ksp = (S)(0.100)2



Practice – Determine the concentration of Ag+

ions in seawater that has a [Cl−] of 0.55 M

Ksp of AgCl = 1.77 x 10−10




write the

dissociation

reaction and Ksp

expression

create an ICE

table defining

the change in

terms of the

solubility of the

solid

[Ag+] [Cl−]

initial 0 0.55

change +S +S

equilibrium S 0.55 + S

AgCl(s) Ag+(aq) + Cl−(aq)

Ksp = [Ag+][Cl−]


4/19/2016

24




139

substitute into

the Ksp

expression,

assume S is

small

find the value of

Ksp from Table

16.2, plug into

the equation,

and solve for S

[Ag+] [Cl−]

Initial 0 0.55

Change +S +S

Equilibrium S 0.55 + S

Ksp = [Ag+][Cl−]

Ksp = (S)(0.55 + S)

Ksp = (S)(0.55)


The Effect of pH on Solubility

• For insoluble ionic hydroxides, the higher the pH,

the lower the solubility of the ionic hydroxide

and the lower the pH, the higher the solubility

higher pH = increased [OH−]

M(OH)n(s) Mn+(aq) + nOH−(aq)

• For insoluble ionic compounds that contain anions

of weak acids, the lower the pH, the higher the

solubility

M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq)

H3O+(aq) + CO3

2− (aq) HCO3− (aq) + H2O(l)



Precipitation

• Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound

• If we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occurQ = Ksp, the solution is saturated, no precipitation

Q < Ksp, the solution is unsaturated, no precipitation

Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate

• Some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions


precipitation

occurs if Q > Ksp

a supersaturated solution will

precipitate if a seed crystal is

added



Selective Precipitation

• A solution containing several different cations

can often be separated by addition of a reagent

that will form an insoluble salt with one of the

ions, but not the others

• A successful reagent can precipitate with more

than one of the cations, as long as their Ksp

values are significantly different


Example 16.12: Will a precipitate form when we mix

Pb(NO3)2(aq) with NaBr(aq) if the concentrations after

mixing are 0.0150 M and 0.0350 M respectively?

write the equation for

the reaction

determine the ion

concentrations of the

original salts

determine the Ksp for

any “insoluble”

product

write the dissociation

reaction for the

insoluble product

calculate Q, using the

ion concentrations

compare Q to Ksp. If

Q > Ksp, precipitation

Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)

Ksp of PbBr2 = 4.67 x 10–6

PbBr2(s) Pb2+(aq) + 2 Br−

(aq)

Pb(NO3)2 = 0.0150 M

Pb2+ = 0.0150 M,

NO3− = 2(0.0150 M)

NaBr = 0.0350 M

Na+ = 0.0350 M,

Br− = 0.0350 M

Q < Ksp, so no precipitation144Tro: Chemistry: A Molecular Approach, 2/e

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25


Practice – Will a precipitate form when we mix

Ca(NO3)2(aq) with NaOH(aq) if the concentrations after

mixing are both 0.0175 M?

Ksp of Ca(OH)2 = 4.68 x 10−6


Practice – Will a precipitate form when we mix

Ca(NO3)2(aq) with NaOH(aq) if the concentrations after

mixing are both 0.0175 M?

write the equation for

the reaction

determine the ion

concentrations of the

original salts

determine the Ksp for

any “insoluble”

product

write the dissociation

reaction for the

insoluble product

calculate Q, using the

ion concentrations

compare Q to Ksp. If

Q > Ksp, precipitation

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)

Ksp of Ca(OH)2 = 4.68 x 10–6

Ca(OH)2(s) Ca2+(aq) + 2 OH−

(aq)

Ca(NO3)2 = 0.0175 M

Ca2+ = 0.0175 M,

NO3− = 2(0.0175 M)

NaOH = 0.0175 M

Na+ = 0.0175 M,

OH− = 0.0175 M

Q > Ksp, so precipitation


Copyright 2011 Pearson Education, Inc.147

Example 16.13: What is the minimum [OH−]

necessary to just begin to precipitate Mg2+ (with

[0.059]) from seawater?

precipitating may just occur when Q = Ksp

Mg(OH)2(s) Mg2+(aq) + 2 OH−

(aq)


Practice – What is the minimum concentration of

Ca(NO3)2(aq) that will precipitate Ca(OH)2 from

0.0175 M NaOH(aq)?

Ksp of Ca(OH)2 = 4.68 x 10−6



Practice – What is the minimum concentration of

Ca(NO3)2(aq) that will precipitate Ca(OH)2 from

0.0175 M NaOH(aq)?


[Ca(NO3)2] = [Ca2+] = 0.0153 M

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)


(aq)

Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.150

Example 16.14: What is the [Mg2+] when Ca2+

(with [0.011]) just begins to precipitate from

seawater?



(aq)


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Example 16.14: What is the [Mg2+] when Ca2+

(with [0.011]) just begins to precipitate from

seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10−6 M

precipitating Ca2+ begins when [OH−] = 2.06 x 10−2 M

when Ca2+ just

begins to

precipitate out, the

[Mg2+] has dropped

from 0.059 M to

4.8 x 10−10 M

Mg(OH)2(s) Mg2+(aq) + 2 OH−

(aq)


Practice – A solution is made by mixing Pb(NO3)2(aq) with

AgNO3(aq) so both compounds have a concentration of

0.0010 M. NaCl(s) is added to precipitate out both AgCl(s)

and PbCl2(aq). What is the [Ag+] concentration when the

Pb2+ just begins to precipitate?



Practice – What is the [Ag+] concentration when

the Pb2+(0.0010 M) just begins to precipitate?


AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)




the Pb2+(0.0010 M) just begins to precipitate?


Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)





the Pb2+(0.0010 M) just begins to precipitate

precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M

precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M

when Pb2+ just

begins to

precipitate out, the

[Ag+] has dropped

from 0.0010 M to

1.6 x 10−9 M



Qualitative Analysis

• An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme

wet chemistry

• A sample containing several ions is subjected to the addition of several precipitating agents

• Addition of each reagent causes one of the ions present to precipitate out


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Qualitative Analysis

Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.158Tro: Chemistry: A Molecular Approach, 2/e


Group 1

• Group one cations are Ag+, Pb2+ and Hg22+

• All these cations form compounds with Cl− that

are insoluble in water

as long as the concentration is large enough

PbCl2 may be borderline

molar solubility of PbCl2 = 1.43 x 10−2 M

• Precipitated by the addition of HCl


Group 2

• Group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,

Pb2+, Sb3+, and Hg2+

• All these cations form compounds with HS− and

S2− that are insoluble in water at low pH

• Precipitated by the addition of H2S in HCl



Group 3

• Group three cations are Fe2+, Co2+, Zn2+, Mn2+,

Ni2+ precipitated as sulfides; as well as Cr3+,

Fe3+, and Al3+ precipitated as hydroxides

• All these cations form compounds with S2− that

are insoluble in water at high pH

• Precipitated by the addition of H2S in NaOH


Group 4

• Group four cations are Mg2+, Ca2+, Ba2+

• All these cations form compounds with PO43−

that are insoluble in water at high pH

• Precipitated by the addition of (NH4)2HPO4


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Group 5

• Group five cations are Na+, K+, NH4+

• All these cations form compounds that are

soluble in water – they do not precipitate

• Identified by the color of their flame


Complex Ion Formation

• Transition metals tend to be good Lewis acids

• They often bond to one or more H2O molecules to form a hydrated ionH2O is the Lewis base, donating electron pairs to form

coordinate covalent bonds

Ag+(aq) + 2 H2O(l) Ag(H2O)2+(aq)

• Ions that form by combining a cation with several anions or neutral molecules are called complex ionse.g., Ag(H2O)2

+

• The attached ions or molecules are called ligandse.g., H2O



Complex Ion Equilibria

• If a ligand is added to a solution that forms a

stronger bond than the current ligand, it will

replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) + 2 H2O(l)

generally H2O is not included, because its complex

ion is always present in aqueous solution

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)


Formation Constant

• The reaction between an ion and ligands to form

a complex ion is called a complex ion

formation reaction


+(aq)

• The equilibrium constant for the formation

reaction is called the formation constant, Kf



Formation Constants


Example 16.15: 200.0 mL of 1.5 x 10−3 M

Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3.

What is the [Cu2+] at equilibrium?

Write the

formation

reaction and Kf

expression.

Look up Kf value

determine the

concentration of

ions in the

diluted solutions

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)


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Example 16.15: 200.0 mL of 1.5 x 10−3 M

Cu(NO3)2 is mixed with 250.0 mL of 0.20 M

NH3. What is the [Cu2+] at equilibrium?

Create an ICE

table. Because

Kf is large,

assume all the

Cu2+ is

converted into

complex ion,

then the system

returns to

equilibrium.

[Cu2+] [NH3] [Cu(NH3)22+]

initial 6.7E-4 0.11 0

change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4

equilibrium x 0.11 6.7E-4

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)


Example 16.15: 200.0 mL of 1.5 x 10-3 M

Cu(NO3)2 is mixed with 250.0 mL of 0.20 M

NH3. What is the [Cu2+] at equilibrium?

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)substitute in

and solve

for x

confirm the

“x is small”

approxima-

tion

[Cu2+] [NH3] [Cu(NH3)22+]

initial 6.7E-4 0.11 0

change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4

equilibrium x 0.11 6.7E-4

2.7 x 10−13 << 6.7 x 10−4, so the approximation is valid



Practice – What is [HgI42−] when 125 mL of 0.0010 M

KI is reacted with 75 mL of 0.0010 M HgCl2?

4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)





Write the

formation

reaction and Kf

expression.

Look up Kf value

determine the

concentration of

ions in the

diluted solutions

Hg2+(aq) + 4 I−(aq) HgI42−(aq)






Create an ICE

table. Because

Kf is large,

assume all the

lim. rgt. is

converted into

complex ion,

then the system

returns to

equilibrium.

[Hg2+] [I−] [HgI42−]

initial 3.75E-4 6.25E-4 0

change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)

equilibrium 2.19E-4 x 1.56E-4


I− is the limiting reagent





substitute in

and solve for

x

confirm the

“x is small”

approximation

2 x 10−8 << 1.6 x 10−4, so the approximation is valid


[HgI42−] = 1.6 x 10−4

[Hg2+] [I−] [HgI42−]

initial 3.75E-4 6.25E-4 0

change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)

equilibrium 2.19E-4 x 1.56E-4


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The Effect of Complex Ion Formation

on Solubility

• The solubility of an ionic compound that

contains a metal cation that forms a complex

ion increases in the presence of aqueous

ligands

AgCl(s) Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10−10


+(aq) Kf = 1.7 x 107

• Adding NH3 to a solution in equilibrium with

AgCl(s) increases the solubility of Ag+

Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.176Tro: Chemistry: A Molecular Approach, 2/e


Solubility of Amphoteric

Metal Hydroxides• Many metal hydroxides are insoluble

• All metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH−

• Some metal hydroxides also become more soluble in basic solutionacting as a Lewis base forming a complex ion

• Substances that behave as both an acid and base are said to be amphoteric

• Some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+


Al3+

• Al3+ is hydrated in water to form an acidic

solution

Al(H2O)63+

(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O

+(aq)

• Addition of OH− drives the equilibrium to the right

and continues to remove H from the molecules

Al(H2O)5(OH)2+(aq) + OH−

(aq) Al(H2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O (l)


Copyright 2011 Pearson Education, Inc.179Tro: Chemistry: A Molecular Approach, 2/e

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