15 Equilibrium part 3.pdf

CHAPTER 15:

APPLICATIONS OF

pH in Solutions of

Weak

Acids

K

for CH

COOH is 1.8 billion times larger than K

.

Thus, H

O

from H

2

O is so small, it can be ignored.

Calculate the pH of a

0.10 M CH

COOH

solution.

CH

COOH(aq) + H

O(

l

) H

O

(aq) + CH

COO

(aq)

Source of hydronium ions:

1. from CH

COOH (0.10 M)

2. from autoionization of H

O

K

= 1.8 x 10

K

= 1.0 x 10

[H

O

]

≠

[CH

pH in Solutions of

Weak

Acids

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.10 ~0 0

x

x

–x

x

0.10 –

x

x

CH

COOH

+ H

O

H

O

+ CH

COO

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

( )

x

0.10

-

x

pH in Solutions of

Weak

Acids

x

0.10

-

x

(

)

»

x

2

0.10

(

)

=

1.8

´

10

x

=

1.8

´

10

x

= [H

O

] = 1.3 x 10

M

pH = –log(1.3 x 10

) = 2.89

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

( )

x

0.10

-

x

The Common Ion Effect

Calculate the pH of a

0.10

M

CH

COOH

solution

when

0.050

M

sodium acetate (CH

COONa)

is added to the solution.

Common ion

CH

COOH(aq) + H

O(

l

) H

O

(aq) + CH

3

COO

(aq)

CH

COONa(aq) Na

(aq) + CH

The Common Ion Effect

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.10 ~0 0.050

x

x

–x

x

0.10 –

x

x

CH

COOH

+ H

O

H

O

+ CH

COO

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

(

0.050

+

x

)

0.10

-

x

The Common Ion Effect

x

( )

(

0.050

+

x

)

0.10

-

x

(

)

»

x

( )

(

0.050

)

0.10

(

)

=

1.8

´

10

x

=

3.6

´

10

x

= [H

O

] = 3.6 x 10

M

pH = –log(3.6 x 10

) = 4.44

K

=

H

O

+

éë

ùû

éë

CH

COO

ùû

CH

COOH

[

]

=

x

( )

(

0.050

+

x

)

0.10

-

x

Percent Dissociation

Before

the addition of the

0.050 M sodium acetate:

%

dis

=

1.3

´

10

-3

(

)

0.10

(

)

´

100%

=

1.3%

[CH

COO

] = 1.3 x 10

M

[CH

COOH] = 0.10 M

Percent dissociation

=

amount dissociated (mol/L)

initial concentration (mol/L)

´

100%

After

the addition of the

0.050 M sodium acetate:

%

dis

=

3.6

´

10

-5

(

)

0.10

(

)

´

100%

=

0.036%

[CH

COO

] = 3.6 x 10

M

The Common Ion Effect

When a compound containing an ion in common with a

dissolved substance is added to a solution at

equilibrium, the equilibrium shifts to the left. This

phenomenon is known as the

common ion effect

.

CH

COOH(aq) + H

O(

l

) H

O

(aq) + CH

COO

(aq)

Addition of CH

COO

The Common Ion Effect Example Problems

Which of the following would cause a decrease

in the percent ionization of nitrous acid

when added to a solution of nitrous acid at equilibrium?

HNO

(aq) + H

O(

l

) H

O

(aq) + NO

2–

(aq)

a) NaNO

b) H

O

c) Ca(NO

)

d) HNO

e) NaNO

A

buffer

is a solution whose

pH changes very little

when acid or base is added.

Most buffers are solutions composed of roughly

equal amounts of:

•

A

weak acid

•

The salt of its

conjugate base

The buffer resists change in pH because

•

Added base, OH

, reacts with the

weak acid

A

buffer

is a solution whose

pH changes very little

when acid or base is added.

When an acid or base is

added to water, the pH

changes drastically

When an acid or base is

added to a buffer

solution, the pH does

not change very much;

How Buffers Work

Most buffers consist of a pair of compounds, one with the ability to react

with OH

(

weak acid

) and the other with the ability to react with H

(its

conjugate base

OR

the

salt of its conjugate base

).

Thus, the components of a buffer solution are

acid-

base

conjugate pairs

which typically have equal concentrations of the weak acid and its salt

(can also be a weak base and a salt of its conjugate acid).

An example is a mixture of

acetic acid

and its salt,

sodium acetate

.

C₂H₃O₂–_(aq) acetate ion (comes from sodium acetate)

+ H+_{(aq) HC}

2H3O2(aq)

acetic acid HC2H3O2(aq)

acetic acid

+ OH–_{(aq) C}

2H3O2–(aq) + H2O(l)

How Buffers Work

Buffer action occurs because the weak acid in a buffer neutralizes base, and the conjugate base in the buffer neutralizes acid. In this way, the pH of the solution is

Does each of the following combinations produce a

buffer solution or not? Explain.

nitric acid sodium nitrate a) HNO₃ and NaNO₃

strong acid salt of its conjugate base

No

hydrofluoric acid sodium fluoride b) HF and NaF

weak acid salt of its conjugate base

Does each of the following combinations produce a

buffer solution or not? Explain.

potassium chloride potassium cyanide c) KCl and KCN

salt salt

No

carbonic acid sodium bicarbonate d) H₂CO₃ and NaHCO₃

weak acid salt of its conjugate base

Does each of the following combinations

produce a buffer solution or not? Explain.

1.

HCl and KCl

2.

H

CO

and NaHCO

3.

H

PO

and NaCl

Focus on the Human Body

Buffers in the Blood

•

Normal blood pH is between

7.35

and

7.45

.

•

The principle buffer in the blood is

carbonic acid

/

bicarbonate

(H

CO

/HCO

).

CO

(

g

) + H

O(

l

)

H

CO

(

aq

)

H₂O

H

O

(

aq

) + HCO

3−

(

aq

)

•

CO

is constantly produced by metabolic

processes in the body.

Focus on the Human Body

Buffers in the Blood

Focus on the Human Body

Buffers in the Blood

Calculating the pH of a Buffer

Calculate the pH of a buffer composed of

0.10

M

CH

COOH/0.10

M

sodium acetate (CH

COONa)

Common ion

CH

COOH(aq) + H

O(

l

) H

O

(aq) + CH

3

COO

(aq)

CH

COONa(aq) Na

(aq) + CH

Calculating the pH of a Buffer

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.10 ~0 0.10

x

x

–x

x

0.10 –

x

x

CH

COOH

+ H

O

H

O

+ CH

COO

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

(

0.10

+

x

)

0.10

-

x

Calculating the pH of a Buffer

x

( )

(

0.10

+

x

)

0.10

-

x

(

)

»

x

( )

(

0.10

)

0.10

(

)

=

1.8

´

10

x

=

K

=

1.8

´

10

x

= [H

O

] = 1.8 x 10

M

pH = –log(1.8 x 10

) = 4.74

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

(

0.10

+

x

)

0.10

-

x

Calculating the pH of a Buffer

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

H

O

+

éë

ùû=

K

[

CH

COOH

]

CH

COO

-éë

ùû

æ

è

ç

ç

ö

ø

÷

÷ =

K

Acid

[

]

Base

Addition of OH

–

to a Buffer

What happens when you add 0.010 mol of solid NaOH to

1.00 L 0.10 M acetic acid–0.10 M sodium acetate solution?

We must take into account the neutralization of the base

before

calculating [H

O

].

Before reaction (mol)

Change (mol)

After reaction (mol)

0.10 0.10

0.09 0.11

CH

COOH

+ OH

H

2

O

+ CH

COO

0.010

–0.010 –0.010 +0.010

Addition of OH

–

to a Buffer

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.09 ~0 0.11

x

x

–x

x

0.09 –

x

x

CH

COOH

+ H

O

H

O

+ CH

COO

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

(

0.11

+

x

)

0.09

-

x

pH of a Buffer After Addition of Strong Base

pH = –log(1.5 x 10

) = 4.82

H

O

+

éë

ùû=

K

[

CH

COOH

]

CH

COO

-éë

ùû

æ

è

ç

ç

ö

ø

÷

÷ =

K

Acid

[

]

Base

[

]

H

O

+

éë

ùû=

K

[

CH

COOH

]

CH

COO

-éë

ùû

æ

è

ç

ç

ö

ø

÷

÷ =

K

Acid

[

]

Base

[

]

=

(

1.8

´

10

)

0.09

0.11

æ

è

Addition of H

3

O

+

to a Buffer

What happens when you add 0.010 mol of HCl to 1.00 L

0.10 M acetic acid–0.10 M sodium acetate solution?

We must take into account the neutralization of the acid

before

calculating [H

O

].

Before reaction (mol)

Change (mol)

After reaction (mol)

0.10 0.10

0.09 0.11

CH

COO

+ H

3

O

H

O

+ CH

COOH

0.010

–0.010 –0.010 +0.010

Addition of H

3

O

+

to a Buffer

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.11 ~0 0.09

x

x

–x

x

0.11 –

x

x

CH

COOH

+ H

O

H

O

+ CH

COO

K

=

H

O

+

éë

ùû

CH

COO

-éë

ùû

CH

COOH

[

]

=

x

( )

(

0.11

+

x

)

0.09

-

x

pH of a Buffer After Addition of a Strong Acid

pH = –log(2.2 x 10

) = 4.66

H

O

+

éë

ùû=

K

[

CH

COOH

]

CH

COO

-éë

ùû

æ

è

ç

ç

ö

ø

÷

÷ =

K

Acid

[

]

Base

[

]

H

O

+

éë

ùû=

K

[

CH

COOH

]

CH

COO

-éë

ùû

æ

è

ç

ç

ö

ø

÷

÷ =

K

Acid

[

]

Base

[

]

=

(

1.8

´

10

)

0.09

0.11

æ

è

pH Changes

pH = 4.66

pH = 4.82

Buffer solution

pH = 4.74

Addition

Henderson–Hasselbach Equation

H

O

éë

ùû=

K

[

Acid

]

Base

[

]

-

log

éë

H

O

ùû=-

log

K

[

Acid

]

Base

[

]

æ

è

çç

ö

ø

÷÷

pH

= -

log

K

-

log

[

Acid

]

Base

[

]

pH

=

pK

+

log

[

Base

]

Acid

Using the Henderson–Hasselbach Equation to

Determine the pH of a Buffer

Calculate the pH of a buffer solution that is 0.45 M in

NH

Cl and 0.15 M in NH

. The K

for NH

is 1.8 x 10

.

K

=

K

K

=

1.0

´

10

1.8

´

10

=

5.6

´

10

-10

pK

= -

log

K

= -

log 5.6

(

´

10

)

=

9.25

pH

=

pK

+

log

[

Base

]

Acid

[

]

pH

=

9.25

+

log

0.15

0.45

æ

è

Using the Henderson–Hasselbach Equation to

Determine the pH of a Buffer

Calculate the pH of a buffer solution that is 0.45 M in

NH

Cl and 0.15 M in NH

. The K

for NH

is 1.8 x 10

.

pH

=

pK

+

log

[

Base

]

Acid

[

]

pK

+

pK

=

14

pH

=

14

-

4.74

+

log

0.15

0.45

æ

è

ç

ö

ø

÷ =

8.78

pH

=

14

-

pK

+

log

[

Base

]

Acid

Buffer Capacity

The

buffer capacity

is the amount of acid or base that a

solution can absorb without a significant change in pH.

Buffer capacity depends on the number of moles of weak

acid and conjugate base are present.

For equal volumes of solution,

the more concentrated the solution,

the greater the buffer capacity.

For solutions having the same concentration,

the greater the volume, the greater the buffer capacity.

Buffer Solutions

The most effective buffering occurs when

pH

=

pK

+

log

[

Base

]

Acid

[

]

=

pK

+

log 1

( )

Base

[

]

Acid

[

]

=

1

pH

=

pK

Preparing a Buffer Solution with a Specific pH

You need a buffer solution that has pH = 7.00. Which of

the following buffer systems should you choose?

H

PO

and H

PO

H

PO

and HPO

HPO

and PO

K

= 7.5 x 10

K

= 6.2 x 10

K

= 4.8 x 10

pK

= 2.12

pK

= 7.21

Acid-Base Titrations are Used to Determine the

Concentration

of An Acid or Base in Solution

•

If we want to know the concentration

of an acid solution, a base of known

concentration is added slowly until the

acid is

neutralized

.

•

When the acid is neutralized:

# of moles of acid = # of moles of base

•

This is called the

end point

of the

End Point of Titration

•

An

indicator

is added to the solution in the flask. The indicator causes the solution to

change color when the solution is neutralized.

•

At the

end point

,

•

The indicator has a permanent color

•

The volume of the solution used to reach the end point is measured

Titration Problems are 3-Step

Solution Stoichiometry

Problems

You need to be able to calculate the molarity or volume

of an acid or base from titration information.

grams of known moles of known moles of unknown grams of unknown

First, write down what you know and what you don

’

t know.

The

known

may be a

reactant

or a

product

.

The

unknown

may be a

reactant

or a

product

.

liters of known liters of unknown Molarity of

solution Molarity of solution These 2 can be exchanged,

Reaction of Acids with Hydroxide Bases

Neutralization reaction

: An

acid

-

base

reaction that

produces a salt and water as products.

H

A(

aq

) + M

OH

(

aq

)

acid

base

OH

(

l

) + MA(

aq

)

H

water

salt

•

The acid H

A

donates a proton (

H

) to the

OH

base

to form H

O.

•

The anion A

from the acid combines with the

Acid + Base Salt + Water

Neutralization Reaction (Double Replacement)

of Acids and Bases

H

Cl(aq) + Na

OH

(aq) NaCl(aq) +

H

O

(

l

)

H

(aq) +

OH

(aq)

H

2

O

(

l

)

Net ionic equation (what the real action is):

from the

A

net ionic equation

contains only the species

involved in a reaction.

H

Cl(

aq

) + Na

OH

(

aq

)

H

—

OH

(

l

) + NaCl(

aq

)

Written as individual ions:

H

(

aq

) +

Cl

(

aq

) + Na

(

aq

) +

OH

(

aq

)

H

—

OH

(

l

) + Na

(

aq

) + Cl

(

aq

)

Omit the spectator ions, Na

and Cl

.

H

(

aq

) +

OH

(

aq

)

H

—

OH

(

l

)

How to

Write and Balance Neutralization

Reactions

Step 1 Write the base and acid formulas:

Mg(OH)

+ HNO

Step 2

Balance the

OH

and

H

:

Mg(

OH

)

+

2

H

NO

Step 3

Balance with

H

O

:

Mg(OH)

+ 2 HNO



salt +

2

H

O

Step 4

Write the

salt

from the remaining ions:

Mg

(OH)

+

2

H

NO



Mg(NO

)

+ 2 H

O

Balancing Neutralization Reactions

H

SO

(aq) + KOH(aq)

??

H

(aq) + OH

(aq) H

2

O(

l

)

Net ionic equation:

H2SO4 is a diprotic acid; it can donate 2 protons.

2

H

(aq) +

OH

(aq)

H

2

O(

l

)

2

2

salt

Practice Balancing Neutralization Reactions

H

PO

+ LiOH



HCl + Ca(OH)



Ba(OH)

+ H

PO



How to Determine the

Molarity of an Acid Solution From Titration

What is the molarity of an HCl solution if

22.5 mL

of a

0.100

M

NaOH

solution are needed to

titrate a

25.0 mL

sample of the acid?

First, write down what you know and what you don

’

t

know.

volume of base

(

NaOH

)

22.5 mL

conc. of base

(

NaOH

)

0.100 M

volume of acid

(

HCl

)

25.0 mL

How to Determine the

Molarity of an Acid Solution From Titration

Step [1]

Determine the number of moles of base used to

neutralize the acid.

22.5 mL NaOH x

1000 mL

1 L

x 0.100 mol NaOH

1 L

=

0.00225 mol NaOH

Moles of

base

Volume of

base

M (mol/L)

conversion

How to Determine the

Molarity of an Acid Solution From Titration

Step [2]

Determine the number of moles of acid that react

from the balanced chemical equation.

HCl(

aq

) + NaOH(

aq

)

H

O(

l

) + NaCl(

aq

)

0.00225 mol NaOH x 1 mol HCl

1 mol NaOH

= 0.00225 mol HCl

Moles of

base

mole–mole

conversion

How to Determine the

Molarity of an Acid Solution From Titration

Step [3]

Determine the molarity of the acid from the

number of moles and the known volume.

M = mol

L

=

25.0 mL solution

0.00225 mol HCl

x 1000 mL

1 L

=

0.0900 M HCl

3 sig. fig. Answer

mL—L

conversion

1. Calculate the Volume of Base

Required to Titrate an Acid

A 20.00 mL sample of 0.200 M HClO

is to be titrated to the

equivalence point using 0.120 M NaOH solution.

Determine the # of milliliters of NaOH solution that will be needed.

This is a simple solution stoichiometry problem. But you weren’t given the

balanced equation for the reaction. You must figure it out!!

HClO

is a

monoprotic

acid (can lose only one proton per

molecule).

1. Calculate the Volume of Base

Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

Known: 20.00 mL of 0.200 M HClO4 solution

Unknown: mL of 0.120 M NaOH

liters of known liters of unknown

Molarity of

solution Molarity of solution

HClO

(aq) + NaOH(aq) NaClO

(aq) + H

O(

l

)

1. Calculate the Volume of Base

Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

Step 1: go from liters of known to moles of known by multiplying by the molarity of the solution.

0.200 M

(

)

20.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

mol = M

´

L

HClO

(aq) + NaOH(aq) NaClO

(aq) + H

O(

l

)

1. Calculate the Volume of Base

Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

Step 2: go from moles of known to moles of unknown by using the coefficients of the balanced equation. Multiply moles of known by a fraction equal to the coefficient of the unknown divided by the coefficient of the known.

0.200 M

(

)

20.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

1

1

æ

è

ç

ö

ø

÷

fraction

=

unknown

known

HClO

(aq) + NaOH(aq) NaClO

(aq) + H

O(

l

)

1. Calculate the Volume of Base

Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

0.200 M

(

)

20.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

1

1

æ

è

ç

ö

ø

÷

0.120 M

=

0.0333 L

Step 3: go from moles of unknown to liters of unknown by dividing by the molarity.

HClO

(aq) + NaOH(aq) NaClO

(aq) + H

O(

l

)

1. Calculate the Volume of Base

Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

0.0333 L

(

)

(

1000 mLL

)

=

33

.

3 mL

Step 4: convert from liters of unknown to mL of unknown.

HClO4(aq) + NaOH(aq) NaClO4(aq) + H2O(l)

2. Calculate the Volume of Base

Required to Titrate an Acid

0.120 mol H

PO

is dissolved in enough water to make 500 mL of

solution. A 20.00 mL sample of the acid solution is to be

titrated to the equivalence point using 0.120 M NaOH

solution. Determine the number of milliliters of NaOH

solution that will be needed.

H3PO4 is a triprotic acid (can lose three protons per molecule).

2. Calculate the Volume of Base Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

Known: 20.00 mL sample of 0.120 mol H3PO4 in 500 mL solution

Unknown: mL of 0.120 M NaOH

liters of known liters of unknown

Molarity of

solution Molarity of solution

H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)

2. Calculate the Volume of Base Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

Step 1: first, determine the concentration of the acid solution. Then go from liters of known to moles of known by multiplying by the molarity of the solution.

0.120 mol

500 mL

1000 mLL

æ

è

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ç

ö

ø

÷

÷

æ

è

ç

ç

ç

ç

ç

ç

ö

ø

÷

÷

÷

÷

÷

÷

20.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

2. Calculate the Volume of Base Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

Step 2: go from moles of known to moles of unknown by using the coefficients of the balanced equation. Multiply moles of known by a fraction equal to the coefficient of the unknown divided by the coefficient of the known.

0.120 mol

500 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

æ

è

ç

ç

ç

ç

ç

ç

ö

ø

÷

÷

÷

÷

÷

÷

20.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

3

1

æ

è

ç

ö

ø

÷

H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)

2. Calculate the Volume of Base Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

Step 3: go from moles of unknown to liters of unknown by dividing by the molarity.

0.120 mol

500 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

æ

è

ç

ç

ç

ç

ç

ç

ö

ø

÷

÷

÷

÷

÷

÷

20.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

3

1

æ

è

ç

ö

ø

÷

0

.

120 M

=

0.120 L

H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)

2. Calculate the Volume of Base Required to Titrate an Acid

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

0.120 L

(

)

(

1000 mLL

)

=

120 mL

Step 4: convert from liters of unknown to mL of unknown.

H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)

3. Calculate Molarity of an Acid

from a Titration With a Base

The following acid solution was titrated to the equivalence point with the base listed. Use the titration data to calculate the molarity of the acid solution.

5.00 mL of dilute H₂SO₄ required 29.88 mL of 1.17 M NaOH solution

H2SO4 is a diprotic acid (can lose two protons per molecule).

3. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

Known: 29.88 mL of 1.17 M NaOH

Unknown: M of 5.00 mL H₂SO₄ solution

liters of known M of unknown

Molarity of

solution Liters of solution

H₂SO₄(aq) + 2NaOHknown (aq) Na₂SO₄(aq) + 2H₂O(l)

3. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

liters of known

Molarity of solution

Step 1: go from liters of known to moles of known by multiplying by the molarity of the solution.

1.17 M

(

)

29.88 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

M of unknown

Liters of solution

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

3. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

liters of known

Molarity of solution

Step 2: go from moles of known to moles of unknown by using the coefficients of the balanced equation. Multiply moles of known by a fraction equal to the coefficient of the unknown divided by the coefficient of the known.

1.17 M

(

)

29.88 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

2

1

æ

è

ç

ö

ø

÷

M of unknown

Liters of solution

H₂SO₄(aq) + 2NaOHknown (aq) Na₂SO₄(aq) + 2H₂O(l)

3. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

liters of known

Molarity of solution

1.17 M

(

)

29.88 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

1

2

æ

è

ç

ö

ø

÷

5.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

=

3

.

50 M

Step 3: go from moles of unknown to molarity of unknown by dividing by the liters.

M of unknown

Liters of solution

H₂SO₄(aq) + 2NaOHknown (aq) Na₂SO₄(aq) + 2H₂O(l)

4. Calculate Molarity of an Acid

from a Titration With a Base

The following acid solution was titrated to the equivalence point with the base listed. Use the titration data to calculate the molarity of the acid solution.

10.00 mL of vinegar (acetic acid) required 35.62 mL of 0.250 M KOH solution

Acetic acid is a monoprotic acid.

4. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

Known: 35.62 mL of 0.250 M KOH

Unknown: M of 10.00 mL acetic acid solution

liters of known M of unknown

Molarity of

solution Liters of solution

4. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

liters of known

Molarity of solution

Step 1: go from liters of known to moles of known by multiplying by the molarity of the solution.

0.250 M

(

)

35.62 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

M of unknown

Liters of solution

4. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

liters of known

Molarity of solution

Step 2: go from moles of known to moles of unknown by using the coefficients of the balanced equation. Multiply moles of known by a fraction equal to the coefficient of the unknown divided by the coefficient of the known.

0.250 M

(

)

35.62 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

1

1

æ

è

ç

ö

ø

÷

M of unknown

Liters of solution

HC2H3O2(aq) + KOH(aq) KC2H3O2(aq) + H2O(l)

4. Calculate Molarity of an Acid

from a Titration With a Base

grams of known moles of known moles of unknown grams of unknown

liters of known

Molarity of solution

0.250 M

(

)

35.62 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

1

1

æ

è

ç

ö

ø

÷

10.00 mL

1000 mLL

æ

è

ç

ç

ö

ø

÷

÷

=

0

.

891 M

Step 3: go from moles of unknown to molarity of unknown by dividing by the liters.

M of unknown

Liters of solution

Calculate Molarity or Volume of an Acid from a

Titration With a Base

What is the molarity of an HCl solution if 18.5 mL of a 0.225

M

NaOH solution are required to neutralize 10.0 mL of the

HCl solution?

(0.416

M

)

HCl(

aq

) + NaOH(

aq

)



NaCl(

aq

) + H

O(

l

)

Calculate the mL of 2.00

M

H

SO

required to neutralize 50.0

mL of 1.00

M

KOH. (12.5 mL)

Titration Curves

At the half way point,

[weak acid] = [conjugate base]

pH

=

pK

+

log

[

Base

]

Acid

[

]

=

pK

+

log 1

( )

Determining the Endpoint

Methyl red

Determining the Endpoint

O

O

OH OH

Determining the Endpoint

Solubility Rules

•

There are simple rules for determining whether

an ionic compound is soluble or insoluble in water

The rules are summarized on the next slide.

•

While these rules are useful, they do not allow us

1. Most nitrate (NO

), acetate (C

2

H

O

), and chlorate (ClO

)

salts are soluble.

2. Most alkali metal (Li

, Na

, K

, Rb

, Cs

) and NH

4+

salts are

soluble.

3. Most Cl

, Br

, and I

salts are soluble (except Ag

, Pb

,

Hg

).

4. Most sulfate (SO

) salts are soluble (except Ba

, Pb

,

Hg

, Ca

, Ag

, Sr

).

5. Most OH

are only slightly soluble (those containing the

cations in Rule 2 are soluble, Ba(OH)

, Sr(OH)

, and

Ca(OH)

are only marginally soluble).

6. Most S

, CO

32–

, CrO

, PO

salts are only slightly

Solubility versus Temperature

The solubility of most solid

and liquid solutes

increases

with increasing temperature.

At a given temperature,

only a certain amount of

a solid will dissolve.

If you put more of the

solute into the solution

Saturated Solutions and Equilibrium

MgF

(s) Mg

(aq) + 2F

(aq)

K

= Mg

éë

ùû

éë ùû

F

=

6.4

´

10

K

is called the

solubility product constant

Calculate the

K

sp

for an Insoluble Salt

A saturated solution of Ca

(PO

)

has

[Ca

] = 2.01 x 10

M and

[PO

] = 1.6 x 10

M.

Calculate K

for Ca

(PO

)

.

Ca

(PO

)

(s) 3Ca

(aq) + 2PO

43–

(aq)

K

= Ca

éë

ùû

PO

-éë

ùû

Calculate the

K

sp

for an Insoluble Salt

A saturated solution of Ag

CrO

prepared by

dissolving solid Ag

CrO

in water has [CrO

] = 6.5 x 10

M.

Calculate K

for Ag

CrO

.

Ag

CrO

(s) 2Ag

(aq) + CrO

42–

(aq)

K

= Ag

éë ùû

CrO

4 2

-éë

ùû=

(

1.3

´

10

)

(

6.5

´

10

)

=

1.1

´

10

[Ag

] =

2

[CrO

Calculate the Solubility

of an Insoluble Salt

Calculate the solubility of copper(II) hydroxide [Cu(OH)

] in g/L.

Cu(OH)

Cu

+ 2OH

K

= Cu

éë

ùû

éë ùû

OH

=

( )

x

2

x

( )

=

1.6

´

10

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0 0

x

x

Calculate the Solubility

of an Insoluble Salt

Calculate the solubility of copper(II) hydroxide [Cu(OH)

] in g/L.

K

= Cu

éë

ùû

éë ùû

OH

=

( )

x

2

x

( )

=

1.6

´

10

4

x

=

1.6

´

10

x

=

1.6

´

10

-19

4

3

=

3.4

´

10

M

Calculate the Solubility

of an Insoluble Salt

Calculate the solubility of copper(II) hydroxide [Cu(OH)

] in g/L.

=

3.4

´

10

mol

1

L

´

97.57

g

1

mol

=

3.3

´

10

-5

g

L

Factors Affecting Solubility: the Common Ion Effect

Calculate the solubility of AgCl in pure water at 25°C.

AgCl

Ag

+ Cl

K

= Ag

éë ùû

éë ùû=

Cl

-1.6

´

10

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0 0

x

x

Factors Affecting Solubility: the Common Ion Effect

x

=

1.6

´

10

K

= Ag

éë ùû

éë ùû=

Cl

-1.6

´

10

x

=

1.6

´

10

=

1.3

´

10

M

Factors Affecting Solubility: the Common Ion Effect

Calculate the solubility of AgCl in a solution that contains

0.10 M AgNO

.

AgNO

Ag

+ NO

AgCl

Ag

+ Cl

Addition of Ag+ should

Factors Affecting Solubility: the Common Ion Effect

AgCl

Ag

+ Cl

K

= Ag

éë ùû

éë ùû=

Cl

-1.6

´

10

Initial

concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.10 0

x

x

Factors Affecting Solubility: the Common Ion Effect

0.10

+

x

(

)

( )

x

=

1.6

´

10

K

= Ag

éë ùû

éë ùû=

Cl

-1.6

´

10

x

=

1.6

´

10

M

The solubility of AgCl in water at 25°C is significantly

Factors Affecting Solubility: the Common Ion Effect

0 0.000002 0.000004 0.000006 0.000008 0.00001 0.000012 0.000014

0 0.00002 0.00004 0.00006 0.00008 0.0001 0.00012 0.00014 0.00016

So

lu

bi

lity

o

f A

gC

l (M

)

Concentration of added AgNO3 (M)

Factors Affecting Solubility: pH

Some salts are more soluble in an acidic solution than in pure water.

S

+ H

3

O

HS

+ H

O

CuS

Cu

+ S

Addition of an acid removes S2– _{and shifts}

equilibrium to the right

S2– _{is the conjugate base of the weak acid HS}–_{. S}2– _{reacts with H}

Factors Affecting Solubility: pH

Some salts are not more soluble in an acidic solution than in pure water.

Cl

+ H

3

O

HCl

+ H

O

AgCl

Ag

+ Cl

There is no equilibrium shift and no enhanced

solubility in an acidic solution.

Factors Affecting Solubility: pH

When a salt dissociates to give the conjugate base of a weak acid, the solubility is greater in an acidic solution than in pure water. The weaker the acid, the greater its solubility is enhanced in an acidic solution.

CN–_{, PO}

43–, S2–, F–, OH–, CO₃2–_{, C}

2O42–, CrO42–

Cl–_{, Br}–_{, I}–_{, NO}

3–, ClO4–

If the salt contains one of these anions, it will have a greater solubility in an acidic solution.

These anions are the conjugate bases of

Will Precipitation Occur?

A common problem in chemistry is to decide whether a

precipitate of an ionic compound will form when

solutions that contain the constituent ions are mixed.

For example, will CaF

precipitate on mixing solutions

of CaCl

and NaF?

CaF

Ca

+ 2F

Ion Product, Q

The

ion product

, Q,

is defined just like the expression for K

.

Initial concentrations

are used in the expression

instead of equilibrium concentrations.

Q

= Ca

éë

ùû

0

F

-éë ùû

Will a precipitate form when 0.150 L of 0.10 M

Pb(NO

3

)

2

and 0.100 L of 0.20 M NaCl are mixed?

•

We know Pb(NO

)

and NaCl are strong electrolytes

and are soluble in water.

•

Possible precipitates are either PbCl

or NaNO

.

Using the general solubility rules, NaNO

is expected

to be soluble in water, whereas, PbCl

is expected to

be insoluble. If there is any precipitate, it will be

PbCl

.

Will a precipitate form when 0.150 L of 0.10 M

Pb(NO

3

)

2

and 0.100 L of 0.20 M NaCl are mixed?

After mixing, the solution has a total volume of 0.150 L + 0.100 L = 0.250 L.

M

mol

L

Moles of Pb2+ _{= (0.10 mol/L)(0.150 L) = 0.015 mol} Moles of Cl– _{= (0.20 mol/L)(0.100 L) = 0.020 mol}

Pb

éë

ùû=

0.015

mol

0.250

L

=

0.060

M

Cl

-éë ùû=

0.020

mol

Will a precipitate form when 0.150 L of 0.10 M

Pb(NO

3

)

2

and 0.100 L of 0.20 M NaCl are mixed?

Q

= Pb

éë ùû

0

Cl

-éë ùû

=

(

0.060

)

(

0.080

)

=

3.8

´

10

K

=

1.6

´

10

Qualitative Analysis by Selective Precipitation

Unknown solution of metal ions

Solution ions of groups II–V Group I precipitate

AgCl, Hg₂Cl₂, PbCl₂

Solution ions of groups III–V Group II precipitate

PbS, CuS, HgS, CdS, Bi2S3, SnS2

Solution ions of groups IV–V Group III precipitate

MnS, FeS, CoS, NiS, ZnS, Al(OH)₃, Cr(OH)₃

Add HCl

Add H2S

Qualitative Analysis by Selective Precipitation

Solution ions of groups IV–V

Solution ions of group V Na+_{, K}+_{, Mg}2+

Group IV precipitate CaCO₃, BaCO₃

Alkali metals Mg2+ _{precipitates as}

Mg(NH₄)PO₄

Alkali metals give characteristic colors

Add (NH₄)₂CO₃

Add (NH₄)₂HPO₄