Ap chem unit 15 presentation

 Solutions of Acids or Bases Containing a
Common Ion
 Buffered Solutions
 Buffering Capacity
 Titrations and pH Curves
 Acid-Base Indicators

Most chemistry in the natural world takes place
in an aqueous solution. One of the most
significant aqueous reactions is one with acids
and bases. Most living systems are very
sensitive to pH and yet are subjected to acids
and bases.
 The main idea of this unit is to understand the
chemistry of a buffered system. Buffered
systems contain chemical components that
enable a solution to be resistant to change in
pH.

Solutions that contain a weak acid and the
salt of it conjugate base create a common ion
system.
 A common ion system can also be a weak
base and the salt of its conjugate acid.
◦ Example: HF and NaF solution or
NH3 and NH4Cl solution.

 The soluble salts of conjugate acids and
conjugate bases are typically strong
electrolytes and therefore dissociate 100%.
 This dissociation increases the
concentration of the conjugate ion and a
shift in equilibrium occurs according to Le
Chatelier. + -
◦ Example:
NH 4Cl(s) ® NH 4(aq) + Cl(aq)
+ -
NH 3(aq) + H 2O(l ) « NH 4(aq) + OH (aq)

 Shift Left!

The shift left in the second equation is known
as the common ion effect.

NH 4Cl(s) ® NH + + Cl(aq)
4(aq)
-

NH 3(aq) + H 2O(l ) « NH + + OH (aq)
4(aq)
-

 This effect makes a solution of NH3 and
NH4Cl less basic than NH3 alone

The shift left in the second equation is known
as the common ion effect.

+ -
NaF(s) ® Na(aq) + F(aq)
+ -
HF(aq) + H 2O(l ) « H 3O(aq) + F(aq)
This effect makes a solution of NaF and HF less
acidic than a solution of HF alone.

Calculations for common ion equilibria is
similar to weak acid calculations except that
the initial concentrations of the anion are not
0.
 Initial concentrations for the equilibrium
calculation must include the concentration of
ions from the complete dissociation of the
salt.

The equilibrium concentration of H+ in a 1.0 M
HF solution is 2.7x10-2M, and the percent
dissociation of HF is 2.7%. Calculate [H+] and
the percent dissociation of HF in a solution
containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M
NaF.
Major species?
ICE:
I C E
HF
F-
H+

The equilibrium concentration of H+ in a 1.0 M
HF solution is 2.7x10-2M, and the percent
dissociation of HF is 2.7%. Calculate [H+] and
the percent dissociation of HF in a solution
containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M
NaF.
[H + ][F - ]
Ka =
[HF]

ANS: .072%

The most important application of a common
ion system is buffering.
 A buffered solution resists a change in its
pH when either hydroxide ions or protons
are added.
◦ Blood is a practical example of a buffered
solution; it is buffered with carbonic acid
and the bicarbonate ion (among others).

 A solution can be buffered at any pH by
choosing the appropriate components.
 Smaller amounts of H+ or OH- can be added
with little pH effect.

A buffered solution contains 0.50 M acetic acid
(Ka= 1.8 x 10-5) and 0.50 M sodium acetate.
Calculate the pH of this solution.

pH = 4.74

 When OH- ions are added to a solution of a
weak acid, the OH- ions react with the best
source of H+ (weak acid) to make water:
- -
OH + HA ® A + H 2O
 The net result is that OH- ions are replaced by
A- ions and water until they are all consumed.

 When H+ ions are added to a solution of a
weak acid, the H+ ions react with the strong
conjugate base.
- +
A + H ® HA
 The net result is that more weak acid is
produced, but free H+ do not accumulate to
make large changes in the pH.

Buffered solutions are simply solutions of weak
acids or bases containing a common ion. The pH
calculations on buffered solutions require exactly
the same procedures introduced in Chapter 14.
 When a strong acid or base is added to a
buffered solution, it is best to deal with the
stoichiometry of the resulting reaction first.
After the stoichiometric calculations are
completed, then consider the equilbrium
calculations.

Calculate the change in pH that occurs when
0.010 mol solid NaOH is added to the buffered
solution in #2 (0.50 M acetic and 0.50 M sodium
acetate). Compare this pH change with the
original (pH=4.74) and the pH that occurs with
0.010 mol of solid NaOH is added to 1.0 L of
water.

 Major species:

 OH- +HC2H3O2  C2H3O2- + H2O

 mole table:
HC2H3O2 OH- C 2 H 3 O2 - H2O
Before
After

[H + ][C2 H 3O2 ]
-
K a = 1.8x10 -5 =
[HC2 H 3O2 ]
 ICE table:
I C E
HC2H3O2
OH-
C2H3O2-

pH= 4.76 (+.02 off original)
pH of just OH-=12.00

Buffered solutions work well as long as the
concentration of the salts and weak
acids/bases in solution are much larger than
the amount of OH- and H+ being added.
 The amount of [H+] being in a buffered
solution is often solved using a
rearrangement of the equilibrium expression:

+ [HA]
[H ] = K a -
[A ]

+ [HA]
 This equation, [H ] = K a -
, can be
changed into another [A ]
useful form by taking the negative log of
both sides:

æ [A - ] ö æ [base]ö
pH = pK a + log ç ÷ = pK a + log ç [acid] ÷
è [HA] ø è ø
This log form of the expression is called the
Henderson-Hasselbalch Equation.

Henderson-Hasselbalch Equation: When using
this equation it is often assumed that the the
equilibrium concentrations of A_ and HA are
equal to their initial concentrations due to the
validity of most approximations.
æ [A - ] ö æ [base]ö
pH = pK a + log ç ÷ = pK a + log ç [acid] ÷
è [HA] ø è ø
 Since the initial concentrations of HA and A-
are relatively large in a buffered solution, this
assumption is generally acceptable.

Calculate the pH of a solution containing 0.75
M lactic acid (Ka=1.4x10-4) and 0.25 M sodium
lactate. Lactic acid (HC3H5O3) is a common
constituent of biologic systems. For example, it
is found in milk and is present in human
muscle tissue during exertion.

pH = 3.38

Buffered solutions can be formed from a weak
base and the corresponding conjugate acid. In
these solutions, the weak base B reacts with
any H+ added:
+ +
B + H ® BH
 The conjugate acid BH reacts with any added
+

OH-: + -
BH + OH ® B + H 2O

A buffered solution contains 0.25 M NH3
(Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the
pH of this solution.

pH=9.05

Calculate the pH of the solution that results
when 0.10 mol gaseous HCl is added to 1.0 L
of the buffered solution from #5 (.25M NH3
and .40 NH4Cl).

pH= 8.73

1. Buffered solutions contain relatively large
concentrations of a weak acid and the
corresponding weak base. They can involve a
weak acid HA and the conjugate base A- or a
weak base B and the conjugate acid BH+.
2. When H+ is added to a buffered solution, it
reacts essentially to completion with the
weak base:
+ - + +
H + A ® HA or H + B ® BH

3. When OH- is added to a buffered solution, it
reacts essentially to completion with the weak
acid present:

OH - + HA ® A- + H 2O or OH - + BH + ® B + H 2O
3. The pH in the buffered solution is determined by
the ratio of the concentrations of the weak acid
and weak base. As long as this ratio remains
virtually constant, the pH will remain virtually
constant. This will be the case as long as the
concentration of the buffering materials (HA and
A- or B and BH+) are large compared with the
amounts of H+ or OH- added.

The buffering capacity of a buffered solution
represents the amount of protons or hydroxide
ions the buffer can absorb without a significant
change in pH.
 The pH of a buffered solution is determined
by the ratio [A-] / [HA]. The capacity of a
buffered solution is determined by the
magnitudes of [HA] and [A-].

Calculate the change in pH that occurs when
0.010 mol of gaseous HCl is added to 1.0 L of
each of the following solutions:
◦ Soln A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
◦ Soln B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
Ka=1.8x10-5.

Soln A: pH=4.74 Soln B: pH=4.56

The optimal buffering occurs when [HA] is
equal to [A-].
 This ratio of 1 resists the most amount of pH
change.
 The pKa of the weak acid to be used in the
buffer should be as close as possible to the
desired pH.
◦ For example, if a buffered solution is needed with a
pH of 4.0. The most effect buffering will occur
when [HA] is equal to [A-] and the pKa of the acid is
close to 4.0 or Ka=1.0x10-4.

A chemist needs a solution buffered at pH=4.30
and can choose from the following acids and
their sodium salts. Calculate the ratio [HA]/[A-]
required for each system to yield a pH of 4.30.
Which system will work best?
a. chloroacetic acid (Ka=1.35x10-3)
b. propanoic acid (Ka=1.3x10-5)
c. benzoic acid (Ka=6.4x10-5)
d. hypochlorous acid (Ka=3.5x10-8)

benzoic acid

An acid-base titration is
often graphed by plotting
the pH of the solution
(y-axis) vs. the amount of
titrant added (x-axis).
 Point (s) is the equivalence
point/stoichiometric point.
This is where [OH-]=[H+].

The net ionic reaction for a strong acid- strong
base titration is: H + + OH - ® H O
(aq) (aq) 2 (l )

To compute [H+] at any point in a titration, the
moles of H+ that remains at a given point must
be divided by the total volume of the solution.
 Titrations often include small amounts. The
mole is usually very large in comparison.
 Millimole (mmol) is often used for titration
amounts.

mol mmol
M= =
L ml

mmol = ml(M )

50.0 ml of 0.200 M HNO3 is titrated with 0.100
M NaOH. Calculate the pH of the solution at
the following points during the titration.
a. NaOH has not been added.
Since HNO3 is a strong acid (complete
dissociation), pH = 0.699

50.0 ml of 0.200 M HNO3 is titrated with 0.100 M
NaOH. Calculate the pH of the solution at the
following points during the titration.
b. 10.0 ml of 0.100 M NaOH has been added.
H+ = 10 mmol – 1.0 mmol of OH- =
9.0mmol
9.0 mmol H+/ 60 ml = 0.15 M
pH=0.82

c. 20.0 mL of 0.100 M NaOH has been added.

pH=0.942

d. 50.0 mL of 0.100 M NaOH has been added.

pH=1.301

e. 100.0 mL of 0.100 M NaOH has been added.

The stoichiometric point/equivalence point is
reached. pH=7

f. 150.0 mL of 0.100 M NaOH has been added.

pH = 12.40

g. 200.0 mL of 0.100 M NaOH has been added.

pH=12.40

The results of these calculations are summarized
by this graph:

The pH changes very gradually until the titration
is close to the equivalence point, then dramatic
change occurs. Near the E.P., small changes
produces large changes in the OH-/H+ ratio.

1. Before the equivalence point, [H+] can be
calculated by dividing the number of
millimoles of H+ remaining by the total
volume of the solution in mL.
2. At the equivalence point, pH=7.0
3. After the equivalence point, [OH-] can be
calculated by dividing the number of
millimoles of excess OH- by the total
volume of the solution. [H+] is calculated
from Kw.

When acids don’t completely dissociate,
calculations used previously need adjusted.
 To calculate [H+] after a certain amount of
strong base has been added, the weak acid
dissociation equilibrium must be used.
 Remember: that a strong base reacts to
completion with a weak acid.

Calculating the pH Curve for a Weak Acid-
Strong Base Titration is broken down into two
steps:
1. A stoichiometric problem. The reaction of
hydroxide ion with the weak acid is
assumed to run to completion, and the
concentrations of the acid remaining and
the conjugate base formed are determined.
2. An equilibrium problem: The position of the
weak acid equilibrium is determined, and
the pH is calculated.

50.0 mL of 0.10 M acetic acid (HC2H3O2,
Ka=1.8x10-5) is titrated with 0.10 M NaOH.
a. NaOH has not been added.

pH = 2.87

b. 10.0 mL of 0.10 M NaOH has been added.

pH= 4.14

c. 25.0 mL of 0.10 M NaOH has been added.

pH=4.74

d. 40.0 mL of 0.10 M of NaOH has been
added.

pH=5.35

e. 50.0 mL of NaOH is added.

pH=8.72

f. 60.0 mL of 0.10 M NaOH is added.

pH = 11.96

g. 75.0 mL of 0.10 M NaOH is added.

pH = 12.30

Some of the differences in the the titration
curves:
 The pH increases more rapidly in the
beginning of the weak acid titration.

curves:
 The titration curve levels off near the halfway
point due to buffering effects.
◦ Buffering happens when [HA]=[A-]. This is exactly
the halfway point of the titration.

curves:
 The value of the pH at the equivalence point
is not 7. The value of the e.p. is greater than
7.

Hydrogen cyanide gas (HCN), a powerful
respiratory inhibitor, is highly toxic. It is a very
weak acid (Ka=6.2x10-10) when dissolved in
water. If a 50.0 mL sample of 0.100 M HCN is
titrated with 0.100 M NaOH, calculate the pH of
the solution.
a. after 8.00 mL of 0.100 M NaOH has been
added.
b. at the halfway point of the titration
c. at the equivalence point of the titration

a) pH=8.49 b) pH=9.21 c)pH=10.96

When comparing the example calculation and
the practice problem with weak acids, two
major conclusions can be made:
1. The same amount of 0.10 M NaOH is
required to reach the equivalence point
even though HCN is a much weaker acid.
◦ It is the amount of acid, not its strength that
determines the equivalence point.

When comparing the example calculation and
the practice problem with weak acids, two
major conclusions can be made:
2. The pH value at the equivalence point is
affected by the acid stregth. The pH at the
e.p. for acetic acid is 8.72. The pH at the
e.p. for hydrocyanic acid is 10.96.
◦ The CN- ion is a much stronger base than C2H3O2-.
The smaller acid strength and stronger conjugate
base produces a higher pH.

The strength of a weak acid has a significant
effect on the shape of its pH curve. The e.p.
occurs at the same point, but the shapes of the
curve is dramatically different.

A chemist has synthesized a monoprotic weak
acid and wants to determine its Ka value. To
do so, the chemist dissolves 2.00 mmol of the
solid acid in 100.0 mL water and titrates the
resulting solution with 0.0500 M NaOH. After
20.0 mL NaOH has been added, the pH is 6.00.
What is the Ka value for the acid?

Ka=1.0x10-6

Find the pH of a solution of 100.0 mL of 0.050
M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-
5).

a. HCl has not been added.

pH is found with Kb equilibrium. pH = 10.96

5).

b. 10.0 mL of HCl is added.

H+ ions react to completion. pH=9.85

5).

c. 25.0 mL of HCl is added.

pH = 9.25

5).

d. 50.0 mL of HCl is added.

pH = 5.36

5).

e. 60.0 mL of HCl is added.

pH= 2.21

There are two common methods for determining
the equivalence point of a titration:
1. Use a pH meter to monitor the pH and then
plot the titration curve. The center of the
vertical region of the pH curve indicates the
equivalence point.
2. Use an acid-base indicator, which marks the
end point of a titration by changing color.

The equivalence point of a titration is defined
by the stoichiometry, but it is not necessarily
the same as the end point where the indicator
changes color).
 Selection of the right indicator for the
titration is very important.

The most common acid-base indicators are
complex molecules that are weak acids (HIn).
 Most exhibit one color when the proton is
attached to the molecule and a different color
when the proton is absent.
◦ Example: Phenolphthalein is colorless in
its HIn form and pink in its In-, or basic
form.

+ -
HIn(aq) « H (aq) + In(aq)
(RED) (BLUE)

[H + ][In - ] Ka [In - ]
Ka = +
=
[HIn] [H ] [HIn]
The color will depend on the ratio of [In-] to
[HIn]. For most indicators, approximately
1/10th of the initial form must be converted to
the final form before a color is perceived by the
human eye.
 the color change will occur at a pH when

[In - ] 1
=
[HIn] 10

Bromthymol blue, an indicator with Ka=1.0x10-
7, is yellow in its HIn form and blue in its In-

form. Suppose we put a few drops of this
indicator in a strongly acidic solution. If the
solution is then titrated with NaOH, at what pH
will the indicator color change first be visible?

pH=6.0

The H-H equation is very useful in determining
the pH at which an indicator changes color.

æ [In - ] ö [In - ] 1
pH = pK a + log ç ÷ , [HIn] = 10
è [HIn] ø
pH = pK a + long( 10 ) = pK a -1
1

 Example:Bromthymol blue (Ka=1x10-7, or
pKa=7), the pH at the color change is pH=7-
1=6

When a basic solution is titrated, the indicator
will initially exist as In- in solution, but as acid
is added more HIn is formed. Color change
will occur at: [In - ] 10
=
[HIn] 1
Substituting this reciprocal into the H-H
equation gives us: pH = pKa + log ( 10 ) = pKa +1
1

◦ Bromthymol blue=pH 7+1=8;
The useful range of bromthymol blue is pH(6-8)

Ap chem unit 15 presentation

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

En vedette

En vedette (10)

Similaire à Ap chem unit 15 presentation

Similaire à Ap chem unit 15 presentation (20)

Plus de bobcatchemistry

Plus de bobcatchemistry (20)

Dernier

Dernier (20)

Ap chem unit 15 presentation