Chapter 15 Applications of Aqueous Equilibria

1
Chapter 15Applications of Aqueous Equilibria

• Acid/base solutions containing common ions
• Buffered Solutions
• Titrations and pH curve
• Acid/Base indicators
• Solubility Equilibria
• Complex Ion Equilibria

2
Common Ion Effect

• The shift in equilibrium that occurs because of
  the addition of an ion already involved in the
  equilibrium reaction.
• AgCl(s) ? Ag(aq) Cl?(aq)

3
QUESTION
Suppose the weak acid HNO2 (Ka 4.0 ?104) was
added to a solution of NaNO2. If the
concentration of acid were 0.10 M and the salt
concentration was 0.060 M, what would be the
H? 1. 2.4 ? 104 M 2. 6.7 ? 104 M 3. 2.0 ?
104 M 4. 4.0 ? 105 M
4
ANSWER
Choice 2 represents the approximate H. The Ka
expression may be used with the assumption that
the dissociation of the acid is very slight. The
presence of NO2 (from the salt) does reduce the
amount of dissociation of the acid. The
assumption that any NO2 from the dissociation
of the acid can be ignored is used in the
calculation so that H Ka (.10)/(.60). Sectio
n 15.1 Solutions of Acids or Bases Containing a
Common Ion
5
A Buffered Solution

• . . . resists change in its pH when either H or
  OH? are added.
• 1.0 L of 0.50 M H3CCOOH
• 0.50 M H3CCOONa
• pH 4.74
• Adding 0.010 mol solid NaOH raises the pH of the
  solution to 4.76, a very minor change.

6
Key Points on Buffered Solutions

• 1. They are weak acids or bases containing a
  common ion.
• 2. After addition of strong acid or base, deal
  with stoichiometry first, then equilibrium.

7
Henderson-Hasselbalch Equation

• Useful for calculating pH when the
  A?/HA ratios are known.

8
Buffered Solution Characteristics

• Buffers contain relatively large amounts of weak
  acid and corresponding base.
• Added H reacts to completion with the weak base.
• Added OH? reacts to completion with the weak
  acid.
• The pH is determined by the ratio of the
  concentrations of the weak acid and weak base.

9
QUESTION
A certain chemical reaction runs very slow at
high pH values and very fast at lower pH values.
To study the reaction, a chemist needs to buffer
the solution at a basic pH. What would be the pH
of a buffer solution that had the following
concentrations 0.42 M NH4Cl and 0.75 M NH3 Kb
of NH3 1.8 ? 105 at 25C. 1. 9.00 2. 4.49 3. 9
.51 4. 11.57
10
ANSWER
Choice 3 provides the correct basic pH of the
solution. Using the Kb value of ammonia allows
the calculation of OH with the given NH3 and
NH4Cl concentrations. Assuming that no
significant NH4 comes from the NH3 H2O
reaction, the OH could be approximated via
OH Kb (NH3)/(NH4) the pH can then be
determined from the pOH using the Kw for
water. Section 15.2 Buffered Solutions
11
Buffering Capacity

• . . . represents the amount of H or OH? the
  buffer can absorb without a significant change in
  pH.

12
QUESTION
Although buffers resist change in pH they are
still affected by the addition of a new base or
acid. A 0.50 L buffer solution containing 0.42 M
NH4Cl and 0.75 M NH3(Kb of NH3 1.8 ? 105 )
has a pH of 9.51 at 25C. What would be the new
pH of the buffer after it is adjusted by the
addition of 0.010 moles of NaOH. Assume no
change in the solution volume. 1. 8.97 2. 9.51 3.
12.00 4. 9.54
13
ANSWER
Choice 4 provides the newly adjusted pH. The
addition of 0.010 moles of new OH causes the
NH4 concentration to decrease and the NH3
concentration to increase. The molarity of both
is first changed to moles (V ? M moles) then
the 0.010 moles of OH is added and subtracted
appropriately. The new number of moles of each
are divided by the solution volume to obtain
molarity. The system can now be treated with the
Kb expression using the new molarities of the
NH4 and NH3 to solve for OH, which can then
be used to obtain pOH and pH. Section 15.2
Buffered Solutions
14
QUESTION
Two separate buffer solutions are prepared using
propanoic acid and calcium propanoate (also used
in some food preservative applications). If
buffer A was 0.10 M in both acid and salt while
buffer B was 0.20 M in both acid and salt,
which of the following would be true? 1. Both
solutions would have the same pH and would
have the same buffer capacity. 2. Solution B
would would have a lower pH and both would have
the same buffer capacity. 3. Solution B would
have a lower pH and a larger buffer capacity. 4.
Both solutions would have the same pH B would
have a larger buffer capacity.
15
ANSWER
Choice 4 offers the accurate comparison of the pH
and buffer capacity for the two solutions. As
the salt and acid have the same concentration,
the pH pKa relationship is the same. However,
because solution B has a higher molarity for
both acid and salt any additional acid or base
additions will have to be much larger before the
ratio changes significantly. Therefore, solution
B has a higher buffering capacity. Section
15.3 Buffer Capacity
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Titration (pH) Curve

• A plot of pH of the solution being analyzed
  as a function of the amount of titrant added.
• Equivalence (stoichiometric) point Enough
  titrant has been added to react exactly with
  the solution being analyzed.
• The next three slide emphasize the plot of
  pH and the equivalence point

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Figure 15.1 The pH curve for the titration of
50.0 mL of 0.200 M HNO3 with 0.100 M NaOH.
18
Figure 15.2 The pH curve for the titration of
100.0 mL of 0.50 M NaOH with 1.0 M HCl.
19
Figure 15.3 The pH curve for the titration of
50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.
20
Comparison of strong and weak acid titration
curves.
21
QUESTION
If 25.0 mL of 0.250 M of HCl were titrated with
0.500 M NaOH, what would you calculate as the
initial pH, pH at the equivalence point and
volume of the total solution at the equivalence
point? 1. 1.000 7.000 25.0 mL 2. 0.600 7.00
50.0 mL 3. 0.600 7.00 37.5 mL 4. 1.000 7.000
12.5 mL
22
ANSWER
Choice 3 provides the correct response for all
three questions. The pH of a strong acid is
taken directly from the molarity. (log 0.250)
The pH of the equivalence point for a strong acid
and strong base titration is 7.00. The mmoles of
HCl initially is 25.0 mL ? 0.250 M, or 6.25
mmoles. Therefore, the base must provide 6.25
mmoles for neutralization. The volume of 0.500 M
NaOH that would provide this could be calculated
from V ? 0.500 M 6.25 V 6.25/0.500 12.5
mL. Section 15.4 Titrations and pH Curves
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QUESTION
What is the pH of the equivalence point of the
titration of 30.0 mL of 0.100 M benzoic acid,
C6H5CO2H (Ka 6.4 ? 105), with 0.100 M NaOH at
25C? 1. 8.596 2. 5.554 3. 8.446 4. I must be
missing a step, I dont get any of these.
24
ANSWER
Choice 3 correctly accounts for the the formation
of the moles of the salt of benzoic acid during
the titration (volume acid ? molarity of the
acid), molarity of the salt (moles / total
volume), determination of the Kb of the conjugate
salt (Kw/Ka), and finally the pH (14
pOH). Section 15.4 Titrations and pH Curves
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Weak Acid - Strong Base Titration

• Step 1 - A stoichiometry problem - reaction is
  assumed to run to completion - then determine
  remaining species.
• Step 2 - An equilibrium problem - determine
  position of weak acid equilibrium and calculate
  pH.

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Acid-Base Indicator

• . . . marks the end point of a titration by
  changing color.
• The equivalence point is not necessarily the same
  as the end point.

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Figure 15.8 The useful pH ranges for several
common indicators. Note that most indicators have
a useful range of about two pH units, as
predicted by the expression pKa ? 1.
28
Figure 15.6 The acid and base forms of the
indicator phenolphthalein. In the acid form
(Hln), the molecule is colorless. When a proton
(plus H2O) is removed to give the base form
(ln-), the color changes to pink.
29
Methyl orange indicator is yellow in basic
solution and red in acidic solution.
30
Figure 15.7 (a) Yellow acid form of bromthymol
blue (b) a greenish tint is seen when the
solution contains 1 part blue and 10 parts
yellow (c) blue basic form.
31
QUESTION
Most acid-base indicators are weak acids. In a
titration of 0.50 M acetic acid (at 25C, Ka
1.8 ? 105) with KOH, which indicator would best
indicate the pH at the equivalence point? The
approximate Ka for each choice is
provided. 1. Bromophenol blue Ka 1 ?
104 2. Methyl Red Ka 1 ? 105 3. Bromothymol
blue Ka 1 ? 107 4. Thymolphthalein Ka 1 ?
1010
32
ANSWER
Choice 4 provides the best choice although there
may also be better choices available than these
four. The equivalence point pH should be as
close as possible to the pKa of the indicator.
As acetic acid is a fairly weak acid and NaOH is
a strong base, the pH at the equivalence point
will be above 7. The only choice above 7 in the
list was thymolphthalein. Without a more
detailed calculation, this would be the best
choice. Section 15.5 AcidBase Indicators
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QUESTION
The acid-base indicator bromcresol purple has an
interesting yellow-to-purple color change. If
the approximate Ka of this indicator is 1.0 ?
106, what would be the ratio of purple A to
yellow HA at a pH of 4.0? 1. 1001 2. 1100 3.
11 4. This choice indicates that I dont know.
34
ANSWER
Choice 2 shows the A/HA ratio at pH 4.0 for
bromcresol purple. The pH can be converted to
H and divided into the Ka value to reveal the
A/HA ratio at pH 4.0. Ka/H
A/HA. Section 15.5 AcidBase Indicators
35
Solubility Product

• For solids dissolving to form aqueous solutions.
• Bi2S3(s) ? 2Bi3(aq) 3S2?(aq)
• Ksp solubility product constant
• and
• Ksp Bi32S2?3

36
Solubility Product

• Solubility s concentration of Bi2S3
  that dissolves, which equals 1/2Bi3 and
  1/3S2?.
• Note Ksp is constant (at a given
  temperature)
• s is variable (especially with a common
• ion present)

37
QUESTION
Lead (II) iodide is used in some camera
batteries. PbI2 has a Ksp of 1.4 ? 108. What
is the molar solubility of this compound? 1. 1.9
? 103 M 2. 2.4 ? 103 M 3. 1.5 ? 103 M 4. 8.4 ?
105 M
38
ANSWER
Choice 3 correctly uses the stoichiometry and
equilibrium expression to determine the molar
solubility of PbI2. Since the dissociation of
PbI2 shows Pb2(aq) 2 I(aq), the Ksp
expression would appear as Ksp X2X2.
Solving for X shows the relationship for PbI2
dissociating in a 11 ratio between PbI2 and
Pb2. Section 15.6 Solubility Equilibria and
the Solubility Product
39
QUESTION
Will a precipitate of BaSO4 form when 10.0 mL of
0.001 0 M barium nitrate are mixed with 20.0 mL
of 0.000 020 M of sodium nitrate? The Ksp of
barium sulfate is 1.5 ? 109. Prove your answer
by reporting the calculated value of Q. 1. Yes
Q 2.0 ? 108 2. Yes Q 4.4 ? 109 3. No Q
7.0 ? 1010 4. No Q 3.0 ? 1017
40
ANSWER
Choice 2 provides the correct prediction. Once
the molarity of each ion (Ba2 and SO42) is
determined they are multiplied to obtain Q. If Q
is larger than Ksp a precipitate will
form. Section 15.7 Precipitation and
Qualitative Analysis
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Equilibria Involving Complex Ions

• Complex Ion A charged species consisting of a
  metal ion surrounded by ligands (Lewis bases).
• Coordination Number Number of ligands
  attached to a metal ion. (Most common are 6 and
  4.)
• Formation (Stability) Constants The
  equilibrium constants characterizing the
  stepwise addition of ligands to metal ions.

42
QUESTION
To separate a solution containing 0.000 10 M
silver and 0.10 M lead ions, as may be done in
qualitative analysis separation schemes, a source
of I may be slowly added to the mixture of ions.
Which will precipitate first AgI (Ksp 1.5 ?
1016 ) or PbI2 (Ksp 1.4 ? 108)? Also, what
would be the concentration of I necessary to see
that first precipitation? 1. AgI I would be
1.5 ? 1012 M 2. AgI I would be 1.4 ? 104
M 3. PbI2 I would be 1.4 ? 107 M 4. PbI2
I would be 1.4 ? 106 M
43
ANSWER
Choice 1 indicates correctly that AgI would
precipitate first when a very small amount of I
is added. Both of the possibilities for
precipitation need to be evaluated. For AgI, the
Ksp AgI would be solved for the I
after inserting the silver ion concentration.
Then the Ksp for PbI2 could also be set to solve
for the I Ksp Pb2I2. Section 15.7
Precipitation and Qualitative Analysis
44
QUESTION
One method that can increase the solubility of
some salts is to form a soluble complex ion.
Which statement best describes the reason behind
this methods result? CuS has a Ksp of 8.5 ?
1045. The reaction for Cu2(aq) 4 NH3(aq) ?
Cu(NH3)42 has an equilibrium constant of
approximately 5.6 ? 1011. What would be the
approximate molar solubility of CuS in a 1.0 M
solution of NH3? 1. 6.9 ? 1017 M 2. 9.2 ? 10
23 M 3. 4.8 ? 1033 M 4. This one seems too
complex.
45
ANSWER
Choice 1 provides a reasonable estimate of the
solubility after the addition of NH3 to help form
a complex with the copper ion. Since the CuS
dissociation and the Cu(NH3)42 complex formation
are going to be added, their K values should be
multiplied to give Ksp ? K Cu2Cu(NH3)42
/NH34. After substitution the expression
becomes 4.8 ? 1033 XX/1.0 4X4. If the
4X term is considered small enough to eliminate
then the math can be manipulated to solve for
X. Section 15.8 Equilibria Involving Complex
Ions