Chapter 8: Applications of Aqueous Equilibria
8.1 Solutions of Acids or Bases Containing a Common Ion
8.2 Buffered Solutions
8.3 Exact Treatment of Buffered Solutions
8.4 Buffer Capacity
8.5 Titrations and pH Curves
8.6 Acid-Base Indicators
8.7 Titration of Polyprotic Acids
8.8 Solubility Equilibria and The Solubility Product
8.9 Precipitation and Qualitative Analysis
8.10 Complex Ion Equilibria
1
pH range where
buffered solutions
play important
roles
CHEMISTRY
BIOLOGY
MEDICINE
2
pH Box
[H3O+] = 10-pH
pH
pH = -log[H3O+]
[H3O+]
Kw = 1 x 10-14
@ 25 °C
pH + pOH
= 14 @ 25 °C
[H3O+][OH-]
= 1 x 10-14
[OH-] = 10-pOH
pOH
pOH = -log[OH-]
[OH-]
3
Common Ion Effect
A shift in the equilibrium POSITION that occurs because of
the addition of an ion already involved in the equilibrium
reaction.
HNO2 + H2O
H3O+ + NO2-
Adding NaNO2 would shift the equilibrium to the left,
decreasing the H3O+ and NO2- produced by the reaction
between HNO2 and H2O. (Le Châtelier’s principle)
4
Which of the following statements is true?
1. HF dissociates more
readily in a solution
of NaF than in pure
water.
2. HF dissociates
equally well in a
solution of KCl and
in pure water.
3. HF does not
dissociate in water to
any extent.
5
Like Example 8.1 - I
Nitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 M solution.
Calculate the [H+], the pH, and the percent dissociation of HNO2 in a
1.0 M solution that is also 1.0 M in NaNO2!
HNO2(aq)
Ka =
Initial Concentration (mol/L)
H+(aq) + NO2-(aq)
[H+] [NO2 -]
-4
[HNO2] = 4.0 x 10
Equilibrium Concentration (mol/L)
[HNO2]0 = 1.0 M
[HNO2] = 1.0 – x
(from dissolved HNO2)
[NO2-]0 = 1.0 M
[NO2-] = 1.0 + x
(from dissolved NaNO2)
[H+]0 ≈ 0
[H+] = x
(neglect the contribution from water)
6
Like Example 8.1 - II
[H+] [NO2-]
=
[HNO2]
Ka =
( x ) ( 1.0 + x )
(1.0 – x )
= 4.0 x 10-4
Assume (1.0 + x) ≈ 1.0 and (1.0 – x) ≈ 1.0 :
x (1.0)
(1.0)
= 4.0 x 10-4
or
x = 4.0 x 10-4 = [H+]
Therefore pH = - log [H+] = - log (4.0 x 10-4) = 3.40
The percent dissociation is:
4.0 x 10-4
1.0
x 100 = 0.040 %
Nitrous acid
alone
[H+] 2.0 x 10-2
pH
1.70
% Diss 2.0
Nitrous acid
+ NaNO2
4.0 x 10-4
3.40
0.040
7
What is a BUFFER?
A buffer is a solution that contains concentrations of both
components of a conjugate acid-base pair.
When small quantities of H+ or OH- are added to the buffer,
they cause a small amount of one buffer component to
convert into the other (HA → A- or A- → HA).
Buffers resist a change in pH. As long as the amounts of H3O+
and OH- are small (compared to the concentrations of the
acid and base in the buffer), the added ions will have little
effect on the pH since they are consumed by the buffer
components.
The ability of a buffer to resist a change of pH is due to the
buffer capacity.
pH equals pKa IF the acid-to-base ratio in the buffer is 1:1
8
9
How Does a Buffer Work
Original
buffered
solution
pH
(H+ or OH- added)
Step 1:
Do stoichiometric calculations to
determine new concentrations.
Assume reaction with H+/OHgoes to completion.
Modified
pH
Step 2:
Do equilibrium
calculations.
10
How Does a Buffer Work
Addition of a strong base to a buffer has the following effect:
OH- + HA
A- + H2O
Final pH of buffer
close to original
Original
buffer pH
+
Ka = [H ] [A ]
[HA]
Added OH- ions
produce A- ions
[H+] = Ka
[HA]
[A-]
11
How a Buffer Works–I
A buffer is made by placing 0.250 mol of acetic acid and 0.250 mol of
sodium acetate in 1 L of solution. What is the pH of 100.0 mL of the
buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added?
What is the pH of 300.00 mL of pure water before and after the same
amount of acid is added?
CH3COOH(aq) + H2O(l)
[H3O+] = Ka x
[CH3COOH]
[CH3COO-]
CH3COO-(aq) + H3O+(aq)
= 1.8 x 10-5 x
(0.250)
(0.250)
= 1.8 x 10-5
pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74 Before HCl added!
Add 1.00 mL conc. HCl to 100.0 mL of buffer:
0.00100 L x 12.0 mol/L = 0.0120 mol H3O+
12
How a Buffer Works–II updates in green
After acid is added, all H3O+ will react with available CH3COO- :
Conc. (moles) CH3COOH(aq) + H2O(l)
CH3COO- + H3O+
Initial
Change
Equilibrium
0.025
+0.012
0.037
----------
0.025
-0.012
0.013
0.012
-0.012
0
Equilibrium must then be reestablished: total vol = 0.101 L
Conc. (M)
CH3COOH(aq) + H2O(l)
CH3COO- + H3O+
Initial
Change
Equilibrium
0.366
-x
0.366 - x
----------
0.129
+x
0.129 + x
0
+x
x
Assuming: 0.366 - x = 0.366 and 0.129 + x = 0.129
[CH3COOH]
(0.366)
[H3O+] = Ka x
=1.8 x 10-5 x
= 5.11 x 10-5
(0.129)
[CH3COO ]
pH = -log(5.11 x 10-5) = 4.29
After the acid is added!
13
How a Buffer Works–III
A buffer is made by placing 0.250 mol of acetic acid and 0.250 mol of
sodium acetate in 1 L of solution. What is the pH of 100.0 mL of the
buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added?
What is the pH of 300.0 mL of pure water after the same amount of acid
is added?
Add 1.00 mL conc. HCl:
0.00100 L x 12.0 mol/L = 0.0120 mol H3O+
to 300.0 mL of water :
0.0120 mol H3O+
= 0.0399 M H3O+
0.3010 L soln.
pH = -log(0.0399 M)
pH = 1.399 Without buffer!
14
How a Buffer Works–IV updates in red
Suppose we add 1.00 mL of a concentrated base instead of an acid. Add
1.00 mL of 12.0 M NaOH to 1 L of the buffer and to pure water, and let’s
see what the impact is:
1.00 mL x 12.0 mol/L OH- = 0.0120 mol OH- → 0.0120 M
INITIAL STOICHIOMETRY
Moles
CH3COOH(aq) + OH-(aq)
CH3COO- + H2O
Initial
Change
Equilibrium
0.250
- 0.0120
0.238
0.250
+0.0120
0.262
0.0120
- 0.0120
0
-------
REESTABLISH EQUILIBRIUM (in 1.001L)
Conc. (M)
CH3COOH(aq) + H2O(l)
Initial
Change
Equilibrium
0.238
-x
0.238-x
CH3COO- + H3O+
----------
0.262
+x
0.262+x
0
+x
x
15
How a Buffer Works–V
Making our normal assumptions:
0.238 - x = 0.238 and 0.262 + x = 0.262
[H3O+] = 1.8 x 10-5 x
0.238
0.262
pH = -log(1.64 x 10-5) = 4.79
= 1.64 x 10-5
After base is added!
By adding 1.00 mL base to 300.0 mL of pure water we would get a
[OH-] concentration of:
0.0120 mol OH= 0.0399 M OH0.3010 L
The [H3O+] concentration is:
Kw
1.0 x 10-14
[H3O+] =
=
= 2.51 x 10-13
[OH-]
0.0399 M
[OH-] =
pH = -log(2.5 x 10-13) = 12.60 Without buffer!
16
How a Buffer Works–VI
In summary:
Buffer alone pH = 4.74
Buffer plus 1.0 mL acid pH = 4.29
Buffer plus 1.0 mL base pH = 4.79
Acid in water pH = 1.399
Base in water pH = 12.60
17
HENDERSON - HASSELBALCH EQUATION
HA(aq) + H2 O(l) ↔ H+(aq) + A-(aq)
Ka =
[H+] [A-]
[HA]
Solving for the hydronium ion concentration gives:
[H+] = Ka x [HA]
[A-]
Taking the negative logarithm of both sides:
( )
( )
-log[H +] = -log Ka + - log [HA]
[A-]
pH = pKa - log [HA]
[A-]
18
pH of a BUFFER
(Henderson –Hasselbalch eqn.)
Generalizing for any conjugate acid-base pair :
pH = pKa + log ( [A ] )
[HA]
when [A-] = [HA], pH= pKa
For acetic acid/sodium acetate solution:
Ka = 1.8 x 10-5 → pKa = 4.74 → pH = 4.74
If [HA] and [A-] are known, the pH can be calculated.
If the pH is known, the ratio of [HA] and [A-] can be calculated.
NOTE: The pH is INDEPENDENT of [HA] and [A- ] only when [HA] = [A- ].
19
The Effect of Added Acetate Ion on pH of Acetic Acid
[CH3COOH]
[CH3COO-]added
% Dissociation
pH
0.10
0.00
1.3
2.89
0.10
0.050
0.036
4.44
0.10
0.10
0.018
4.74
0.10
0.15
0.012
4.92
NOTE:
Remember, sodium acetate solution is basic !
Increasing concentration of acetate increases pH of solution!
20
The ammonia-ammonium ion buffer has a pH of
about 9.2 and can be used to keep solutions in the
basic pH range. What mass of ammonium chloride
must be added to 400.0 mL of a 3.00 M ammonia
solution to prepare a buffer?
Kb (NH3) = 1.8 x 10-5
1.
2.
3.
4.
5.
1.20 g
20.44 g
21.65 g
62.94 g
64.19 g
21
Preparing a Buffer
Problem: The ammonia-ammonium ion buffer has a pH of about 9.2
and can be used to keep solutions in the basic pH range. What mass of
ammonium chloride must be added to 400.0 mL of a 3.00 M ammonia
solution to prepare a buffer?
Plan: We want to add sufficient ammonium ion to equal the aqueous
ammonia concentration.
Solution:
NH
+H O
NH + + OH3 (aq)
Kb =
[NH4+] [OH-]
[NH3]
2
(l)
4 (aq)
(aq)
= 1.8 x 10-5
NH3 = 3.00 mol x 0.4000 L = 1.20 mol…same amount of NH4+ needed
L
NH4Cl = 53.49 g/mol
Therefore mass = 1.20 mol x 53.49 g/mol
= 64.19 g NH4Cl 22
Instructions for making a buffer say to mix 60.0
mL of 0.100 M NH3 with 40.0 mL of 0.100 M
NH4Cl. What is the pH of this buffer?
Kb (NH3) = 1.8 x 10-5
1.
2.
3.
4.
2.7 x 10-5
3.7 x 10-10
4.57
9.43
23
Like Example 8.3 – I
Problem: Instructions for making a buffer say to mix 60.0 mL of
0.100 M NH3 with 40.0 mL of 0.100 M NH4Cl. What is the pH of this
buffer?
The combined volume is 60.0 mL + 40.0 mL = 100.0 mL
Moles of ammonia = VNH3 x MNH3 = 0.060 L x 0.100 M = 0.0060 mol
Moles of ammonium ion = VNH4Cl x MNH4Cl = 0.040 L x 0.100 M
= 0.0040 mol
[NH3] =
0.0060 mol
= 0.060 M ;
0.100 L
[NH4+] =
Concentration (M) NH3 (aq) + H2O(l)
Starting
0.060
Change
-x
Equilibrium
0.060 – x
0.0040 mol
0.100 L
= 0.040 M
NH4+(aq) + OH-(aq)
0.040
0
+x
+x
0.040 + x
x 24
Like Example 8.3 - II
Substituting into the equation for Kb:
Kb =
[NH4+] [OH-]
[NH3]
= 1.8 x 10-5 =
(0.040 + x) (x)
(0.060 – x)
Assume : 0.060 – x = 0.060 and 0.040 + x = 0.040
Kb = 1.8 x 10-5 =
Check assumptions:
0.040 (x)
0.060
x = 2.7 x 10-5
0.040 + 0.000027 = 0.040 or 0.068%
0.060 – 0.000027 = 0.060 or 0.045%
[OH-] = 2.7 x 10-5 ; pOH = - log[OH-] = - log (2.7 x 10-5) = 4.57
pH = 14.00 – pOH = 14.00 – 4.57 = 9.43
25
Like Example 8.3 - III
OR, using the H-H equation:
pH = pKa + log
K
Ka = Kw = 5.6 x 10-10
b
pH = 9.25 + log
pH = 9.25 + log
[A ]
([HA]
)
-
pKa = -log 5.6 x 10-10 = 9.25
(
(
[NH3]
[NH4+]
0.060
0.040
)
)
pH = 9.25 + log 1.5 = 9.43
26
pH and Capacity of Buffered Solutions
The pH of a buffered solution is determined by the ratio[A-]/[HA].
The buffer capacity of a buffered solution is determined by the
magnitudes of [HA] and [A-]
The HIGHER the concentration of buffer components
the LARGER the buffering capacity.
The more concentrated the buffer is the more it will resist pH change.
27
The Relation Between Buffer
Capacity and pH Change
28
Example 8.5 - I
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 acetic acid/acetate buffer
solutions. For acetic acid, Ka = 1.8 x 10-5
Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Solution B: 0.050 M HC2H3O2 and 0.0500 M NaC2H3O2
Use the Henderson-Hasselbalch equation for initial pH:
pH = pKa + log
( [H[C CHHOO] ] )
2
3
2
2
3
-
2
Since [C2H3O2-] = [H C2H3O2]
The equation becomes:
pH = pKa + log (1) = pKa = -log(1.8 x 10-5) = 4.74
Adding 0.010 mol of HCl will cause a shift in the equilibrium due to:
H+(aq) + C2H3O2-(aq)
HC2H3O2 (aq)
29
Example 8.5 - II
For Solution A:
Before reaction
After reaction
H+
+
0.010 M
0
C2H3O2-
H C2H3O2
5.00 M
4.99 M
5.00 M
5.01 M
Calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log
For Solution B:
Before reaction
After reaction
( [H[C CHHOO] ] ) = 4.74 + log ( 4.99
)
5.01
2
3
2
2
-
3
H+
2
+
0.010 M
0
The new pH is: pH = 4.74 + log
C2H3O20.050 M
0.040 M
= 4.74 – 0.0017
= 4.74
H C2H3O2
0.050 M
0.060 M
(0.040
) = 4.74 – 0.18 = 4.56
0.060
30
Summary: Characteristics of Buffered Solutions
Buffered solutions contain relatively large concentrations of a weak acid and
its corresponding base. They can involve a weak acid HA and its conjugate
base A- or a weak base B and its conjugate acid BH+.
When H+ is added to a buffered solution, it reacts essentially to completion
with the weak base present:
H + + AHA
or
H+ + B
BH+
When OH- is added to a buffered solution, it reacts essentially to completion
with the weak acid present.
OH- + HA
A- + H2O or OH- + BH+
B + H2O
The pH of the buffered solution is determined by the ratio of the
concentrations of the weak base and weak acid. As long as this ratio
remains virtually constant, the pH will remain virtually constant. This will
be the case as long as the concentrations of the buffering materials (HA/A- or
B/BH+) are large compared with the amounts of H+ or OH- added.
31
Review
TITRATIONS
• Volumetric analysis - technique by which one solution is used to
analyze another solution
• Titrant is delivered from a buret.
• The sample and an indicator are in an Erlenmeyer flask.
• The indicator changes color at the equivalence point, the point at
which the amounts of titrant and sample are stoichiometrically equal.
• The concentration of one of the solutions must be known exactly (in
order to determine moles, millimoles (1 mmol = 10-3 mol), or
equivalents added).
mol = mmol → mL * M = mmol
L
mL
• The concentration of an unknown sample is calculated from the
volume and concentration of the titrant used to reach the equivalence
point, along with the volume of the sample.
32
ACID-BASE TITRATIONS
There are three types of titration curves:
•
strong acid/strong base
•
weak acid/strong base
•
weak base/strong acid
33
STRONG ACID / STRONG BASE
The curve is symmetrical.
The equivalence point is at
pH = 7.00 (think about effect of
salts on pH/acidity).
The range of pH values is
proportional to original
concentration of analyte at the
starting point.
A titration curve is characterized
by starting point, equivalence
point, and final point.
34
SA/SB titration curve:
HCl titrated by NaOH
HCl + NaOH
NaCl + H2O
NaCl+NaOH
Note:
strong acid - strong base
system…
NaCl
neutral salt (NaCl) at
equivalence point!
HCl
pH calculations along the curve:
1)Stoichiometry with added titrant
2)pH based on what’s left in the beaker
35
Calculating pH along the Titration Curve
Initial pH: 100 mL 0.030 M HCl
pH = -log (0.030 M) = 1.52
Region #1: Before the equivalence point
100 mL 0.030 M HCl + 25 mL 0.030 M NaOH
pH determined by excess acid and new total volume
Moles
HCl(aq) +
NaOH(aq)
Initial
3.0 mmol
0.75 mmol
Change
-0.75 mmol -0.75 mmol
Equilibrium 2.25 mmol 0
NaCl
+
0
+0.75 mmol
0.75 mmol
New total volume = 100 mL + 25 mL = 125 mL
New [H+] = 2.25 mmol / 125 mL = 0.018 M
pH = -log (0.018 M) = 1.74
H2O
-------
36
Calculating pH along the Titration Curve
Region #2: At the equivalence point
100 mL 0.030 M HCl + 100 mL 0.030 M NaOH
pH determined by ions present at eq. pt. and new total volume
Moles
HCl(aq) +
Initial
3.0 mmol
Change
-3.0 mmol
Equilibrium 0 mmol
NaOH(aq)
3.0 mmol
-3.0 mmol
0 mmol
NaCl
+
0
+3.0 mmol
3.0 mmol
H2O
-------
New total volume = 100 mL + 100 mL = 200 mL
Only 0.015 M NaCl is present at the equivalence point
pH = 7 at equivalence point for SA/SB
37
Calculating pH along the Titration Curve
Region #3: After the equivalence point
100 mL 0.030 M HCl + 150 mL 0.030 M NaOH
pH determined by excess base and new total volume
Moles
HCl(aq) +
Initial
3.0 mmol
Change
-3.0 mmol
Equilibrium 0 mmol
NaOH(aq)
4.5 mmol
-3.0 mmol
1.5 mmol
NaCl
+
0
+3.0 mmol
3.0 mmol
H2O
-------
New total volume = 100 mL + 150 mL = 250 mL
[OH-] = 1.5 mmol / 250 mL = 0.0060 M
pOH = -log (0.0060 M) = 2.22
pH = 14 - pOH = 14 - 2.22 = 11.78
38
39
SB/SA TITRATION
CURVE :
titration of
100 mL of
0.5 M NaOH
by 1.0 M HCl
40
Titration Curve Calculations for
Weak Acid/Strong Base
A stoichiometry 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.
An equilibrium problem. The position of the weak acid
equilibrium is determined (I.C.E. table) and the pH is calculated.
41
Consider the salt that would be formed
by the titration of a weak acid with a
strong base. What do you expect the pH
to be at the equivalence point?
1. < 7
2. > 7
3. = 7
42
Remember
Effects of Salts on pH and Acidity
WEAK ACIDS
For any salt whose cation has neutral properties (cation
from a strong acid - such as Na+ or K+) and whose anion
is the conjugate base of a weak acid, the aqueous
solution will be basic.
Examples: NaF, KCN, NaC2H3O2, Na3PO4, etc.
43
TITRATION of WEAK ACID
The curve is asymmetrical
The equivalence point is NOT at
pH = 7.00 (pH will be > 7.00)
The range of pH values is
proportional to original
concentration of analyte at the
starting point and the Ka value of
the titrated acid!
A titration curve is characterized
by starting point, buffer region,
equivalence point, and final
point.
halfway point (or midpoint)
in the buffer region
Titration of 0.1 M acetic acid
with 0.1 M NaOH
44
What is the pH of 20.00 mL of 0.250 M
nitrous acid? (HNO2; Ka = 4.0 x 10-4)
1.
2.
3.
4.
5.
2.000
2.849
3.398
4.000
5.699
45
Calculating the pH During a Weak Acid-Strong Base
Titration – I (Like Ex 8.7)
Problem: Calculate the pH during the titration of 20.00 mL of 0.250 M
nitrous acid (HNO2; Ka = 4.0 x 10-4) after adding the following volumes of
0.150 M NaOH: (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 33.33 mL, and
(e) 40.00 mL.
Solution:
Stoichiometry
HNO2 (aq) + NaOH(aq)
Weak Acid Dissoc.
(a)
Ka =
H+(aq) + NO2-(aq)
HNO2 (aq) + H2O(l)
[H+] [NO2-]
[HNO2]
H2O(l) + NaNO2 (aq)
=
x2
0.250 M
= 4.00 x 10-4
x2 = 1.00 x 10-4
x = 1.00 x 10-2
pH = -log(1.0 x 10-2) = 2.000 starting point
46
Calculating the pH During a WA/SB Titration – II
(b) 15.00 mL of 0.150 M NaOH is added; new total volume = 35.00 mL
15.00 mL x 0.150 mmol/mL = 2.25 mmol of OH20.00 mL x 0.250 mmol/mL = 5.00 mmol of HNO2
Moles
Initial
Change
Equilibrium
HNO2 (aq) + OH-(aq)
5.00 mmol 2.25 mmol
-2.25 mmol -2.25 mmol
2.75 mmol 0
Concentration (M)
HNO2 (aq) + H2O(l)
Initial
0.0786 M
---Change
-x
---Equilibrium
0.0786 - x
---pH = pKa + log
-
[NO ]
([HNO
)]
2
H2O (l) + NO2-(aq)
---0
---+2.25 mmol
---2.25 mmol
H3O+(aq) + NO2-(aq)
0
0.0643 M
+x
+x
x
0.0643 + x
= 3.40 + log(0.0643/0.0786)
2
pH = 3.40 - 0.0872 = 3.31 after 15.0 mL of NaOH added
47
NOTE: this pH is actually before the halfway point of the titration!
Calculating the pH During a WA/SB Titration – III
(c) 20.00 mL of 0.150 M NaOH is added; new total volume = 40.00 mL
20.00 mL x 0.150 mmol/mL = 3.00 mmol of OH20.00 mL x 0.250 mmol/mL = 5.00 mmol of HNO2
Moles
Initial
Change
Equilibrium
HNO2 (aq) + OH-(aq)
5.00 mmol 3.00 mmol
-3.00 mmol -3.00 mmol
2.00 mmol
0
Concentration (M)
HNO2 (aq) + H2O(l)
Initial
0.0500 M
---Change
-x
---Equilibrium
0.0500 - x
---pH = pKa + log
-
[NO ]
([HNO
)]
2
H2O (l) + NO2-(aq)
---0
---+3.00 mmol
---3.00 mmol
H3O+(aq) + NO2-(aq)
0
0.0750 M
+x
+x
x
0.0750 + x
= 3.40 + log(0.0750/0.0500)
2
pH = 3.40 + 0.176 = 3.58 after 20.0 mL of NaOH added
48
NOTE: this pH is actually after the halfway point of the titration!
Calculating the pH During a WA/SB Titration – IV
(d) 33.33 mL of 0.150 M NaOH is added; new total volume = 53.33 mL
33.33 mL x 0.150 mmol/mL = 5.00 mmol of OH20.00 mL x 0.250 mmol/mL = 5.00 mmol of HNO2
Moles
Initial
Change
Equilibrium
HNO2 (aq) + OH-(aq)
5.00 mmol 5.00 mmol
-5.00 mmol -5.00 mmol
0
0
H2O (l) + NO2-(aq)
---0
---+5.00 mmol
---5.00 mmol
Weak Base Dissociation!!
Concentration (M)
NO2- (aq) + H2O(l)
Initial
0.0938 M
---Change
-x
---Equilibrium
0.0938 - x
----
OH-(aq) + HNO2 (aq)
0
0
+x
+x
x
x
[OH-] [HNO2]
x2
=
= 2.50 x 10-11 x2 = 2.34 x 10-12
[NO2 ]
0.0938 M
x = 1.53 x 10-6
pH = 14 – pOH = 14 – 5.815 = 8.185 after 33.33 mL of NaOH added
49
NOTE: this pH is AT the equivalence point of the titration!
Kb =
Calculating the pH During a WA/SB Titration – V
(e) 40.00 mL of 0.150 M NaOH is added; new total volume = 60.00 mL
40.00 mL x 0.150 mmol/mL = 6.00 mmol of OH20.00 mL x 0.250 mmol/mL = 5.00 mmol of HNO2
Moles
Initial
Change
Equilibrium
HNO2 (aq) + OH-(aq)
5.00 mmol 6.00 mmol
-5.00 mmol -5.00 mmol
0
1.00 mmol
H2O (l) + NO2-(aq)
---0
---+5.00 mmol
---5.00 mmol
Excess of 1.00 mmol of NaOH which will control the pH.
[OH-] = 1.00 mmol / 60.00 mL = 0.0167 M
pOH = -log (0.0167 M) = 1.778
pH = 14 - pOH = 14 – 1.778 = 12.222 when all of the acid is
neutralized and there is an excess of NaOH
50
51
Influence of Ka value
on shape of titration
curve.
Buffer region contains
halfway point (midpoint)
52
53
Consider the salt that would be formed
by the titration of a strong acid with a
weak base. What do you expect the pH
to be at the equivalence point?
1. < 7
2. > 7
3. = 7
54
Remember
Effects of Salts on pH and Acidity
WEAK BASES
A salt whose cation is the conjugate acid of a weak base
produces an acidic solution when dissolved in water.
Examples: NH4Cl, AlCl3, Na2CO3, K2S, Na2C2O4, etc.
55
Figure 8.5: pH curve for 100 mL of 0.05 M NH3 and 0.1 M HCl
B + H+ → BH+
Like the weak acid
titration, the curve is
asymmetrical.
Buffer region
(halfway point, midpoint)
The equivalence point
is NOT at pH = 7.00
(pH will be < 7.00)
56
NH4Cl
pH = 5.27 at
equivalence
point
57
The Color Change of the Indicator Bromthymol Blue
58
Figure 8.8 : Useful pH ranges for indicators (pKa ± 1)
59
For titration of
strong acid with
strong base
EITHER
phenolphtalein OR
methyl red can be
used
•Color change will be sharp
•Several suitable indicators
available
60
For titration of acetic
acid (weak acid) by a
strong base
NaOH + acetate
phenolphtalein is OK,
methyl red is not !!
Na acetate
•The weaker the acid, the
smaller the vertical area
around the equivalence point
•Look for useful pH range with
midpoint as close to pH at
equivalence point of titration
•Much less flexibility
compared to SA/SB titrations
Buffer point
Acetic acid
61
SOLUBILITY and SOLUBILITY PRODUCT
When a solution becomes saturated (no more solute will dissolve), a
precipitate forms and we can calculate the quantity of material that
remains in solution. We are working with the:
“Solubility Product”
The equilibrium constant that is used for these calculations is called the
solubility product constant:
Ksp
Example : Lead chromate
Ksp =
[Pb2+][CrO42-]
[PbCrO4]
PbCrO4 (s)
Pb2+(aq) + CrO42-(aq)
Since the activity of a solid is 1.0,
we can move it to the other side of the equal
sign and combine it with the constant yielding
the solubility product constant Ksp
Ksp = [Pb2+][CrO42-]
62
Writing Ion-Product Expressions for Slightly Soluble Ionic
Compounds
Problem: Write the ion-product expression for (a) silver bromide;
(b) strontium phosphate; (c) aluminum carbonate; (d) nickel(III) sulfide.
Plan: Write the equation for a saturated solution, then write the
expression for the solubility product.
Solution:
(a) silver bromide:
AgBr(s)
Ag+(aq) + Br -(aq)
Ksp = [Ag+] [Br -]
(b) strontium phosphate: Sr3(PO4)2(s)
3 Sr2+(aq) + 2 PO43-(aq)
Ksp = [Sr2+]3[PO43-]2
(c) aluminum carbonate:
Al2(CO3)3 (s)
3+
2
Ksp = [Al ] [CO32-]3
(d) nickel(III) sulfide:
Ni2S3 (s) + 3 H2O(l)
2 Al3+(aq) + 3 CO32-(aq)
2 Ni3+(aq) + 3 HS -(aq) + 3 OH-(aq)
Ksp =[Ni3+]2[HS-]3[OH-]3
63
64
65
Determining Ksp from Solubility
Problem: Lead chromate is an insoluble compound that at one time was
used as the pigment in the yellow stripes on highways. It’s solubility is
4.33 x 10-6 g/100 mL water. What is the Ksp?
Plan: We write an equation for the dissolution of the compound to see
the number of ions formed, then write the ion-product expression.
Solution:
PbCrO4 (s)
Pb2+(aq) + CrO42-(aq)
Molar solubility of PbCrO4 =
4.33 x 10-6 g 1000 ml
x
1L
100 mL
x
1mol PbCrO4
323.2 g
= 1.34 x 10 -8 M PbCrO4
1 Mole PbCrO4 = 1 mole Pb2+ and 1 mole CrO42Therefore [Pb2+] = [CrO42-] = 1.34 x 10-8 M
Ksp = [Pb2+] [CrO42-] = (1.34 x 10 -8 M)2 = 1.54 x 10-16
66
What is the solubility of PbCrO4 in water if
the Ksp is equal to 2.00 x 10-16?
1.
2.
3.
4.
5.
6.
1.41 x 10-16 M
2.00 x 10-16 M
1.41 x 10-8 M
2.00 x 10-8 M
2.82 x 10-8 M
4.00 x 10-8 M
67
Determining Solubility from Ksp
Problem: Calculate the solubility of PbCrO4 in water if the Ksp is equal to
2.00 x 10-16.
Plan: We write the dissolution equation and the ion-product expression.
Solution:
Pb2+(aq) + CrO42-(aq)
PbCrO4 (s)
Ksp = 2.00 x 10-16 =[Pb2+][CrO42]
Concentration (M)
PbCrO4
Pb2+
CrO42-
Initial
Change
Equilibrium
----------------------------
0
+x
x
0
+x
x
Ksp = [Pb2+] [CrO42-] = (x)(x) = 2.00 x 10-16
x = 1.41 x 10-8
Therefore the solubility of PbCrO4 in water is 1.41 x 10-8 M 68
Like Example 8.12
The Ksp value for the mineral fluorite, CaF2 is 3.4 x 10-11 . Calculate
the solubility of fluorite in units of grams per liter.
Concentration (M)
Ca2+(aq) + 2 F-(aq)
CaF2 (s)
Initial
Change
Equilibrium
0
+x
x
0
+2x
2x
Substituting into Ksp:
[Ca2+][F-]2 = Ksp
3.4 x 10-11
(x) (2x)2 = 3.4 x 10-11 x = 3
4
4x3 = 3.4 x 10-11
x = 2.0 x 10-4
The solubility is 2.0 x 10-4 moles CaF2 per liter of water. To get mass
we must multiply by the molar mass of CaF2 (78.1 g/mol).
2.0 x 10-4 mol CaF2 x
78.1 g CaF2
1 mol CaF2
= 1.6 x 10-2 g CaF2 per L
69
Relationship Between Ksp and Solubility at 25 °C
No. of Ions
2
Formula
PbSO4
Cation:Anion
1:1
Ksp
Solubility (M)
1.6 x 10-8
1.3 x 10-4
2
MgCO3
1:1
3.5 x 10-8
1.9 x 10-4
2
BaCrO4
1:1
2.1 x 10-10
1.4 x 10-5
3
Ca(OH)2
1:2
6.5 x 10-6
1.2 x 10-2
3
BaF2
1:2
1.5 x 10-6
7.2 x 10-3
3
CaF2
1:2
3.2 x 10-11
2.0 x 10-4
3
Ag2CrO4
2:1
2.6 x 10-12
8.7 x 10-5
70
A solution of silver nitrate is added to a
solution of the slightly soluble salt, silver
chromate (Ksp = 2.6 x 10-12). The concentration
of chromate ion in solution will ___________.
1. Increase
2. Decrease
3. Stay the same
71
Calculating the Effect of a Common Ion on Solubility
Problem: What is the solubility of silver chromate in 0.0600 M silver
nitrate solution? Ksp = 2.6 x 10-12
Plan: From the equation and the ion product expression for Ag2CrO4, we
predict that the addition of silver ion will decrease the solubility.
Solution: Writing the equation and ion-product expression:
Ag2CrO4 (s)
2 Ag+(aq) + CrO42-(aq)
Concentration (M)
Ag2CrO4 (s)
Initial
Change
Equilibrium
-------------------------
Ksp = [Ag+]2[CrO42-]
2 Ag+(aq)
+
0.0600
+2x
0.0600 + 2x
CrO42-(aq)
0
+x
x
Assuming that Ksp is small, 0.0600 M + 2x ≈ 0.600 M
Ksp = 2.6 x 10-12 = (0.0600)2(x)
x = 7.22 x 10-10 M
Therefore, the solubility of silver chromate here is 7.22 x 10-10 M
72
Predicting the Effect on Solubility of Adding Strong Acid
Problem: Write balanced equations to explain whether addition of H3O+
from a strong acid affects the solubility of:
(a) iron(II) cyanide, (b) potassium bromide, and (c) aluminum hydroxide
Plan: Write the balanced dissolution equation and note the anion. Anions
of weak acids react with H3O+ and shift the equilibrium position toward
more dissolution. Strong acid anions do not react, so added acid has no
effect.
Solution: (a) Fe(CN)2 (s)
Fe2+(aq) + 2 CN-(aq) Increases solubility
CN- ion reacts with H3O+ to form the weak acid HCN, so CN- would be
removed from the product side of the equation; shifts rxn to the right.
(b) KBr(s)
K+(aq) + Br -(aq) No effect
Br- is the anion of a strong acid and K+ is the cation of a strong base.
(c) Al(OH)3 (s)
Al3+(aq) + 3 OH-(aq)
Increases solubility
OH is the anion of water, a very weak acid, so it reacts with the
added acid to produce water in a simple acid-base reaction; shifts 73
rxn to the right.
Predicting the Formation of a Precipitate: Qsp vs. Ksp
The solubility product constant, Ksp, can be compared to the ion product
constant, Qsp , to understand the characteristics of a solution with respect
to forming a precipitate.
Qsp = Ksp : When a solution becomes “saturated,” no more solute will
dissolve. No changes will occur.
Qsp > Ksp : Precipitates will form until the solution becomes
saturated.
Qsp< Ksp : Solution is unsaturated; no precipitate will form.
74
Will a precipitate form when a
concentrated solution containing barium
nitrate is added to a concentrated solution
of sodium chromate?
1. Yes
2. No
3. Not enough info
75
Predicting Whether a Precipitate Will Form–I
Problem: Will a precipitate form when 0.100 L of a solution containing
0.55 M barium nitrate is added to 200.0 mL of a 0.100 M solution of
sodium chromate?
Plan: Determine if the solutions will yield soluble ions. Calculate the new
concentrations, adding the two volumes together to get the total volume of
the solution. Calculate the ion product constant (Qsp) and compare it to
the solubility product constant to see if a precipitate will form.
Solution: Both Na2CrO4 and Ba(NO3)2 are soluble, so we will have Na+,
CrO42-, Ba2+ and NO3- ions present in 0.300 L of solution. After
referencing the table of solubility rules (Ch 4), and we find that BaCrO4 is
insoluble, so we calculate it’s ion-product constant and compare it to
the solubility product constant of 2.1 x 10-10:
For Ba2+: (0.100 L Ba(NO3)2) x (0.55 M) = 0.055 mol Ba2+
0.055 mol Ba2+
[Ba2+] =
= 0.183 M in Ba2+
0.300 L
76
Predicting Whether a Precipitate Will Form–II
For CrO42- :
(0.100 M Na2CrO4)(0.200 L) = 0.0200 mol CrO42-
[CrO42-] =
0.0200 mol CrO420.300 L
= 0.667 M in CrO42-
Qsp = [Ba2+] [CrO42-] =(0.183 M Ba2+)(0.667 M CrO42-) = 0.121
Since Ksp = 2.1 x 10-10 and Qsp = 0.121,
Qsp >> Ksp so a precipitate
will form.
77
FYI
Figure 8.12 : Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+
78
FYI
The General Procedure for Separating
Ions in Qualitative Analysis
79
Separating Ions by Selective Precipitation–I
Problem: A solution consists of 0.10 M AgNO3 and 0.15 M CuNO3.
Calculate the [I -] that would separate the metal ions as their iodides.
Kspof AgI = 8.3 x 10-17; Kspof CuI = 1.0 x 10-12
Plan: Since the two iodides have the same formula type (1:1), we compare
their Ksp values and we see that CuI is about 100,000 times more soluble
than AgI. Therefore, AgI precipitates first, and we solve for [I -] that will
give a saturated solution of AgI.
Solution: Writing chemical equations and ion product expressions:
AgI(s)
CuI(s)
H2 O
H2 O
Ag+(aq) + I -(aq)
Ksp = [Ag+][I -]
Cu+(aq) + I -(aq)
Ksp = [Cu+][I -]
Calculating the quantity of iodide needed to give a saturated solution
of CuI:
[I -] =
Ksp
[Cu+]
=
1.0 x 10-12
0.15 M
= 6.7 x 10-11 M
80
Separating Ions by Selective Precipitation–II
Thus, the concentration of iodide ion that will give a saturated solution
of copper(I) iodide is 6.7 x 10-11 M, and this will not precipitate the
copper(I) ion, but should remove most of the silver ion. Calculating the
quantity of silver ion remaining in solution we get:
[Ag+] =
Ksp
[I -]
=
8.3 x 10-17
6.7 x 10-11
= 1.2 x 10-6 M
Since the initial silver ion was 0.10 M, most of it has been removed,
and essentially none of the copper(I) was removed, so the separation
was quite complete. If the iodide was added as sodium iodide, you
would have to add only a few nanograms of NaI to remove nearly all
of the silver from solution:
6.7 x 10-11 mol I - x
1 mol NaI
mol I -
x
149.9 g NaI
mol NaI
= 10.0 ng NaI
81
Complex ion: A charged
species consisting of a
metal ion surrounded by
ligands
Ligand: A neutral molecule or ion having a lone pair that
82
can be used to form a covalent bond to a metal ion
The Stepwise Exchange of NH3 for H2O
in M(H2O)42+
Formation (stability) constants: equilibrium constants
for the addition of ligands to metal ions
83
Formation Constants (Kf) of Some Complex Ions at 25oC
Complex Ion
Ag(CN)2Ag(NH3)2+
Ag(S2O3)23AlF63Al(OH)4Be(OH)42CdI42Co(OH)42Cr(OH)4Cu(NH3)42+
Kf
3.0 x 1020
1.7 x 107
4.7 x 1013
4 x 1019
3 x 1033
4 x 1018
1 x 106
5 x 109
8.0 x 1029
5.6 x 1011
Complex Ion
Fe(CN)64Fe(CN)63Hg(CN)42Ni(OH)42Pb(OH)3 Sn(OH)3 Zn(CN)42Zn(NH3)42+
Zn(OH)42-
Kf
3 x 1035
4.0 x 1043
9.3 x 1038
2 x 1028
8 x 1013
3 x 1025
4.2 x 1019
7.8 x 108
3 x 1015
84