Chapter 8
Applications of Aqueous
Equilibria
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
A base swirling in a solution
containing phenolphthalein
Le Châtelier’s principle for the dissociation
equilibrium for HF
HF(aq)
H+(aq) + F-(aq)
Molecular model: F-, Na+, HF, H2O
Like Example 8.1 (P 278-9) - 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)
H+(aq) + NO2-(aq)
[H+] [NO2 -]
Ka = [HNO ] = 4.0 x 10-4
2
Initial Concentration (mol/L)
Equilibrium Concentration (mol/L)
[HNO2]0 = 1.0 M
(from dissolved HNO2)
[NO2-]0 = 1.0 M
(from dissolved NaNO2)
[H+]0 = 0
(neglect the contribution from water)
[HNO2] = 1.0 – x
[NO2-] = 1.0 + x
[H+] = x
Like Example 8.1 (P 274-5) - II
[H+] [NO2-]
Ka =
[HNO2] =
( x ) ( 1.0 + x )
-4
=
4.0
x
10
(1.0 – x )
Assume x is small as compared to 1.0:
X (1.0)
= 4.0 x 10-4
(1.0)
or
x = 4.0 x 10-4
= [H+]
Therefore pH = - log [H+] = - log ( 4.0 x 10-4 ) = _________
The percent dissociation is:
4.0 x 10-4
1.0
x 100 = _______ %
Nitrous acid
alone
[H+]
2.0 x 10-2
pH
1.70
% Diss
2.0
Nitrous acid
+ NaNO2
4.0 x 10-4
_______
_______
Example 8.2 (P279-82) - I
A buffered solution contains 0.50 M acetic acid (HC2H3O2,
Ka = 1.8 x 10-5) and o.50 M sodium acetate (NaC2H3O2). Calculate the
pH of this solution, and the pH when 0.010 M of solid NaOH is added
to 1.0 L of this buffer and to pure water.
H+(aq) + C2H3O2 (aq)
HC2H3O2 (aq)
Ka = 1.8 x 10-5 =
Initial Concentration (mol/L)
[HC2H3O2]0 = 0.50
[C2H3O2-]0 = 0.50
[H+]0 =~ 0
[H+] [C2H3O2-]
[HC2H3O2]
Equilibrium Concentration (mol/L)
X mol/L of HC2H3O2
dissociates to reach
equilibrium
[HC2H3O2] = 0.50 – x
[C2H3O2-] = 0.50 +x
[H+] = x
Example 8.2 (P279-82) - II
+][C H O -]
[H
( x ) ( 0.50 + x) ~ (x) (0.50)
2 3 2
-5
Ka = 1.8 x 10 =
=
=
[HC2H3O2]
0.50 - x
0.50
x = 1.8 x 10-5
The approximation by the 5% rule is fine:
[H+] = x = 1.8 x 10-5 M
and
pH = 4.74
To calculate the pH and concentrations after adding the base:
Before
reaction:
OH- + HC2H3O2
0.010 mol
0.50 mol
After
reaction: 0.010 – 0.010 0.50 – 0.10
= 0 mol
= 0.49 mol
H2O + C2H3O2-
0.50 mol
-
0.50 + 0.10
= 0.51 mol
Note that 0.01 mol of acetic acid has been converted to acetate ion
by the addition of the base.
Example 8.2 (P279-82) - III
Initial Concentration (mol/L)
[HC2H3O2]0 = 0.49
[C2H3O2-]0 = 0.51
[H+]0 = 0
Ka = 1.8 x 10-5 =
Equilibrium concentration (mol/L)
X mol/L of HC2H3O2
Dissociates to reach
equilibrium
[H+][C2H3O2-]
[HC2H3O2]
x = ______________
[HC2H3O2] = 0.49 – x
[C2H3O2-] = 0.51 + x
[H+] = x
= (x)(0.51+ x) =
0.49 - x
(x)(0.51)
0.49
and pH = ___________
If the base is added to pure water without the buffer being present
we get an entirely different solution:
If the 0.01 mol of NaOH is added to 1.0 L of pure water the
Concentration of hydroxide ion is 0.01 M.
-14
Kw
1.0
x
10
+
[H ] =
=
= __________ and the pH = ______
-2
[OH-]
1.0 x 10
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 equilibrium
calculations. This procedure can be
represented as follows:
OH- ions are not allowed to accumulate
but are replaced by A- ions.
When the OH- is added, the concentrations of HA
and A- change, but only by small amounts. Under
these conditions the [HA]/[A-] ratio and thus the
[H+] stay virtually constant.
How Does a Buffer Work
Lets add a strong base to a weak acid and see what happens:
OH- + HA
A- + H2O
Final pH of buffer
close to original-
Original
buffer pH
[H+] [A-]
Ka =
[HA]
Added OH- ions
Replaced by A- ions
[H+]
[HA]
= Ka [A-]
The Effect of Added Acetate Ion on the Dissociation
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
[CH3COOH]dissoc
% Dissociation =
x 100
[CH3COOH]init
Human blood is
a buffered
solution
CO2 (g), H2CO3 (aq) and
HCO3-(aq) are the buffering
components in Blood that
hold the pH to a range
that will allow Hemoglobin
to transport oxygen from
the lungs to the cells of the
body for metabolism.
Source: Visuals Unlimited
Pure water at pH 7.00
Molecular model: HC2H3O2, C2H3O2-
How a Buffer Works–I
A buffer consists of a solution that contains “high” concentrations of
the acidic and basic components. This is normally a weak acid and the
anion of that weak acid, or a weak base and the corresponding cation of
the weak base. When small quantities of H3O+ or OH- are added to the
buffer, they cause a small amount of one buffer component to convert
into the other. As long as the amounts of H3O+ and OH- are small as
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.
Consider a buffer made from acetic acid and sodium acetate:
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
Ka =
[CH3COO-] [H3O+]
[CH3COOH]
or
[H3
O+]
= Ka x
[CH3COOH]
[CH3COO-]
How a Buffer Works–II
Let’s consider a buffer made by placing 0.25 mol of acetic acid and
0.25 mol of sodium acetate per liter of solution. What is the pH of the
buffer? And what will be the pH of 100.00 mL of the buffer before and
after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? What
will be the pH of 300.00 mL of pure water if the same acid is added?
[H3
O+]
= Ka x
[CH3COOH]
= 1.8 x
10-5
x
(0.25)
= 1.8 x 10-5
[CH3COO-]
(0.25)
pH = -log[H3O+] = -log(1.8 x 10-5) = pH = _____ Before acid added!
1.00 mL conc. HCl
1.00 mL x 12.0 mol/L = 0.012 mol H3O+
Added to 300.00 mL of water :
0.012 mol H3O+ = 0.0399 M H O+
3
301.00 mL soln.
pH = -log(0.0399 M)
pH = _____ Without buffer!
How a Buffer Works–III
After acid is added:
Conc. (M)
CH3COOH(aq) + H2O(aq)
Initial
0.250
---Change
+0.012
---Equilibrium
0.262
---Solving for the quantity ionized:
Conc. (M)
CH3COOH(aq) + H2O(aq)
Initial
Change
Equilibrium
0.262
-x
0.262 - x
----------
CH3COO- + H3O+
0.250
0
-0.012
0.012
0.238
0.012
CH3COO- + H3O+
0.238
+x
0.238 + x
0
+x
x
Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238
[CH3COOH]
+
[H3O ] = Ka x
=1.8 x 10-5 x (0.262) = 1.982 x 10-5
(0.238)
[CH3COO-]
pH = -log(1.982 x 10-5) = 5.000 - 0.297 = _____After the acid is added!
How a Buffer Works–IV
Suppose we add 1.0 mL of a concentrated base instead of an acid. Add
1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what
the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OHThis will reduce the quantity of acid present and force the equilibrium
to produce more hydronium ion to replace that neutralized by the
addition of the base!
Conc. (M)
Initial
Change
Equilibrium
CH3COOH(aq) + H2O(aq)
0.250
---- 0.012
---0.238
----
CH3COO- + H3O+
0.250
0
+0.012 +0.012
0.262
+0.012
Assuming: Again, using x as the quantity of acid dissociated we get:
our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238
[H3O+] = 1.8 x 10-5 x
0.238
0.262
= 1.635 x 10-5
pH = -log(1.635 x 10-5) = 5.000 - 0.214 = _____ After base is added!
How a Buffer Works–V
By adding the 1.00mL base to 300.00 mL of pure water we would get a
hydroxide ion concentration of:
0.012
mol
OH
=
= 3.99 x 10-5 M OH301.00 mL
The hydrogen ion concentration is:
-14
Kw
1
x
10
+
-10
[H3O ] =
=
=
2.506
x
10
[OH-]
3.99 x 10-5 M
[OH-]
This calculates out to give a pH of:
pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the
base in pure water!
In summary:
Buffer alone pH = 4.74
Buffer plus 1.0 mL base pH = 4.79
Buffer plus 1.0 mL acid pH = 4.70
Base alone pH = 9.59
Acid alone pH = 1.40
The Relation Between Buffer
Capacity and pH Change
A digital pH meter shows the pH of the
buffered solution to be 4.74
When 0.01 mol NaOH is added to 1.0 L of
pure water, the pH jumps to 12.00
Molecular model: Cl-, NH4+
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 400.00 mL of a 3.00 M ammonia
solution to prepare a buffer ?
Plan: The conjugate pair is the ammonia-ammonium ion pair which has
an equilibrium constant Kb = 1.8 x 10 -5. The reaction equation with water
can be written along with the Kb expression, since we want to add
sufficient ammonium ion to equal the aqueous ammonia concentration.
Solution: The reaction for the ammonia-ammonium ion buffer is:
NH3 (aq) + H2O(l)
NH4+(aq) + OH-(aq)
[NH4+] [OH-]
Kb =
= 1.8 x 10-5
[NH3]
[NH4+] = 3.00 mol x 0.400 L = 1.20 mol
L
NH4Cl = 53.49 g/mol
Therefore mass =NH4Cl
= 1.20 mol x 53.49g/mol
mass = ________ g NH4Cl
The Henderson-Hasselbalch Equation
Take the equilibrium ionization of a weak acid:
HA(aq) + H2O(aq) = H3O+(aq) + A-(aq)
Ka =
Solving for the hydronium ion concentration gives:
[H3O+] = Ka x
[H3O+] [A-]
[HA]
[HA]
[A-]
Taking the negative logarithm of both sides:
( )
-log[H3O +] = -log Ka - log [HA]
[A-]
( )
pH = -log Ka - log [HA]
[A-]
Generalizing for any conjugate acid-base pair :
( )
[base]
pH = log Ka + log [acid]
Henderson-Hasselbalch
equation
Like Example 8.3 (P 285-7) -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 = VolNH3 x MNH3 = 0.060 L x 0.100 M = 0.0060 mol
Moles of Ammonium ion = VolNH4Cl x MNH4Cl = 0.040 L x 0.100 M =
= 0.0040 mol
0.0060 mol
0.0040 mol
+
[NH3] = 0.100 L = 0.060 M ; [NH4 ] = 0.100 L = 0.040 M
Concentration (M) NH3 (aq) + H2O(l)
Starting
0.060
Change
-x
Equilibrium
0.060 – x
NH4+(aq) + OH-(aq)
0.040
0
+x
+x
0.040 – x
x
Like Example 8.3 (P 285-7) - II
Substituting into the equation for Kb:
[NH4+] [OH-]
(0.040 + x) (x)
-5
Kb =
= 1.8 x 10 =
[NH3]
(0.060 – x)
Assume : 0.060 – x =~ 0.060 ; 0.040 + x =~ 0.040
Kb = 1.8 x
10-5
0.040 (x)
=
0.060
Check assumptions:
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) = 5 – 0.43
pOH = 4.57
pH = 14.00 – pOH = 14.00 – 4.57 = ___________
pH Box
pH
[H3O+] = 10
-pH
pH = -log[H3O+]
Kw = 1 x 10-14
@ 25oC
[H3O+][OH-]=
=1 x 10-14
pH + pOH =
= 14 @ 25oC
pOH
[H3O+]
[OH-]
= 10
-pOH
pOH = -log[OH-]
[OH-]
Summary: Characteristics of Buffered Solutions
Buffered solutions contain relatively large concentrations of a weak acid and its
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+.
When H+ is added to a buffered solution, it reacts essentially to completion with
the weak base present:
H+ + A HA
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 and A- or G and BH+) are large
compared with the amounts of H+ or OH- added.
Exact Treatment of Buffer Solutions
We can use several relationships to calculate the exact solution to
buffered solution problems:
Charge – balance equation:
[Na+] + [H+] = [A-] + [OH-]
Material – Balance equation:
[A-]0 + [HA]0 = [HA] + [A-]
Since [A-]0 = [Na+] and Kw = [OH-][H+] , we can rewrite the charge
balance equation, and solve for [A-] :
+] 2 – K
[H
w
[A ] = [A ]0 +
[H+]
From the mass balance equation solved for [HA] we get:
[HA] = [A-]0 + [HA]0 – [A-]
Substituting the expression for [A-], and substituting into the Ka
expression for HA we obtain:
+] 2 – K
[H
w
[H+]{ [A-]0 +
}
+]
+
[H
[H ][A ]
Ka =
=
+] 2 – K
[HA]
[H
w
[HA]0 [H+]
Example 8.4 (P 289-90) - I
Calculate the pH of a buffered solution containing 3.0 x 10-4 M HOCl
(Ka = 3.5 x 10-8) and 1.0 x 10-4 M NaOCl.
[H+] [OCl-]
Ka =
[HOCl]
3.5 x
10-8
= 3.5 x 10-8
[H+] [OCl-]
=
=
[HOCl]
Let x = [H+] then:
[OCl-] = 1.0 x 10-4 + x
[HOCl] = 3.0 x 10-4 - x
(x)(1.0 x 10-4 + x)
(3.0 x 10-4 – x )
Assuming x is small compared to 1.0 x 10-4 and solving for x we have:
[H+]
-11
1.05
x
10
-7 M = 1.1 x 10-7 M
=x=
=
1.05
x
10
1.0 x 10-4
Since this is close to that of water we must use the equation that uses
water, and takes it’s ionization into account.
Example 8.4 (P 289-90) - II
Ka = 3.5 x 10-8 =
+]2 – 1.0 x 10-14
[H
[H+]{[OCl-]0 +
[H+]
[HOCl]0 Where:
}
[H+]2 – 1.0 x 10-14
[H+]
[OCl-]0 = 1.0 x 10-4 M
[HOCl]0 = 3.0 x 10-4 M
We expect [H+] to be close to 1.0 x 10-7, so [H+]2 to be about 1.0 x 10-14
+]2 – 1.0 x 10-14
[H
-4
[OCl ]0 = 1.0 x 10 M >>>
[H+]
+]2 – 1.0 x 10-14
[H
[HOCl]0 = 1.0 x
M >>>
[H+]
The expression becomes:
+][OCl-]
+](1.0 x 10-4)
[H
[H
3.5 x 10-8 =
=
[HOCl]
(3.0 x 10-4)
10-4
Example 8.4 (P 289-90) - III
+][OCl-]
[H
3.5 x 10-8 =
[HOCl]
[H+](1.0 x 10-4)
=
(3.0 x 10-4)
[H+] = 1.05 x 10-7 M = 1.1 x 10-7 M
Using this result, we can check the magnitude of the neglected term:
[H+]2
– 1.0 x
[H+]
10-14
=
(1.05 x 10-7)2 – 1.0 x 10-14
1.05 x 10-7
This result suggests that the approximation was fine!
= 9.8 x 10-9
pH and Capacity of Buffered Solutions
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-]
Example 8.5 (P 290-2) - 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 solutions:
Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Solution B: 0.050 M HC2H3O2 and 0.0500 M NaC2H3O2
For Acetic acid, Ka = 1.8 x 10-5
Use the Henderson-Hasselbalch equation for initial pH:
[C2H3O2-]
pH = pKa + log{
}
[H C2H3O2]
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)
H C2H3O2 (aq)
Original solution and new solution
Original solution and new solution
Example 8.5 (P 290-2) - II
For Solution A:
H+
Before reaction
After reaction
0.010 M
0
+
C2H3O2-
5.00 M
4.99 M
H C2 H 3 O 2
5.00 M
5.01 M
Calculate the new pH using the Henderson-Hasselbalch equation:
-]
4.99
[C
H
O
2
3
2
pH = pKa + log (
) = 4.74 + log ( 5.01 ) = 4.74 – 0.0017
[H C2H3O2]
= 4.74
For Solution B:
Before reaction
After reaction
H+
0.010 M
0
+
C2H3O2-
H C2H3O2
0.050 M
0.040 M
0.050 M
0.060 M
The new pH is: pH = 4.74 + log( 0.040 ) = 4.74 – 0.18 = 4.56
0.060
Solution A and Solution B
A buret valve
Source: American Color
Figure 8.1: The pH curve for the titration of
50.0 ml of Nitric acid with 0.10M NaOH
Vol NaOH added (mL)
Figure 8.2: The pH curve for the titration of 100.0
ml of 0.50 M NaOH with 0.10 M HCl.
Weak acid
Figure 8.3: The pH curve for the titration of 50.0
ml of Acetic acid with 0.10 M NaOH
Treat the stoichiometry
Figure 8.4: The pH curves for the
titrations of 50.0
Calculating the pH During a Weak Acid-Strong Base
Titration–I
Problem: Calculate the pH during the titration of 20.00 mL of 0.250 M
nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of
0.150 M NaOH : (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 35.00 mL.
Plan: (a) We just calculate the pH of a weak acid. (b)-(d) We calculate
the amounts of acid remaining after the reaction with the base, and the
anion concentration, and plug these into the Henderson-Hasselbalch eq.
Solution:
HNO2 (aq) + NaOH(aq)
H2O(l) + NaNO2 (aq)
HNO2 (aq) + H2O(l)
(a)
[H3O+] [NO2-]
x (x)
Ka =
=
[HNO2]
0.250 M
H3O+(aq) + NO2-(aq)
= 4.5 x 10-4
x2 = 1.125 x 10-4
x = 1.061 x 10-2
pH = -log(1.061 x 10-2) = 2.000 - 0.0257 = _______ no base added
Calculating the pH During a Weak Acid-Strong Base
Titration–II
(b) 15.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M
HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will
neutralize (15.00 mL x 0.150 mmol/mL = 2.25 mmol of HNO2), leaving
2.75 mmol HNO2, and generating 2.25 mmol of nitrite anion.
Concentration (M)
Initial
Change
Equilibrium
HNO2 (aq) + H2O(l)
0.00275
-x
0.00275 - x
(
-]
[NO
2
pH = pKa + log
[HNO2]
)
----------
H3O+(aq) + NO2-(aq)
0
+x
x
0.00225
+x
0.00225 + x
= 3.35 + log(0.00225/0.00275)
pH = 3.35 -0.0872 = ________ with 15.0 mL of NaOH added
Calculating the pH During a Weak Acid-Strong Base
Titration–III
(c) 20.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M
HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will
neutralize (20.00 mL x 0.150 mmol/mL = 3.00 mmol of HNO2), leaving
2.00 mmol HNO2, and generating 3.00 mmol of nitrite anion.
Concentration (M)
HNO2 (aq) + H2O(l)
H3O+(aq) + NO2-(aq)
Initial
0.00200
---x
-]
[NO
2
pH = pKa + log
= 3.35 + log(0.00300/0.00200)
[HNO2]
(
)
0.00300
pH = 3.35 + 0.176 = ____________ with 20.00 mL of base added
(d) 35.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M
HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will
neutralize (35.00 mL x 0.150 mmol/mL = 5.25 mmol of HNO2), leaving
no HNO2, and generating 5.00 mmol of nitrite anion. There will be an
excess of 0.25 mmol of NaOH which will control the pH.
Calculating the pH During a Weak Acid-Strong Base
Titration–IV
(d) continued
Since all of the HNO2 has been neutralized, we only have to look at
the concentration of hydroxide ion in the total volume of the solution to
calculate the pH of the resultant solution.
combined volume = 20.00 mL + 35.00 mL = 55.00 mL
0.000250 mol OH[OH ] =
=0.004545 M
0.05500 L
-14
Kw
1
x
10
+
-12
[H3O ] =
=
=
2.200
x
10
[OH-]
0.004545
pH = -log (2.200 x 10-12) = 12.000 - 0.342 = ______ when all of the acid
neutralized, and there
is an excess of NaOH
Summary: Titration Curve Calculations
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, and the pH is calculated.
Figure 8.5: The pH curve for the titration of
100.0 ml of 0.050 M NH3 with ).10 M HCl
Figure 8.6: The indicator phenolphthalein is pink
in basic solution and colorless in acidic solution.
Figure 8.7: (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.
Figure 8.8: The useful pH ranges for
several common indicators
Colors and Approximate pH Range of
Some Common Acid-Base Indicators
Figure 8.9:
pH curve of
0.10 M HCI
being titrated
with 0.10 M
NaOH
Figure 8.10:
pH of 0.10 M
HC2H3O2
being titrated
with NaOH
Figure 8.11: A summary of the important
equilibria at various points in the titration of a
triprotic acid
An X ray of the upper gastrointestinal
Source: Photo Researchers, Inc.
Precipitation of bismuth
Equilibria of Slightly Soluble Ionic Compounds
When a solution becomes saturated and a precipitate forms we move into
the area of insoluble material in solution, and we begin to calculate the
quantity of material that remains in solution. We are working with what
we call the:
“Solubility Product”
The equilibrium constant that is used for these calculations is called the
Solubility-product constant: Ksp
Example : Lead chromate
Qc =
[Pb2+][CrO42-]
[PbCrO4]
PbCrO4 (s)
Pb2+(aq) + CrO42-(aq)
Since the concentration of a solid is constant,
we can move it to the other side of the equals
sign and combine it with the constant yielding
the solubility product constant Ksp
[PbCrO4] x Qc = [Pb2+][CrO42-] = Ksp
Like Example 8.12 (P 320)
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)
CaF2 (s)
Starting
Change
Equilibrium
Substituting into Ksp: [Ca2+][F-]2 = Ksp
(x) (2x)2 = 3.4 x 10-11
4x3 = 3.4 x 10-11
Ca2+(aq) + 2 F-(aq)
0
+x
x
x=
0
+2x
2x
3
3.4 x 10-11
4
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).
78.1 g CaF2
-4
2.0 x 10 mol CaF2 x 1 mol CaF
= _________ g CaF2 per L
2
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)(s)
3 Sr2+(aq) + 2 PO43-(aq)
Ksp = [Sr2+]3[PO43-]2
(c) Aluminum carbonate:
Al2(CO3)3 (s)
2 Al3+(aq) + 3 CO32-(aq)
Ksp = [Al3+]2[CO32-]3
(d) Nickel(III) sulfide:
Ni2S3 (s) + 3 H2O(l)
2 Ni3+(aq) + 3 HS -(aq) + 3 OH-(aq)
Ksp =[Ni3+]2[HS-]3[OH-]3
Table 8.5: Ksp Values at 25 C for
Common Ionic Solids
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 -6g/100mL 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)
-6g
4.33
x
10
Molar solubility of PbCrO4 =
x 1000 ml x 1mol PbCrO4
1L
100 mL
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 = ________________
Determining Solubility from Ksp
Problem: Lead chromate used to be used as the pigment for the yellow
lines on roads, and is a very insoluble compound. 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: Writing the dissolution equation, and the ion-product
expression: PbCrO4 (s)
Pb2+(aq) + CrO42-(aq)
Ksp = 2.00 x 10-16 =[Pb2+][CrO42]
Concentration (M)
PbCrO4
Pb2+
Initial
Change
Equilibrium
----------------------------
0
+x
x
Ksp = [Pb2+] [CrO42-] = (x)(x ) = 2.00 x 10-16
CrO420
+x
x
x = 1.41 x 10-8
Therefore the solubility of PbCrO4 in water is _______________ M
Relationship Between Ksp and Solubility at 25oC
No. of Ions
2
Formula
MgCO3
Cation:Anion
Ksp
Solubility (M)
1:1
3.5 x 10-8
1.9 x 10-4
2
PbSO4
1:1
1.6 x 10-8
1.3 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
The Effect of a Common Ion on Solubility
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
PbCrO4(s)
Pb2+(aq) + CrO42-(aq;
added)
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)
Concentration (M)
Initial
Change
Equilibrium
2 Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-]
Ag2CrO4 (s)
2 Ag+(aq)
-------------------------
0.0600
+2x
0.0600 + 2x
+
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 is 7.22 x 10-10 M
CrO42-(aq)
0
+x
x
Test for the Presence of a Carbonate
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
(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-(a) Increases solubility
We noted earlier that CN- ion reacts with water to form the weak acid
HCN, so it would be removed from the solubility expression.
(b) KBr(s)
K+(aq) + Br -(aq) No effect This occurs since 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
The 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.
The Chemistry of Limestone Formation
Gaseous CO2 is in equilibrium with aqueous CO2 in natural waters:
CO2 (g)
H2O(l)
CO2 (aq)
The concentration of CO2 is proportional to the partial pressure
of CO2 (g) in contact with the water (Henry’s law; section 13.3):
[CO2 (aq)] (proportional to) PCO2
The reaction of CO2 with water produces H3O+:
CO2 (aq) + 2 H2O(l)
H3O+(aq) + HCO3-(aq)
Thus, the presence of CO2 (aq) forms H3O+, which increases the
solubility of CaCO3:
CaCO3 (s) + CO2 (aq) + H2O(l)
Ca2+(aq) + 2 HCO3-(aq)
Predicting the Formation of a Precipitate: Qsp vs. Ksp
The solubility produce 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, and the solution is called “saturated.” There
will be no changes that will occur.
Qsp > Ksp : Precipitates will form until the solution becomes
saturated.
Qsp< Ksp : Solution is unsaturated, and no precipitate will form.
Predicting Whether a Precipitate Will Form–I
Problem: Will a precipitate form when 0.100 L of a solution
containing 0.055 M barium nitrate is added to 200.00 mL of a 0.100 M
solution of sodium chromate?
Plan: We first see if the solutions will yield soluble ions, then we
calculate the concentrations, adding the two volumes together to get the
total volume of the solution, then we calculate the product constant
(Qsp), and compare it to the solubility product constant to see if
a precipitate will form.
Solution: Both Na2CrO4 and Ba(NO3) are soluble, so we will have Na+,
CrO42-, Ba2+ and NO3- ions present in 0.300 L of solution. We change
partners, look up solubilities, and we find that BaCrO4 would be
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] [0.55 M] = 0.055mol Ba2+
2+
0.055
mol
Ba
2+
[Ba ] =
= ___________ M in Ba2+
0.300 L
Predicting Whether a Precipitate Will Form–II
Solution cont.
For CrO42- : [0.100 M Na2CrO4] [0.200 L] = 0.0200 mol CrO42[CrO42-] =
0.0200 mol CrO420.300 liters
= 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 and a precipitate
will form.
Figure 8.12: Separation of Cu2+ and Hg2+
Figure 8.13: Separation of common cations by
precipitation
The Stepwise Exchange of NH3 for H2O
in M(H2O)42+
Formation Constants (Kf) of Some Complex Ions
at 25oC–I
Complex Ion
Kf
Ag(CN)2Ag(NH3)2+
Ag(S2O3)23AlF63Al(OH)4Be(OH)42CdI42Co(OH)42Cr(OH)4Cu(NH3)42+
Fe(CN)64Fe(CN)63-
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
3 x 1035
4.0 x 1043
Formation Constants (Kf) of
Some Complex Ions at 25oC–II
Complex Ion
Hg(CN)42Ni(OH)42Pb(OH)3 Sn(OH)3 Zn(CN)42Zn(NH3)42+
Zn(OH)42-
Kf
9.3 x 1038
2 x 1028
8 x 1013
3 x 1025
4.2 x 1019
7.8 x 108
3 x 1015
Calculating the Concentrations of Complex Ions–I
Problem: A chemist converts Ag(H2O)2+ to the more stable form
Ag(NH3)2+ by mixing 50.0 L of 0.0020 M Ag(H2O)2+ and 25.0 L of
0.15 M NH3. What is the final [Ag(H2O)2+]? Kf Ag(NH3)2+ = 1.7 x 107.
Plan: We write the equation and the Kf expression, set up the table
for the calculation, then substitute into Kf and solve.
Solution: Writing the equation and Kf expression:
Ag(H2O)2+(aq) + 2 NH3 (aq)
Ag(NH3)2+(aq) + 2 H2O(l)
Kf =
[Ag(NH3)2+]
[Ag(H2O)2+][NH3]2
= 1.7 x 107
Finding the initial concentrations:
[Ag(H2O)2+]init = 50.0 L (0.0020 M) = 1.3 x 10-3 M
50.0 L + 25.0 L
[NH3]init = 25.0 L (0.15 M) = __________ M
50.0 L + 25.0 L
Calculating the Concentration of Complex Ions–II
We assume that all of the Ag(H2O)2+ is converted Ag(NH3)2+, so we set
up the table with x = [Ag(H2O)2+] at equilibrium.
Ammonia reacted = [NH3]reacted = 2(1.3 x 10-3 M) = 2.6 x 10-3 M
Concentration (M) Ag(H2O)2+(aq) 2NH3 (aq)
Initial
Change
Equilibrium
1.3 x 10-3
~(-1.3 x 10-3)
x
Ag(NH3)2+ 2 H2O(aq)
5.0 x 10-2
0
---~(-2.6 x 10-3) ~(+1.3 x 10-3) ---4.7 x 10-2
1.3 x 10-3
----
[Ag(NH3)2+]
-3
1.3
x
10
Kf =
=
= 1.7 x 107
[Ag(H2O)2+][NH3]2
x(4.7 x 10-2)2
x = _________________ M = [Ag(H2O)2+]
The Amphoteric Behavior of
Aluminum Hydroxide
Separating Ions by Selective Precipitation–I
Problem: A solution consists of 0.10 M AgNO3 and 0.15 M Cu(NO3)2.
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)
H2O
H2O
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:
-12
Ksp
1.0
x
10
[I ] =
=
= _________________ M
+
[Cu ]
0.15 M
Separating Ions by Selective Precipitation–II
Thus, the concentration of iodide ion that will give a saturated solution
of copper(I) iodide is 1.0 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 molNaI
mol I -
x
149.9 g NaI
mol NaI
= _____ ng NaI
The General Procedure for Separating
Ions in Qualitative Analysis
Separation into Ion Groups
Ion Group 1: Insoluble chlorides
Ag+, Hg22+, Pb2+
Ion Group 2: Acid-insoluble sulfides
Cu2+, Cd2+, Hg2+, As3+, Sb3+, Bi3+, Sn2+, Sn4+, Pb2+
Ion Group 3: Base-insoluble sulfides and hydroxides
Zn2+, Mn2+, Ni2+, Fe2+, Co2+ as sulfides,
and Al3+, Cr3+ as hydroxides
Ion Group 4: Insoluble phosphates
Mg2+, Ca2+, Ba2+
Ion Group 5: Alkali metal and ammonium ions
Na+, K+, NH4+
Tests to Determine the Presence of
Cations in Ion Group 5
Na+ ions
K+
ions
OH - + NH4+
NH3 + H2O
plus litmus paper
A
Qualitative
Analysis
Scheme for
Ag+, Al3+,
Cu2+ and
Fe3+
Figure 8.14:
Separation of
Group I ions
The Acid Rain Problem–I
Normally the pH of precipitation is controlled by the reaction of carbon
dioxide with water to form carbonic acid, which keeps the pH of “clean”
rain in the slightly acid range, of about 5.6:
CO2 (g) + H2O(l)
H3O+(aq) + HCO3-(aq)
H2CO3 (aq)
Sulfurous acid is produced by the reaction of sulfur dioxide with water,
and even though it is a weak acid, it does add to the acidity of rain.
A greater worry is the oxidation of the SO2 to form SO3 which reacts
with water to form the strong acid, sulfuric acid.
SO2 (g) + H2O(l)
H2SO3 (aq)
2 SO2 (g) + O2 (g)
2 SO3 (g)
SO3 (g) + H2O(l)
H2SO4 (aq)
The Acid Rain Problem–II
Another strong acid is formed from the nitrogen oxides that are produced
in all internal combustion engines, that is nitric acid, which is not only a
strong acid, but also a strongly oxidizing acid.
N2 (g) + O2 (g)
2 NO(g)
2 NO(g) + O2 (g)
3 NO2 (g) + H2O(l)
2 NO2 (g)
2 HNO3 (aq) + NO(g)
(Final step in the
Ostwald process)
These acids together make the problem of acid rain so severe in the
eastern United States, and far worse in other parts of the world. The
average pH of rain in the eastern U.S. back in 1984 was 4.2. Sweden and
Pennsylvania share second place with a pH of 2.7, but the record was in
Wheeling, West Virginia where the pH was 1.8. Some areas of
California also reach a pH of 1.6. The problems are global in nature.
Formation of Acidic Precipitation