George Mason University
General Chemistry 212
Chapter 19
Ionic Equilibria
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7th edition, 2011, McGraw-Hill
Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George
Mason are intended for those students enrolled in a science
/engineering oriented curricula, with particular emphasis on
chemistry, biochemistry, and biology The material on these slides is
taken primarily from the course text but the instructor has modified,
condensed, or otherwise reorganized selected material.
Additional material from other sources may also be included.
Interpretation of course material to clarify concepts and solutions to
problems is the sole responsibility of this instructor.
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1
Ionic Equilibria in Aqueous Solutions


Equilibria of Acid-Base Buffer Systems

The Common Ion Effect

The Henderson-Hasselbalch Equation

Buffer Capacity and Range

Preparing a buffer
Acid-Base Titration Curves

Acid-Base Indicators

Strong-Acid-Strong Base Titrations

Weak Acid-Strong Base Titrations

Weak Base-Strong Acid Titrations

Polyprotic Acid Titrations

Amino Acids as Polyprotic Acids
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2
Ionic Equilibria in Aqueous Solutions


Equilibria of Slightly Soluble Ionic Compounds

The Solubility-Product Constant

Calculations involving Ksp

The Effect of a Common Ion

The Effect of pH

Qsp vs. Ksp
Equilibria involving Complex Ions

Formation of Complex Ions

Complex Ions and Solubility

Amphoteric Hydroxides
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3
Ionic Equilibria in Aqueous Solutions

The simplest acid-base equilibria are those in which
a single acid or base solute reacts with water

In this chapter, we will look at:


solutions of weak acids and bases through
acid/base ionization

The reactions of salts with water

Titration curves
All of these processes involve equilibrium theory
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4
Ionic Equilibria in Aqueous Solutions

Equilibria of Acid-Base Buffer Systems

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Buffer

An Acid-Base Buffer is a species added to a
solution to minimize the impact on pH from
the addition of [H3O+] or [OH-] ions

Small amounts of acid or base added to an
unbuffered solution can change the pH by
several units

Since pH is a logarithmic term, the change in
[H3O+] or [OH-] can be several orders of
magnitude
5
Ionic Equilibria in Aqueous Solutions

The Common Ion Effect

Buffers work through a phenomenon known as the:
Common Ion Effect

The common ion effect occurs when a given ion is
added to an “equilibrium mixture of a weak acid or
weak base that already contains that ion

The additional “common ion” shifts the equilibrium
away from its formation to more of the
undissociated form; i.e., the acid or base dissociation
decreases

A buffer must contain an “acidic” component that
can react with the added OH- ion, and a “basic”
component than can react with the added [H3O+]
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6
Ionic Equilibria in Aqueous Solutions
The buffer components cannot be just any acid
or base
 The components of a buffer are usually the
conjugate acid-base pair of the weak acid (or
base) being buffered
Ex: Acetic Acid & Sodium Acetate
Acetic acid is a weak acid, slightly dissociated

CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
Acid
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Base
Conjugate
Conjugate
Base
Acid
7
Ionic Equilibria in Aqueous Solutions

Ex: Acetic Acid & Sodium Acetate (Con’t)

Now add Sodium Acetate (CH3COONa), a strong
electrolyte (Acetate ion is the conjugate base)
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
(added)
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
The added Acetate ions shift the equilibrium to the left
forming more undissociated acetic acid

This lowers the extent of acid dissociation, which
lowers the acidity by reducing the [H3O+] and
increasing the pH

Similarly, if Acetic Acid is added to a solution of
Sodium Acetate, the Acetate ions already present act
to suppress the dissociation of the acid
8
Ionic Equilibria in Aqueous Solutions
When a small amount of [H3O+] is add to acetic acid/acetate buffer, that
same amount of acetate ion (CH3COO-) combines with it, increasing the
concentration of acetic acid (CH3COOH). The change in the [HA]/[A-]
ratio is small; the added [H3O+] is effectively tied up; thus, the change
in pH is also small
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9
Ionic Equilibria in Aqueous Solutions

Essential Features of a Buffer

A buffer consists of high concentrations of the
undissociated acidic component [HA] and the
conjugate base [A-] component

When small amounts of [H3O+] or [OH-] are
added to the buffer, they cause small amounts of
one buffer component to convert to the other
component
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10
Ionic Equilibria in Aqueous Solutions
Ex. When a small amount of H3O+ is added to an
Acetate buffer, that same amount of CH3COOcombines with it, increasing the amount of
undissociated CH3COOH tying up potential
[H3O+]
Similarly, a small amount of OH- added
combines with undissociated CH3COOH to form
CH3COO- & H2O tying up potential [OH-]
In both cases, the relative changes in the
amount of buffer components is small, but the
added [H3O+] or [OH-] ions are tied up as
undissociated components; thus little impact
on pH
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11
Ionic Equilibria in Aqueous Solutions

The equilibrium perspective
CH3COOH + H 2O = H 3O+
+ CH 3COO-
[H + ][A - ] [H 3O + ][CH 3COO - ]
Ka =
=
HA
[CH 3COOH]
[CH 3COOH]
 HA
[H 3O+ ] = K a × - = K a ×
 A 
[CH 3COO- ]
 The [H3O+] (pH) of the solution depends directly on the
buffer-component concentration ratio

If the ratio [HA]/[A-] goes up, [H3O+] goes up (pH down)

If the ratio [HA]/[A-] goes down, [H3O+] goes down
([OH-] goes up) and pH increases (less acidic)

When a small amount of a strong acid is added, the increased
amount of [H3O+] reacts with a stoichiometric amount of
acetate ion from the buffer to form more undissociated
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acetic acid
12
Practice Problem
What is the pH of a buffer solution consisting of 0.50 M
CH3COOH (HAc) and 0.50 M CH3COONa (NaAc)?
Ka (CH3COOH) = 1.8 x 10-5
x = [CH 3COOH]dissoc = [H3O+ ]
NaAc is a sodium salt :  it dissociates completely
[H3O+ ] from autoionization of H 2O is neglible and is neglected
+
H2O(l)

Ac-(aq)
+
H3O+
Concentration
HAc(aq)
Initial Ci
0.50
-
0.50
0

-X
-
+X
+X
Equilibrium Cf
0.50 - X
-
0.50 + X
X
If x is small, i.e., < 5% CH3COOHinit , then:
[CH 3COOH] = 0.50 - x  0.50
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[CH3COO- ] = 0.50 + x  0.50
(Con’t)
13
Practice Problem (Con’t)
[H 3O + ][CH 3COO- ]  x  ×[CH 3COO- ]
Ka =
=
[CH 3COOH]
[CH 3COOH]
[CH 3COOH]
x = [H 3O ] = K a ×
[CH 3COO- ]
+
x = [H 3O ] = 1.8  10
+
-5
0.5
×
= 1.8×10-5 M
0.5
pH = - log[H3O+ ] = - log(1.8  10-5 )
pH = 4.74
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14
Practice Problem (Con’t)
Calculate the pH of the previous buffer solution after adding
0.020 mol of solid NaOH to 1 L of the solution
Set up reaction table for the stoichiometry of NaOH (strong
base) to Acetic acid (weak acid)
0.020mol
Molarity NaOH (M) =
= 0.020 mol / L
1L
+
OH-(aq)

Ac-(aq)
H2O(l)
Concentration
HAc(aq)
+
Initial Ci
0.50
0.020
0.50
-

- 0.020
-0.020
+ 0.020
-
Equilibrium Cf
0.48
0.0
0.52
-
Set up reaction table for acid dissociation using new
concentrations
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+
H2O

Ac-(aq)
+
H3O+(aq)
Concentration
HAc(aq)
Initial Ci
0.48
-
0.52
0

-X
-
+X
+X
Equilibrium Cf
0.48 - X
-
0.52 + X
X
(Con’t)
15
Practice Problem (Con’t)
Assuming x is small:
[CH3COOH] = 0.48 M - x = 0.48 M and [CH 3COO- ] = 0.52 M + x = 0.52 M
x = [H 3O + ] = K a ×
[CH 3COOH]
0.48
-5
=
(1.8×10
)
×
0.52
[CH 3OO- ]
[H 3O+ ] = 1.7 ×10-5 M
pH = -log[H 3O + ] = -log(1.7 ×10-5 )
pH = 4.77
The addition of the strong base increased the concentration
of the basic buffer [CH3COO-] component to 0.52 M at the
expense of the acidic buffer component [CH3COOH] (0.48 M)
The concentration of the [H3O+] (x) decreased, but only
slightly, thus, the pH increased only slightly from 4.74 to 4.77
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16
Ionic Equilibria in Aqueous Solutions

The Henderson-Hasselbalch Equation

For any weak acid, HA, the dissociation equation and Ka
expression are:
+
HA + H2O
Ka
H 3O + A
[H 3O+ ][A - ]
=
[HA]
The key variable that determines the [H3O+] is the ratio of
acid species (HA) to base species (A-)
[H 3O + ] = k a ×
[HA] (acid species)
[A - ] (base species)
[HA]
-log[H 3O ] = - logK a - log [A ]
[A - ]
pH = pK a + log
[HA]
[base]
pH = pK a + log
[acid]
+
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Note the change in sign of
“log” term and inversion of
acid-base terms
17
Practice Problem
What is the pH of a buffer formed by adding 0.15 mol of
acetic acid and 0.25 mol of sodium acetate to 1.5 L of water?
Ans: Both the acid and salt forms are added to the solution at
somewhat equal concentrations
Neither form has much of a tendency to react
Dissociation of either form is opposed in bi-directional
process
The final concentrations will approximately equal the
initial concentrations
[HA] = 0.15 mol / 1.5 L = 0.10 M
[A- ] = 0.25 mol / 1.5 L = 0.167 M
K a for acetic acid = 1.7 × 10-5 (from table "C")
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pK a = - log(K a ) = - log(1.7  10-5 ) = 4.77
(Con’t)
18
Practice Problem
Example Problem (con’t)
From the Henderson-Hasselbalch equation
 [A - ] 
pH = Pk a + log 

[HA]


 0.167M 
pH = 4.77 + log 
 = 4.77 + log(1.67) = 4.77 + 0.22
 0.10M 
pH = 4.99
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19
Ionic Equilibria in Aqueous Solutions

Buffer Capacity

A buffer resists a pH change as long as the concentration
of buffer components are large compared with the
amount of strong acid or base added

Buffer Capacity is a measure of the ability to resist pH
change

Buffer capacity depends on both the absolute and
relative component concentrations
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
Absolute - The more concentrated the components of
a buffer, the greater the buffer capacity

Relative – For a given addition of acid or base, the
buffer-component concentration ratio changes less
when the concentrations are similar than when they
are different
20
Ionic Equilibria in Aqueous Solutions

Ex. Consider 2 solutions

Solution 1 – Equal volumes of 1.0M HAc and 1.0 M Ac-

Solution 2 – Equal volumes of 0.1M HAc and 0.1M Ac-

Same pH (4.74) but 1.0 M buffer has much larger buffer
capacity
(Con’t)
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21
Ionic Equilibria in Aqueous Solutions
Ex. Buffer #1
[HA] = [A-] = 1.000 M
Add 0.010 mol of OH- to 1.00 L of buffer
[HA] changes to 0.990 M
[A - ]init
1.000 M
=
= 1.000
[HA]init 1.000 M
[A-] changes to 1.010 M
[A - ]final
1.010 M
=
= 1.020
[HA]final
0.990 M
1.020 - 1.000
Percent change =
×100 = 2%
1.000
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22
Ionic Equilibria in Aqueous Solutions
Buffer #2
[HA] = 0.250 M
[A-] = 1.75 M
Add 0.010 mol of OH- to 1.00 L of buffer
[HA] changes to 0.240 M
[A - ]init
1.750 M
=
= 7.000
[HA]init
0.250 M
[A-] changes to 1.760 M
[A - ]final
1.760 M
=
= 7.330
[HA]final
0.240 M
7.330 - 7.000
Percent change =
× 100 = 4.7%
7.000
Buffer-component concentration ratio is much
larger when the initial concentrations of the
components are very different
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23
Ionic Equilibria in Aqueous Solutions

A buffer has the highest capacity when the
component acid / base concentrations are equal:

A
 Base =   = 1
 Acid   HA 
 [A - ] 
pH = pK a + log 
= pK a + log1 = pK a + 0 = pK a

 [HA] 
A buffer whose pH is equal to or near the pKa
of its acid component has the highest buffer
capacity
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24
Ionic Equilibria in Aqueous Solutions

Buffer Range
 pH range over which the buffer acts effectively is
related to the relative component concentrations
 The further the component concentration ratio is
from 1, the less effective the buffering action
 If the [A-]/[HA] ratio is greater than 10 or less than
0.1 (or one component concentration is more than
10 times the other) the buffering action is poor
 10 
pH = pK a + log   = pK a + 1
 1 
 1 
pH = pK a + log   = pK a - 1
 10 

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Buffers have a usable range within:
± 1 pH unit or pKa value of the acid components
25
Ionic Equilibria in Aqueous Solutions

Equilibria of Acid-Base Systems

Preparing a Buffer

Choose the Conjugate Acid-Base pair

Driven by pH

Ratio of component concentrations close to
1 and pH ≈ pKa
Con’t
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26
Ionic Equilibria in Aqueous Solutions
Ex. Assume you need a biochemical buffer whose pH is 3.9
1. pKa of acid component should be close to 3.9
Ka = 10-3.9 = 1.3 x 10-4
2. From a table of pKa values select buffer possibilities
Lactic acid
(pKa = 3.86)
Glycolic acid (pKa = 3.83)
Formic Acid (pKa = 3.74)
3.
To avoid common biological species, select Formic Acid
Buffer components of Formic Acid
Formic Acid – HCOOH
(Acid)
Formate Ion – HCOO(Conjugate Base)
Obtain soluble Formate salt – HCOONa
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Con’t
27
Ionic Equilibria in Aqueous Solutions
4.
Calculate Ratio of Buffer Component
Concentrations ([A-]/[HA]) that gives desired pH
 [A- ] 
pH = Pka + log 

[HA]


 [HCOO- ] 
3.9 = 3.74 + log 

[HCOOH]


 [HCOO- ] 
log 
 = 3.9 - 3.74 = 0.16
 [HCOOH] 
Thus :
 [HCOO- ] 
0.16
=
10
= 1.4


 [HCOOH] 
1.4 mol of HCOONa is needed for every mole
HCOOH
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Con’t
28
Ionic Equilibria in Aqueous Solutions

Preparing a Buffer (Con’t)

Determine the Buffer Concentrations

The higher the concentration of the components,
the higher the buffer capacity

Assume 1 Liter of buffer is required and you have a
stock of 0.40 M Formic Acid (HCOOH)

Compute moles and then grams of Sodium Formate
(CHOONa) needed to produce 1.4/1.0 ratio
Moles of HCOOH = 1.0 L ×
1/23/2015
0.40 mol HCOOH
= 0.40 mol HCOOH
1.0 L soln
Moles of HCOONa = 0.40 mol HCOOH ×
1.40 mol HCOONa
= 0.56 mol HCOONa
1.0 mol HCOOH
Mass of HCOONa = 0.56 mol HCOONa ×
68.01 g HCOONa
= 38 g HCOONa Con’t
1 mol HCOONa
29
Ionic Equilibria in Aqueous Solutions

Preparing a Buffer (Con’t)

Mix the solution and adjust the pH

The prepared solution may not be an ideal
solution (see Chapter 13, Section 6)
The desired pH (3.9) may not exactly match
the actual value of the buffer solution

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The pH of the buffer solution can be
adjusted by a few tenths of a pH unit by
adding strong acid or strong base.
30
Practice Problem
Example of preparing a buffer solution without using the
Henderson-Hasselbalch equation
How many grams of Sodium Carbonate (Na2CO3) must be
added to 1.5 L of a 0.20 M Sodium Bicarbonate (NaHCO3)
solution to make a buffer with pH = 10.0?
Ka (HCO3-) = 4.7 x 10-11
HCO 3 - (aq) + H 2O(l)
(From Appendix C)
H 3O + (aq) + CO 3 2- (aq)
Conjugate Pair - HCO3- (acid) and CO32- (base)
Convert pH to [H3O+]
[H3O+ ] = 10-pH = 10-10 = 1×10-10 M
Con’t
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31
Practice Problem (Con’t)
3.
Use Ka to compute [CO32-]
[H 3O + ][CO 3 2- ]
Ka =
[HCO 3- ]
2[HCO
0.20 
-11 
3 ]
[CO3- ] = K a
=
(4.7

10
)
= 0.094 M

+
-10 
[H 3O ]
 1.0  10 
4.
Compute the amount of CO32- needed for 1.5 L of solution
Moles CO3 2-
5.
0.094 mol CO3 2= 1.5 L soln ×
= 0.14 mol CO 321 L soln
Compute the mass (g) of Na2CO3 needed
105.99 g Na 2CO 3
Mass (g) of Na 2CO 3 = 0.14 mol Na 2CO 3 ×
= 14.8 g
1mol Na 2CO 3
Con’t
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32
Practice Problem (Con’t)
6.
Dissolve the 14.8 g Na2CO3 in slightly less than 1.5 L of the
NaHCO3, say 1.3 L
7.
Add additional NaHCO3 to make 1.5 L; adjust pH to 10.0
with a strong acid
8.
Check for a useful buffer range:
The concentration of the acidic component [HCO3-]
must be within a factor of “10” of the concentration of
the basic component [CO32-]

 HCO3-  =  1.5 L  × 0.20 M HCO3-

= 0.30 mol HCO3-
0.30 mol HCO3- vs. 0.14 mol CO320.30 / 0.14 = 2.1 << 10
Therefore, the buffer solution is reasonable
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33
Ionic Equilibria in Aqueous Solutions

Acid-Base Titration Curves

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Acid-Base Indicators

Weak organic acid (HIn) that has a different color
than its conjugate base (In-)

The color change occurs over a relatively narrow
pH range

Only small amounts of the indicator are needed;
too little to affect the pH of the solution

The color range of typical indicators reflects a
100 - fold range in the [HIn]/In-] ratio

This corresponds to a pH range of 2 pH units
34
Mixing Acids & Bases



Acids and bases react through neutralization reactions
The change in pH of an acid mixed with a base is tracked
with an Acid-Base Titration Curve
Titration:
 Titration Curve: Plot of pH vs. the volume of the “Strong”
acid or “Strong” base being added via buret
Solution in buret is called the “titrant”
Equivalence Point: Point in a titration curve where
stoichiometric amounts of acid and base have been mixed
(point of complete reaction)



3 Important Cases:
Strong Acid + Strong Base (& vice versa)
Weak Acid + Strong Base
Weak Base + Strong Acid
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35
Titration Curves

Titration Curve: Plot of measured pH versus Volume of acid
or base added during a “Neutralization” experiment

All Titration Curves have a characteristic
“Sigmoid (S-shaped) profile


Beginning of Curve:
pH changes slowly

Middle of Curve:
pH changes very rapidly

End of Curve:
pH changes very slowly again
pH changes very rapidly in the titration as the equivalence
point (point of complete reaction) is reached and right after
the equivalence point
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36
Neutralization Reactions

Acids and Bases react with each other to form salts (not
always) and water (not always) through
Neutralization Reactions

Neutralization reaction between a strong acid and strong
base
HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
acid

base
salt
water
Neutralization of a strong acid and strong base reaction lies
very far to the right (Kn is very large); reaction goes to
completion; net ionic equation is
H3O+(aq) + OH-(aq)  2 H2O(l)
Kn

[H 2O]2
1
1
14
=
=
=
=
1×10
Kw
[H 3O + ][OH - ] [H 3O + ][OH - ]
H3O+ and OH- efficiently react with each other to form water
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37
Titration Curves
Curve for Titration of a Strong Acid by a Strong Base
1/23/2015
38
Practice Problem
Calculate pH after the addition of 10, 20, 30 mL NaOH during
the titration of 20.0mL of 0.0010 M HCL(aq) with 0.0010 M
NaOH(aq)
Mixture
Strong Acid + Strong Base
H 3O +(aq) +
OH - (aq)
€
2H 2O(l)
Initial pH HCl soln: [HCl] = [H3O+] = 0.0010 M
pH = - log[H3O+ ] = - log(0.0010) = 3.0
Con’t
1/23/2015
39
Practice Problem
Add 10 mL 0.0010 M NaOH
1 mol H3O+ disappears for each mole OH- added
moles = Molarity(mol / L) × Volume(L)
Initial moles of H 3O+ = 0.0010 mol / L x 0.020L = 0.00002 mol H 3O+
- moles OH-added
= 0.0010 mol / L x 0.010L = 0.00001 mol OH-
moles of H 3O+ remaining
= 0.00001 mol H 3O+
Con’t
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40
Practice Problem (Con’t)

Calculate [H3O+] & pH, taking “Total Volume” into account
+
amount(mol)
of
H
O
remaining
+
3
[H 3O ] =
original volume of acid + volume of added base
+
0.000010
mol
H
O
3
[H 3O + ] =
= 0.000033M
0.020L + 0.010L
pH = - log[H 3O+ ] = - log(0.0000333) = 3.48
Con’t
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41
Practice Problem (Con’t)

pH at “Equivalence Point”
Equal molar amounts of HCL & NaOH have been
mixed
20 mL x 0.0010 M HCl reacts with
20 mL x 0.0010 M NaOH
At the equivalence point all H3O+ has been
neutralized by the addition of all the NaOH, except
for [H3O+] from the autoionization of water, i.e.,
negligible
[H3O+] = pure water = 1 x 10-7 M
Con’t
1/23/2015
42
Practice Problem (Con’t)

After the equivalence point

After the equivalence point, the pH calculations are
based on the moles of excess OH- present

A total of 30 mL of NaOH has been added
Total moles of OH- added = 0.0010 mol / L x 0.030L = 0.00003 mol OH- moles H 3O+ consumed
= 0.0010 mol / L x 0.020L = 0.00002 mol H 3O+-
moles of excess OH-
= 0.00001 mol OH-
amount excess OH 0.000010 moles OH [OH ] =
=
original volume acid + volume base added
0.020 L + 0.030 L
-
[OH - ] = 0.00020M
pOH = - log[OH - ] = - log(0.00020) = 3.70
1/23/2015
pH = pK w - pOH = 14.0 - 3.7 = 10.3
43
Neutralization Reactions

Neutralization reactions between weak acids and strong
bases (net ionic equation shown below)
C6H5COOH(aq)
benzoic acid
+
NaOH  C6H5COONa(aq)
base
+
benzoate “salt”
H2O(l)
water

The above equilibrium lies very far to the right

The two equilibria below can be added together to provide
the overall neutralization reaction shown above
(1) C6H5COOH(aq) + H2O(l)  C6H5COO-(aq) + H3O+(aq)
Ka = 1.7 x 10-5
(from Appendix C)
(2) H3O+(aq) + OH-(aq)  2 H2O(l)
Recall : K w = [H 3O + ][OH - ] = 1.0 × 10-14
1/23/2015
1
= 1.0×1014
Kw
44
Neutralization Reactions

The overall neutralization equilibrium constant (Kn)
is the product of the two intermediate equilibrium
constants
(Ka & 1/Kw)
KN = Ka x 1/Kw = (1.7 x 10-5)(1 x 1014) = 1.9 x 109
(very large!)

Weak acids react completely with strong bases
1/23/2015
45
Titration Curves
Curve for Titration of a Weak Acid by a Strong Base
1/23/2015
46
Weak Acid-Strong Base Titration Curve

Consider the reaction between Propanoic Acid (weak acid)
and Sodium Hydroxide (NaOH) (strong base)

Ka for CH3CH2COOH (HPr) - 1.3 x 10-5

The Titration Curve (see previous slide)

The bottom dotted curve corresponds to the strong acidstrong base titration

The Curve consists of 4 regions, the first 3 of which differ
from the strong acid case

The initial pH is “Higher”


1/23/2015
The weak acid dissociates only slightly producing
less [H3O+] than with a strong acid
The gradually arising portion of the curve before the
steep rise to the equivalence point is called the “buffer
region”
47
Weak Acid-Strong Base Titration Curve

As HPr reacts with the strong base, more and more
of the conjugate base (Pr-) forms creating an
(HPr/Pr-) buffer

At the midpoint of the “Buffer” region, half the
original HPr has reacted
 HPr 
= [Pr - ]
[Pr - ]
or
=1
[HPr]
 [Pr - ] 
pH = Pk a + log 
 = pK a + log(1) = pK a
 [HPr] 
pH at midpoint of “Buffer” region is common
method of estimating pKa of an “unknown” acid
1/23/2015
48
Weak Acid-Strong Base Titration Curve
The pH at the “equivalence point” is greater
than 7.00
 The weak-acid anion (Pr-) acts as a weak
base to accept a proton from H2O forming
OH The additional [OH-] raises the pH
 Beyond the equivalence point, the ph
increases slowly as excess OH- is added

1/23/2015
49
Practice Problem
Calculate the pH during the titration of 40 ml of 0.1000 M of
the weak acid, Propanoic acid (HPr); Ka = 1.3 x 10-5 after the
addition of the following volumes of 0.1000 M NaOH
0.00 mL
30.0 mL
40.0 mL
50.0 mL
a. 0.0 mL NaOH added
HPr(aq) + H 2O
H 3O+ (aq) + Pr - (Aq)
[HPr] = [HPr]init - [HPr]dissoc  [HPr]init
[H 3O + ] = [Pr - ] = x
[H 3O + ][Pr - ]
(X)(X)
Ka =
= 1.3  10-5 
[HPr]
[HPr]init
X2 = [H 3O+ ]2 = K a ×  HPr init = 1.3 × 10-5 × 0.1000 M
[H 3O + ] = 1.14  10-3 M
1/23/2015
pH = - log[H3O ] =
2.94
Con’t
50
Practice Problem (Con’t)
b. 30.00 mL 0.1000 M NaOH added
Original moles of HPr = 0.0400L x 0.1000 M = 0.004000 mol HPr
moles of NaOH added = 0.03000 L x 0.1000 M = 0.003000 mol OH -
For each mol of NaOH that reacts, 1 mol Pr- forms
+
OH-(aq)
→
Pr-
+
H2O(l)
Amount (mol)
HPr(aq)
Initial
0.004000
0.003000
0
-
Change
- 0.003000
-0.003000
+ 0.003000
-
Final
0.001000
0
0.003000
-
Last line shows new initial amounts of HPr & Pr- that
react to attain a “new” equilibrium
At equilibrium:
[H 3O + ][Pr - ]
Ka =
[HPr]
[H 3O + ] = K a
1/23/2015
[HPr]
0.001000 mol
=
= 0.3333
[Pr ]
0.003000 mol
[HPr]
-5
-6
×
=
(1.3
x
10
)
(0.3333)
=
4.3
x
10
M
[Pr - ]
pH = - log[H 3O + ] = - log(4.3  10-6 ) = 5.37
Con’t
51
Practice Problem (Con’t)
c. 40.00 mL NaOH (0.00400 mol) added
 Reaction is at “equivalence point”
 All the HPr has reacted, leaving 0.004000 mol Pr Calculate concentration of Pr
[Pr - ] =

0.004000 mol
= 0.05000 M
0.04000 L + 0.004000 L
Kw
1.0  10-14
-10
Kb =
=
=
7.7

10
Ka
1.3  10-5
Calculate Kb
 [HPr] = [OH-]
x x [OH - ]2

[HPr][OH - ]
Kb =
=

[Pr ]
[Pr - ]
 Pr 
K w = [H 3O + ][OH - ]
[OH - ] = K b x [Pr - ]
[H 3O ] 
+
Kw
K b  [Pr - ]
=
1.0  10-14
(7.7  10-10 ) (0.05000)
= 1.6  10-9 M
pH = - log[H 3O + ] = - log(1.6  10-9 ) = 8.80
1/23/2015
Con’t
52
Practice Problem (Con’t)
d. 50 mL of 0.1000 M NaOH added
 Beyond equivalence point, just excess OH is being
added
 Calculation of pH is same as for strong acid-strong base
titration
 Calculate the moles of excess OH from the initial
concentration (0.1000 M) and the remaining volume
(50.0 mL - 40.0 mL)
Moles excess OH - = (0.1000 M)(0.05000 L - 0.04000 L) = 0.001000 mol
moles of excess OH 0.001000 mol
[OH ] =
=
= 0.01111 M
total volume
0.04L + 0.05L
Kw
moles of excess OH [H 3O+ ] =
where
[OH
]
=
[OH - ]
total volume
-
Kw
1.0  10-14
-13
[H 3O ] =
=
=
9.0

10
[OH - ]
0.01111
+
pH = - log[H 3O + ] = - log(9.0  10-13 ) = 12.05
1/23/2015
53
Weak Base-Strong Acid Titration

Neutralization reactions between weak bases and strong
acids (net ionic equation shown below)
NH3(aq) + H3O+(aq)  NH4+(aq) + H2O(l)
KN=?
base
acid
salt
water

Above equilibrium also lies very far to the right; the two
equilibria below can be added together to provide the
overall neutralization reaction shown above
NH3(aq)
+ H2O(aq)  NH4+ + OH-(aq)
H3O+(aq) + OH-(aq) 
2 H2O(l)
Kb = 1.8 x 10-5
1/Kw = 1.0 x 1014
KN = Kb X 1/Kw = (1.8 x 10-5)(1 x 1014) = 1.8 x 109 (very large!)

Weak bases react completely with strong acids
1/23/2015
54
Weak Base-Strong Acid Titration
Curve for Titration of a Weak Base by a Strong Acid
1/23/2015
55
Polyprotic Acid Titrations





Polyprotic Acids have more than one ionizable proton
Except for Sulfuric Acid (H2SO4), the common polyprotic
acids are all weak acids
Successive Ka values for a polyprotic acid differ by several
orders of magnitude
The first H+ is lost much more easily than subsequent ones
All “1st” protons (H+) are removed before any of the “2nd”
protons
H 2SO 3 (aq) + H 2O(l)
K a1 = 1.4 x 10-2
HSO 3 - (aq) + H 2O(l)
1/23/2015
K a2 = 6.5 x 10-8
HSO 3 - (aq) + H 3O + (aq)
and
pK a1 = 1.85
SO 3 2- (aq) + H 3O + (aq)
and
pK a2 = 7.19
56
Polyprotic Acid Titrations



Each mole of H+ is titrated separately
In a diprotic acid two OH- ions are required to completely
remove both H+ ions
All H2SO3 molecules lose one H+ before any HSO3- ions lose
a H+ to form SO3H 2SO 3 
 HSO 3 
 SO 3 21 mol OH -
-
1 mol OH -
Titration curves looks like two
weak acid-strong base curves
joined end-to-end
HSO3- is the conjugate base
of H2SO3 (Kb = 7.1x10-13)
HSO3- is also an acid
(Ka=6.5x10-8) dissociating to
form its conjugate base SO32-
1/23/2015
57
Slightly Soluble Compounds

Most solutes, even those called “soluble,” have a
limited solubility in a particular solvent

In a saturated solution, at a particular
temperature, equilibrium exists between the
dissolved and undissolved solute

Slightly soluble ionic compounds, normally called
“insoluble,” reach equilibrium with very little of the
solute dissolved

Slightly soluble compounds can produce complex
mixtures of species

Discussion here is to assume that the small amount
of a slightly soluble solute that does dissolve,
dissociates completely
1/23/2015
58
Equilibria - Slightly Soluble Ionic Compounds
Solubility Rules for Ionic Compounds in Water
1/23/2015
59
Equilibria - Slightly Soluble Ionic Compounds


Equilibrium exists between solid solute and the aqueous
ions
PbSO4(s) ⇄ Pb2+(aq)
+ SO42-(aq)
Set up “Reaction Quotient”
Qc  PbSO 4 (s) =  Pb 2+  SO 42- 
[Pb 2+ ][SO4 2- ]
Qc =
[PbSO4 (s)]
Note: concentration of a solid - [PbSO4(s)] = 1


Define Solubility Product
Qsp = Qc [PbSO 4 (s)] = [Pb 2+ ][SO 4 2- ]
When solid PbSO4 reaches equilibrium with Pb2+ and SO42at saturation, the numerical value of Qsp attains a constant
value called the Solubility-Product constant (Ksp)
1/23/2015
Qsp at saturation = K sp = Solubility - Product constant
60
Equilibria - Slightly Soluble Ionic Compounds

The Solubility Product Constant (Ksp)

Value of Ksp depends only on temperature, not individual
ion concentrations

Saturated solution of a slightly soluble ionic compound,
MpXq, composed of ions Mn+ and Xz-, the equilibrium
condition is:
n+ p
z- q
Qsp = [M ] [X ] = K sp
1/23/2015
61
Equilibria - Slightly Soluble Ionic Compounds

Single-Step Process
Cu(OH)2 (s)
Cu 2+ (aq) + 2OH - (aq)
K sp = [Cu 2+ ][OH - ]2

Multi-Step Process
MnS(s)
Mn 2+ (aq) + S2- (aq)
S 2- (aq) + H 2O(l)
HS - (aq) + OH - (aq)
MnS(s) + H 2O(l)
Mn 2+ (aq) + HS - (aq) + OH - (aq)
K sp = [Mn 2+ ][HS - ][OH - ]
1/23/2015
62
Equilibria - Slightly Soluble Ionic Compounds
Solubility-Product Constants
(Ksp) of Selected Ionic
Compounds at 25oC
Relationship Between Ksp and Solubility at 25 oC
1/23/2015
63
Practice Problem
Determine Ksp from solubility
The solubility of PbSO4 at 25oC is 4.25 x10-3 g /100 mL
solution. Calculate the Ksp
PbSO 4 (s) = Pb 2+ (aq) + SO 4 2- (aq)
Convert solubility to molar solubility
Molar Solubility =
0.00425 g PbSO4
1 mol PbSO4
1000 ml
x
x
100 mL
1L
303.3 g PbSO 4
= 1.40  10-4 M PbSO 4
Some PbSO4 dissolves into Pb2+ & SO42[Pb 2+ ] = SO4 2- 
[Pb 2+ ] = SO4 2-  = 1.40  10-4 M
1/23/2015
K sp = [Pb 2+ ][SO4 2- ] = (1.40  10-4 )2 = 1.96  10-8
64
Practice Problem
Determine Solubility from Ksp
Determine the solubility of Calcium Hydroxide, Ca(OH)2
Ksp for Ca(OH)2 = 6.5 x 10-6
Ca(OH)2 (s)
Ca 2+ (aq) + 2OH - (aq)
Set up reaction table (S = molar solubility)
⇄
Ca2+(aq)
+
2OH-(aq)
Concentration (M)
Ca(OH)2(s)
Initial

0
0
Change

+S
+2S
Equilibrium

S
2S
K sp = [Ca 2+ ][OH - ]2 = (S)(2S)2 = (S)(4S 2 ) = 4S 3 = 6.5 x 10-6
2+
[Ca ] = S =
1/23/2015
3
6.5x10-6
= 1.2 x 10-2 M = 0.012 M
4
[OH- ] = 2S = 2  1.2  10-2 = 4.2  10-2 M = 0.042 M
65
Equilibria - Slightly Soluble Ionic Compounds

The Effect of the Common Ion

The presence of a “Common Ion” decreases the solubility
of a slightly soluble ionic compound
PbCrO 4 (s) €
Pb 2+ (aq) + CrO 4 2- (aq)
K sp = [Pb 2+ ][CrO4 2- ] = 2.3 x 10-13

1/23/2015
Add some Na2CrO4, a soluble salt, to the saturated
PbCrO4 solution

Concentration of CrO42- increases

Some of excess CrO42- combines with Pb2+ to form
PbCrO4(s)

Equilibrium shifts to the “Left”

This shift “reduces” the solubility of PbCrO4
66
Practice Problem
What is the solubility of Ca(OH)2 in 0.10 M Ca(NO3)2?
Ksp Ca(OH)2 = 6.5 x 10-6
The addition of the common ion, Ca2+, should lower the
solubility of Ca(OH)2 Note: (Compare to result from slide 65)
Ca(OH)2 (s)
Ca 2+ (aq) + 2OH - (aq)
⇄
+
Ca2+(aq)
2OH-(aq)
Concentration (M)
Ca(OH)2(s)
Initial

0.10
0
Change

+S
+2S
Equilibrium

0.10 + S = 0.1
2S
Since Ksp is small, S << 0.10 then, 0.10 M + S  0.1 M
K sp = [Ca 2+ ][OH - ]2 = 6.5 x 10-6  (0.1)(2S)2
6.5  10-6
4S =
= 6.5  10-5
0.10
2
4.0 x 10-3 M
 100 = 4.0% < 5%
0.10 M
1/23/2015
[Ca2+ ]this prob vs. [Ca 2+ ]prev prob
2+
S = [Ca ] =
6.5 x 10-5
= 4.03  10-3 M
4
Assumption : S << 0.10 is valid
( 0.00403 M vs. 0.012 M)
67
Equilibria - Slightly Soluble Ionic Compounds

The Effect of pH on Solubility

The Hydronium Ion (H3O+) of a strong acid increases the
solubility of a solution containing the anion of a weak
acid (HA-)

Adding some strong acid to a saturated solution of
Calcium Carbonate (CaCO3) introduces large amount of
H3O+ ion, which reacts immediately with the CaCO3 to
form the weak acid HCO3CaCO 3 (s) € Ca 2+ (aq) + CO 3 2- (aq)
CO 3 2- (aq) + H 3O + € HCO 3 - (aq) + H 2O(l)

Additional H3O+ reacts with the HCO3- to form carbonic
acid, H2CO3, which immediately decomposes to H2O and
CO2
HCO 3 - (aq) + H 3O +
1/23/2015
H 2CO 3 (aq) + H 2O(l)
CO 2 (g) + 2H 2O(l)
68
Equilibria - Slightly Soluble Ionic Compounds

The equilibrium shifts to the “Right” and more CaCO3
dissolves – increased solubility

The overall reaction is:
CaCO 3 (s) € Ca
2+
H 3O +
H 3O +
+ CO 3  HCO 3 
2-
-
H 2 CO 3  CO 2 (g) + H 2O + Ca 2+

Adding H3O+ to a saturated solution of a compound with a
strong acid anion – AgCl

Chloride ion, Cl-, is the conjugate base of a strong acid
(HCl)

It coexists with water, i.e. does not react with water

There is “No” effect on the equilibrium
1/23/2015
69
Practice Problem
Write balanced equations to explain whether addition of H3O+
from a strong acid affects the solubility of the following ionic
compounds
Lead Bromide (PbBr2)
PbBr2 €
Pb 2+ + 2Br -
Bromide (Br - ) is the anion of the strong acid HBr
Bromide ion does not react with water
Addition of an Acid has No effect
1/23/2015
70
Practice Problem

Cu(II) Hydroxide (Cu(OH)2
Cu(OH)2 (s) €
Cu 2+ (aq) + 2OH - (aq)
OH - is anion of H 2O, a very weak acid
It reacts with added H 3O + to form water (H 2O)
OH - (aq) + H 3O + (aq)  2H 2O
more Cu(OH)2 dissolves to replace OH -
Increases Solubility
1/23/2015
71
Practice Problem
Iron(II) Sulfide (FeS)
Some of the FeS dissolves in water (H 2O) to form HS - and OH -
FeS(s) + H 2O(l) €
Fe 2+ (aq) + HS - (aq) + OH - (aq)
Additional H 3O + reacts with both HS - and OH - (both weak acid anions)
HS - (aq) + H 3O + (aq)  H 2S(aq) + H 2O(l)
OH - (aq) + H 3O +  2H 2O(l)
More FeS dissolves to replace the HS - & OH -
Additional H 3O + Increases Solubility
1/23/2015
72
Practice Problem

Calcium Fluoride (CaF2)
CaF2 (s) €
2+
-
Ca (aq) + 2F (aq)
F - is the conjugate base of a weak acid (HF)
F - reacts with added H 3O + to form the acid and water
F - (aq) + H 3O + (aq)  HF(aq) + H 2O(l)
More CaF2 dissolves to replace the F-
The solubility of CaF2 increases
1/23/2015
73
Equilibria - Slightly Soluble Ionic Compounds

Predicting Formation of a Precipitate

Qsp = Ksp when solution is “saturated”

Qsp > Ksp solution momentarily “supersaturated”


1/23/2015
Additional solid precipitates until Qsp = Ksp
again
Qsp < Ksp solution is “unsaturated”

No precipitate forms at that temperature

More solid dissolves until Qsp = Ksp
74
Practice Problem
Does CaF2 precipitate when 0.100 L of 0.30 M Ca(NO3)2 is
mixed with 0.200 L of 0.060 M NaF?
Ksp (CaF2) = 3.2 x 10-11
Ions present: Ca2+ Na+ NO3FNote: All sodium & nitrate salts are soluble
moles Ca 2+ = 0.30 M Ca 2+ x 0.100L = 0.030 mol Ca 2+
2+
[Ca ]init
0.030 mol Ca 2+
=
= 0.10 M Ca 2+
0.100 L + 0.200 L
moles F  = 0.060 M F  x 0.200L = 0.012 mol F 
-
[F ]init
0.012 mol F =
= 0.040 M F 0.100 L + 0.200 L
2
Qsp = [Ca 2+ ]init [F - ]init
= (0.10)(0.040)2 = 1.6  10-4
Qsp > K sp (1.6  10-4 > 3.2  10-11 )
1/23/2015
 CaF2 will precipitate until Qsp = K sp
75
Equilibria - Slightly Soluble Ionic Compounds

Selective Precipitation of Ions

Separation of one ion in a solution from another

Exploit differences in the solubility of their compounds

The Ksp of the less soluble compound is much smaller
than the Ksp of the more soluble compound

Add solution of precipitating ion until the Qsp value of the
more soluble compound is almost equal to its Ksp value

The less soluble compound continues to precipitate while
the more soluble compound remains dissolved, i.e., the
Ksp of the less soluble compound is always being
exceeded, i.e. precipitation is occurring

At equilibrium, most of the ions of the less soluble
compound have been removed as the precipitate
1/23/2015
76
Practice Problem
A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2
Calculate the [OH-] that would separate the metal ions as
their hydroxides
Ksp of Mg(OH)2 = 6.3 x 10-10 Ksp of Cu(OH)2 = 2.2 x 10-20
Mg(OH)2 is 1010 more soluble than Cu(OH)2; thus Cu(OH)2
precipitates first
Solve for [OH-] that will just give a saturated solution of
Mg(OH)2
This concentration of [OH-] is more than enough for the less
soluble CuCl2 to reach its Ksp and precipitate completely.
Con’t
1/23/2015
77
Practice Problem
Write the equations and ion-product expressions
Mg(OH)2 (s) € Mg 2+ (aq) + 2OH - (aq)
Cu(OH)2 (s) €
Cu 2+ (aq) + 2OH - (aq)
K sp = [Mg 2+ ][OH - ]2
K sp = [Cu 2+ ][OH - ]2
Calculate [OH-] of saturated Mg(OH)2 solution
-
[OH ] =
K sp
2+
[Mg ]
=
6.3  10-10
= 5.6  10-5 M
0.20
This is the maximum amount of [OH-] that will not precipitate
Mg2+ ion
Calculate [Cu2+] remaining in solution
K sp
2.2 x 10-20
-12
[Cu ] =
=
=
7.0
x
10
M
- 2
-5 2
[OH ] (5.6 x 10 )
2+
Since initial [Cu2+] was 0.10M, virtually all Cu2+ is precipitated
1/23/2015
78
Equilibria - Slightly Soluble Ionic Compounds

Equilibria Involving Complex Ions
 The product of any Lewis acid-base reaction is called an
“Adduct”, a single species that contains a new covalent
bond
A + :B
A - B (Adduct)

A complex ion consists of a central metal ion covalently
bonded to two or more anions or molecules, called
ligands
Ionic ligands
– OH- Cl- CNMolecular ligands – H2O CO NH3
Ex Cr(NH3)63+
Cr3+ is the central metal ion


1/23/2015
NH3 molecules are molecular ligands
Metal acts as Lewis Acid by accepting electron pair;
ligands acts as Lewis base by donating electron pair
All complex ions are Lewis adducts
79
Equilibria - Slightly Soluble Ionic Compounds

Complex Ions
 Acidic Hydrated metal ions are complex ions with water
molecules as ligands
 When the hydrated cation is treated with a solution of
another ligand, the bound water molecules exchange for
the other ligand
M(H 2O)4 2+ (aq) + 4NH 3 (aq)
M(NH 3 )4 2+ (aq) + 4H 2O(l)

At equilibrium

Water is constant in aqueous reactions
Incorporate in Kc to define Kf (formation constant)

[M(NH 3 )4 2+ ][H 2O]4
Kc =
[M(H 2O)4 2+ ][NH 3 ]4
Kc
[M(NH 3 )4 2+ ]
Kf =
=
4
[H 2 O]
[M(H 2 O)4 2+ ][NH 3 ]4
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80
Equilibria - Slightly Soluble Ionic Compounds

Complex Ions (Con’t)
 The actual process is stepwise, with ammonia molecules
replacing water molecules one at a time to give a series
of intermediate species, each with its own formation
constant (Kf)
M(H 2O)4 2+ (aq) + NH 3 (aq)
K f1




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M(H 2O)3 (NH 3 )2+ (aq) + H 2O(l)
[M(H 2O)3 (NH 3 )2+
=
[M(H 2O)4 2+ ][NH 3 ]
The sum of the 4 equations gives the overall equation
The product of the individual formation constants gives
the overall formation constant
The Kf for each step is much larger than “1” because
“ammonia” is a stronger Lewis base than H2O
K f = K f1 x K f2 x K f3 x K f4
Adding excess NH3 replaces all H2O and all M2+ exists as
M(NH3)42+
81
Practice Problem
What is the final [Zn(H2O)42+] after mixing 50.0 L of
0.002 M Zn(H2O)42+ and 25 L of 0.15 M NH3
K f of Zn(NH3 )42+ = 7.8  108
Zn(H 2O)4 2+ + 4NH 3 (aq)
Zn(NH 3 )4 2+ + 4H 2O(l)
[Zn(NH 3 )4 2
Kf =
[Zn(H 2O)4 2 ][NH 3 ]4
Determine Limiting Reagent
Moles Zn(H 2O)42+ = 50.0 L×0.002 mol / L = 0.100 mol
Moles NH 3 = 25 L× 0.15 mol / L
= 3.75 mol
Moles NH3 (3.75) >> 4 Moles Zn(H 2O)42+ (4  0.1)
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thus, Zn(H 2O)42+ is limiting
Con’t
82
Practice Problem (Con’t)

Find the initial reactant concentrations:
2+
[Zn(H 2O)4 ]init
[NH 3 ]init
50.0 L  0.0020 M
=
= 1.3  10-3 M
50.0 L + 25.0 L
25.0 L  0.15 M
=
= 5.0  10-2 M
50.0 L + 25.0 L
Con’t
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83
Practice Problem (Con’t)
Assume most Zn(H2O)42+ is converted to Zn(NH3)42+
[NH3 ]reacted  4(1.3  10-3 M)  5.2  10-3 M
[Zn(NH 3 )42+ ]  1.3  10-3 M
Concentration (M)
Zn(H2O)42+(aq) +
4NH3(aq)
⇄ Zn(NH3)42+(aq)
+
4H2O(l)
Initial
1.3 x 10-3
5.0 x 10-2
0

Change
(-1.3 x 10-3)
4 x -1.3 x 10-3
(-5.2 x 10-3)
(+1.3 x 10-3)

Equilibrium
X
4.5 x 10-2
1.3 x 10-3

Solve for x, the [Zn(H2O]42+ remaining at equilibrium
[Zn(NH 3 )4 2+ ]
1.3  10-3
8
Kf =
= 7.8  10 
2+
4
[Zn(H 2O)4 ][NH 3 ]
(x) (4.5  10-2 )4
x = [Zn(H 2O)4 2+ ]  4.1  10-7 M
Kf is very large;  [Zn(H2O)42+ should be “low”
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84
Complex Ions – Solubility of Precipitates

Increasing [H3O+] increases solubility of slightly soluble
ionic compounds if the anion of the compound is that of a
weak acid

A Ligand increases the solubility of a slightly soluble ionic
compound if it forms a complex ion with the cation
ZnS(s) + H 2O(l)
Zn 2+ (aq) + HS - (aq) + OH - (aq)
K sp = 2.0  10-22

When 1.0 M NaCN is added to the above solution, the CNions act as ligands and react with the small amount of
Zn2+(aq) to form a complex ion
Zn 2+ (aq) + 4CN - (aq)
Zn(CN)4 2- (aq)
K f = 4.2  1019
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85
Complex Ions – Solubility of Precipitates

Add the two reactions and compute the overall K
ZnS(s) + H 2O(l) €
Zn 2+ (aq) + HS - (aq) + OH - (aq)
Zn 2+ (aq) + 4CN - (aq) €
ZnS(s) + 4CN - (aq) + H 2O(l)
Zn(CN)4 2- (aq)
Zn(CN)4 2- + HS - (aq) + OH - (aq)
K overall = K sp  K f = (2.0  10-22 ) (4.2  1019 ) = 8.4  10-3

The overall equilibrium constant, Koverall , is more than a
factor of 1019 larger than the original Ksp (2.0 x 10-22)

This reflects the increased amount of ZnS in solution as
Zn(CN42-)
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86
Practice Problem
Calculate the solubility of Silver Bromide (AgBr) in water (H2O) and
1.0 M of photographic “Hypo” – Sodium Thiosulfate (Na2S2O3),
which forms the “complex” ion Ag(S2O3)23K sp (AgBr) = 5  10-13
K f (Ag(S2O3 )2 3- ) = 4.7  1013
a. Solubility AgBr in Water
K sp = 5  10-13
AgBr(s)
Ag + (aq) + Br - (aq)
K sp = 5.0  10-13 = [Ag + ][Br - ]
S = [AgBr]dissolved = [Ag + ] = [Br - ]
K sp = 5.0  10-13 = [Ag + ][Br-] = S x S
S = 7.1  10-7 M
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Con’t
87
Practice Problem (con’t)
Solubility AgBr in 1.0 M Hypo - Sodium Thiosulfate (Na2S2O3)
Write the overall equation:
AgBr(s) € Ag + (aq) + Br - (aq)
Ag + (aq) + 2S 2O 3 2- (aq) € Ag(S 2O 3 )2 3- (aq)
AgBr(s) + 2S 2O 3 2- (aq)
Ag(S 2O 3 )2 3- (aq) + Br - (aq)
Calculate Koverall (product of Ksp & Kf):
K overall
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[Ag(S 2O 3 )23- ][Br - ]
-13
13
=
=
K

K
=
(5.0
×
10
)(4.7
×
10
) = 24
sp
f
2[(S 2O 3 ) ]
88
Practice Problem (Con’t)

Set Up the Reaction Table
S = [AgBr]dissolved = [Ag(S 2O 3 )2 3- ]
+
2S2O32-(aq)
⇄
Ag(S2O3)23-(aq)
+
Br-(l)
Concentration (M)
AgBr(s)
Initial

1.0
0
0
Change

-2S
+S
+S
Equilibrium

1.0 - 2S
S
S
K overall
[Ag(S 2O 3 )2 3- ][Br-]
S2
=
=
 24
22
[(S 2O 3 ) ]
(1.0 M - 2S)
S
=
1.0 - 2S
24 = 4.9
S(1 + 2 × 4.9) = 4.9
S = 4.9 - 2S(4.9)
S =
S + 2S(4.9) = 4.9
4.9
= 0.45 M
1 + 9.8)
S = Solubility Ag(S 2O 3 )2 3- or AgBr = 0.45 M
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89
Ionic Equilibria in Aqueous Solutions

Equilibria Involving Complex Ions

Amphoteric Oxides & Hydroxides (Recall Chapter 8)

Some metals and many metalloids form oxides or
hydroxides that are amphoteric; they can act as acids or
bases in water

These compounds generally have very little solubility in
water, but they do dissolve more readily in acids or bases

Ex. Aluminum Hydroxide
Al(OH)3(s) ⇆ Al3+(aq) + 3OH-(ag)
Ksp = 3 x 10-34 (very insoluble in water)

In acid solution, the OH- reacts with H3O+ to form water
3H3O+(ag) + 3OH-(aq)  6H2O(l)
Al(OH)3(s) + H3O+(aq)  Al3+(aq) + 6H2O(l)
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90
Ionic Equilibria in Aqueous Solutions

Equilibria Involving Complex Ions

Aluminum Hydroxide in basic solution
Al(OH)3(s) + OH-(aq)  Al(OH)4-(aq)
1/23/2015

The above reaction is actually a much more
complex situation, involving multiple species

When dissolving an aluminum salt, such as Al(NO3),
in a strong base (NaOH), a precipitate forms initially
and then dissolves as more base is added

The formula for hydrated Al3+ is Al(H2O)63+

Al(H2O)63+ acts as a “weak polyprotic acid and
reacts with added OH- in a stepwise removal of the
H2O ligands attached to the hydrated Al
91
Ionic Equilibria in Aqueous Solutions

Amphoteric Aluminum Hydroxide in basic solution
Al(H2O)63+(aq) + OH-(aq) ⇆ Al(H2O)5OH2+(aq)
+ H2O(l)
Al(H2O)52+(aq) + OH-(aq) ⇆ Al(H2O)4(OH)2+(aq) + H2O(l)
Al(H2O)4+(aq) + OH-(aq) ⇆ Al(H2O)3(OH)3(s)
+ H2O(l)

Al(H2O)3(OH)3(s) is more simply written Al(OH)3(s)

As more base is added, a 4th H+ is removed from a H2O ligand
and the soluble ion Al(H2O)2(OH)4-(aq) forms
Al(H2O)3(OH)3(s) + OH-(aq) ⇆
Al(H2O)2(OH)4-(aq)
The ion is normally written as Al(OH)4-(aq)
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92
Equation Summary
[HA - ][H 3O + ]
Ka =
[HA]
B(aq)
A-(aq)
Initial Ci
-
-
-
0

-
-
+X
+X
Equilibrium Cf
-
-
X
X
€
+
H2O(l)
A(aq)
HA + H2O
+

Concentration-
H 3O+ + A-
[H 3O + ][A - ]
Ka =
[HA]
[base]
pH = pK a + log
[acid]
[A - ]
When :
=1
[HA]
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 [A - ] 
pH = pK a + log 
= pK a + log1 = pK a + 0 = pK a

 [HA] 
93
Equation Summary
[BH + ][OH - ]
Kb =
[B]
K w = K a  K b = 1.0  10-14
pK w = pH + pOH = 14
Qsp = [M n+ ]p [Xz- ]q = K sp
2H 2 O
Kw
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+
+ OH-
[H 3 O + ][OH - ]
+
=
=
[H
O
][OH
] = 1× 10
3
2
[H 2 O]
H3O
Kn =
H3O
+
[H 2 O]2
[H 3 O + ][OH - ]
+ OH =
-14
2H 2 O
1
1
14
=
=
1×10
Kw
[H 3 O + ][OH - ]
94