Analytical Chemistry Daniel Harris

1. Chemical equilibriumChapter 6
The rates of the forward and
reverse reactions are equal
Chemical equilibrium is achieved when:
The amount of the reactants and
products remain constant
Dr. Sa’ib Khouri
AUM-JORDAN
QUANTITATIVE CHEMICAL ANALYSIS
By Daniel C. Harris

2. 2
Chemical Kinetics and Chemical Equilibrium
A + B C + D
kf
kr
Forward Rate = kf [A] [B]
Reverse Rate = kr [C] [D]
At Equilibrium: Forward Rate = Reverse Rate
kf [A] [B] = kr [C] [D]
kf
kr
[C] [D]
[A] [B]
= K =

3. This chapter introduces equilibriums for the solubility of ionic
compounds, complex formation, and acid-base reactions.
The symbol [ ] stands for the concentration at
equilibrium state
Capital letter stands for a chemical species, and
small letter denotes stoichiometry coefficient
K >> 1 Favor products
K << 1 Favor reactants

4. 1. Concentrations of solutes should be expressed as moles per liter.
Homogeneous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g) 2NO2 (g)
e.g.
2. Concentrations of gases should be expressed in bars.
3. Concentrations of pure solids, pure liquids, and solvents
are omitted because they are unity.
Practically, no units for the equilibrium constant
Heterogeneous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
e.g.
K =
PNO2
2
PN2O4
K = PCO2

5. 5
Kp =
2
PNO PO
2
PNO
2
2
PO2 = Kp
PNO
2
2
PNO
2
= 347 atm
Example
The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO2
= 0.400 atm and PNO = 0.270 atm?
2NO2 (g) 2NO (g) + O2 (g)
Solution
PO2 = 158 x
(0.400)2
(0.270)2

6. Equilibrium constant for reverse reaction
Equilibrium constant for forward reaction
Manipulating Equilibrium Constants
When the equation for a reversible reaction is written in the
opposite direction, the equilibrium constant becomes the
reciprocal of the original equilibrium constant.

7. If n reactions are added, the overall equilibrium constant is
the product of n individual equilibrium constants.
If two reactions are added, the new K is the product
of the two individual values:
Equilibrium constant for sum of reactions:

8. The third reaction can be obtained by reversing the second reaction and
adding it to the first reaction:
Solution

9. Enthalpy, Entropy, and Free Energy
Enthalpy change, ΔH: the heat absorbed or released when the reaction
takes place.
The standard enthalpy change, ΔHº : the heat absorbed or released when all
reactants and products are in their standard states.
Endothermic reaction: positive value of ΔH
Exothermic reaction: negative value of ΔH
e.g.
Equilibrium and Thermodynamics
Equilibrium is controlled by the thermodynamics of a chemical reaction
DH = Hproducts – Hreactants

10. Entropy, S, of a substance is a measure of its “disorder,”
The greater the disorder, the greater the entropy. Gas has higher entropy
than a liquid
Ions in aqueous solution are normally more disordered than in their
solid salt:
ΔS is the change in entropy (entropy of products minus entropy of reactants)
Free Energy
A chemical reaction is driven toward the formation of products (the reaction is
clearly favored) by a negative value of ΔH (heat given off) or a positive value
of ΔS (more disorder) or both.
When ΔH is positive and ΔS is negative, the reaction is clearly disfavored
DS = Sproducts – Sreactants

11. Gibbs free energy, ΔG,
A reaction is favored if ΔG is negative
(At constant pressure)
e.g. For the dissociation of HCl
(favors)
(disfavors)

12. ΔG is negative, so the reaction is favored when all species are in
their standard state.
The equilibrium constant depends on ΔG
R is the gas constant
8.314 J/ (Kmol)
T is temperature (Kelvin)
For the last example

13. Because the equilibrium constant is large, HCl gas molecules
are very soluble in water and are nearly completely ionized to
cations and anions when dissolve in water.
We say that a reaction is spontaneous under standard
conditions if ΔG⁰ is negative
Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset as the
system reaches a new equilibrium position.

14. 14
Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
Add
NH3
Equilibrium
shifts left
to offset
stress

15. Example

16. According to the principle of Le Châtelier, the reaction should go
back to the left to partially offset the increase in dichromate, which
appears on the right side of Reaction

17. When the temperature of a system is changed
This term is independent of T

19. The equilibrium constant for the reaction in which a solid salt
dissolves to give its constituent ions in solution (Solid is omitted
from the equilibrium constant)
e.g. Mercury (I) chloride (Hg2Cl2, also called mercurous chloride)
The reaction in water is
The solubility product constant, Ksp, is
A saturated solution is one that contains excess, undissolved solid. The
solution contains all the solid capable of being dissolved under the prevailing
conditions.

21. 21
Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a
saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a
saturated solution.

22. 22
What is the solubility of silver chloride in g/L?
AgCl (s) Ag+
(aq) + Cl-
(aq)
Ksp = [Ag+][Cl-]Initial (M)
Change (M)
Equilibrium (M)
0.00
+s
0.00
+s
s s
Ksp = s2
s = Ksp
s = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M
Solubility of AgCl =
1.3 x 10-5 mol AgCl
1 L soln
143.35 g AgCl
1 mol AgCl
x
= 1.9 x 10-3 g/L
Ksp = 1.6 x 10-10
Example

24. 24
If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a
precipitate form?
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q = [Ca2+]0[OH-]0
2
= 0.100 x (4.0 x 10-4)2 = 1.6 x 10-8
Q < Ksp No precipitate will form
Example
Ca(OH)2 (s) Ca2+
(aq) + 2OH- (aq)

25. Exactly 200 mL of 0.0040 M BaCl2 are added to exactly 600 mL of 0.0080 M
K2SO4. Will a precipitate form?
Example

28. Example
If the solubility of Ag2CrO4 (FM=332 g/mol) is 0.0279 g/L. Calculate Ksp?
Solution
[Ag2CrO4] = 0.0279
𝑔
𝐿
·
𝑚𝑜𝑙
332 𝑔
= 8.40 x 10-5 𝑚𝑜𝑙
𝐿
Ag2CrO4
2Ag+
(aq) + CrO4
2-
(aq)
Ksp
2S S
Ksp = [Ag+]2 [CrO4
2-]
= (2S)2 ∙ S = 4S3
= 4 (8.40 x 10-5 )3
= 2.37 x 10-12

29. Example
Calculate the solubility of Ba(IO3)2 (FM=487 g/mol), Ksp=1.57 x 10-9
Ba(IO3)2
Ba2+
(aq) + 2IO3
-
(aq)
Ksp
S 2S
Solution
Ksp = [Ba2+] [IO3
-]2
1.57 x 10-9 = (S) ∙ (2S)2 = 4S3
S3 =
1.57 x 10−9
4
= 3.93 x 10-10 S= (3.93 x 10-10)1/3
S = 7.32 x 10-4 mol/L (molar solubility)
Solubility = 7.32 x 10-4
mol
L
·
487 g
mol
= 0.357
g
L

30. This application of Le Chatelier’s principle is called the common ion
effect. A salt will be less soluble if one of its constituent ions is already
present in the solution.
Common Ion Effect
For the ionic solubility reaction

31. Example
Calculate the solubility of Ba(IO3)2 in the presence of 0.10 M Ba(NO3)2.
Ksp (Ba(IO3)2) = 1.57 x 10-9
Solution

33. Bronsted-Lowry definition does not require the formation of H3O+
(Extended to non-aqueous solutions or gas phase)
Acid: proton donor
Base: proton acceptor
acid base salt
Protic Acids and Bases
The word protic refers to chemistry involving transfer of H+
from one molecule to another
In aqueous chemistry:
an acid is a substance that increases [H3O+] (hydronium ion) when added to
water. (e.g. HCl in H2O).
a base is a substance that decreases [H3O+] , therefore, a base increases [OH-
]
in aqueous solution. (e.g. NaOH in H2O).
HCl(g) + NH3(g) → NH4Cl(s)
Most salts with a single positive and negative charge dissociate completely into ions in water

34. Conjugate Acids and Bases
The products of a reaction between an acid and a base are also classified as acids
and bases:
Conjugate acids and bases are related by the gain or loss of one proton

35. The Nature of H+
and OH-
The proton does not exist by itself in water. The simplest formula found in some
crystalline salts is H3O+
For example, crystals of perchloric acid monohydrate contain pyramidal
hydronium (also called hydroxonium) ions:
A more accurate formula would be

36. In aqueous solution, H3O+
is tightly associated with
three molecules of water through exceptionally strong
hydrogen bonds (dotted lines), and one H2O (at the top)
is held by weaker ion-dipole attraction (dashed line)
To emphasize the chemistry of water, we will write H+
as H3O+
For example, water can be either an acid or a base.
Water is an acid with respect to methoxide:
But with respect to hydrogen bromide, water is a base:

37. Water undergoes self-ionization, called autoprotolysis (also called
autoionization), in which it acts as both an acid and a base:

38. Protic solvents have a reactive H+
, and all protic solvents undergo autoprotolysis.
Example: Acetic acid:

39. As the concentration of H+
increases, the concentration of OH
-
necessarily
decreases, and vice versa.
An approximate definition of pH is the negative logarithm of the H+
concentration

40. pH of various substances

41. Although pH generally falls in the range 0 to 14, these are not the limits of pH
A pH of -1, for example, means -log[H+
] = -1; or [H+
] = 10 M
This concentration is attained in a concentrated solution of a strong acid such as HCl.

42. Acids and bases are commonly classified as strong or weak, depending on
whether they react nearly “completely” or only “partially” to produce H+
or
OH-
Even though the hydrogen halides HCl, HBr,
and HI are strong acids, HF is not a strong
acid.
HF does completely give up its proton to H2O:
Here the hydronium ion remains tightly
associated with F-
through a hydrogen bond,
even in dilute solution
Thus, HF does not behave as a strong acid

43. All weak acids, denoted HA, react with water by donating a proton to H2O:
Dissociation of weak acid:
The equilibrium constant is called Ka (the acid dissociation constant)
A weak base, B, reacts with water by abstracting a proton from H2O:
Base hydrolysis:
The equilibrium constant Kb (the base hydrolysis constant)

44. Common Classes of Weak Acids and Bases
Acetic acid is a typical weak acid;
Most carboxylic acids are weak acids, and most carboxylate anions
are weak bases.

45. Methylamine is a typical weak base;
weak bases weak acids
Amines are nitrogen-containing compounds:

46. Polyprotic acids and bases
Compounds that can donate or accept more than one proton.
e.g. Oxalic acid (diprotic)

47. Phosphate is tribasic

48. Carbonic acid (H2CO3), a very important diprotic carboxylic acid is formed by the
reaction of CO2 with H2O:

49. Relation Between Ka and Kb
A relation between Ka and Kb of a conjugate acid-base pair in aqueous
solution.
e.g. The acid HA and its conjugate base A-
;

50. For a diprotic acid, we can derive relationships between each
of two acids and their conjugate bases:
The final results are

52. HW

53. HW
Ksp of La(OH)3 is 2.0 x 10-21

54. →
(Ksp of La(OH)3 is 2.0 x 10-21
)
Solution

57. Solution