1. CHEMICAL EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16

2. OVERVIEW THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
CHEMICAL EQUILIBRIUM
THE EQUILIBRIUM STATE
EQUILIBRIUM-CONSTANT EXPRESSIONS
TYPES OF EQUILIBRIUM CONSTANTS IN ANALYTICAL CHEMISTRY
APPLYING THE ION-PRODUCT CONSTANT FOR WATER

3. OVERVIEW SOLUBILITY-PRODUCT CONSTANTS LE CHATELIER’S PRINCIPLE

4. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
Electrolytes:
It forms ions when dissolved in water (or certain other solvents) and thus produce solutions that conduct electricity.
Strong electrolytes ionize essentially completely in a solvent, but weak electrolytes ionize only partially.
These characteristics mean that a solution of a weak electrolyte will only conduct electricity as well as a solution containing an equal concentration of a strong electrolytes.

5. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
A salt is produced in the reaction of an acid with a base.
Examples include NaCl, Na2SO4, NaOOCH3
Table 1: Classification of Electrolytes

6. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
Acids and Bases:
According to Bronsted-Lowry concept, an acid donates protons and a base accepts protons.
An acid donates protons only in the presence of a proton acceptor (a base). Likewise, a base accepts protons only in the presence of a proton donor (an acid).
A conjugate base is formed when an acid loses a proton.
Acetate ion is the conjugate base of acetic acid.

7. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
acid1 ↔ base1 + proton
We refer to acid1 and base1 as a conjugate pair.
Every base accepts a proton to produce a conjugate acid.
base2 + proton ↔ acid2
Neutralization reaction:
acid1 + base2 ↔ base1 + acid2
Many solvents are proton donors or proton acceptors & can induce basic or acidic behavior in solutes dissolved in them.

8. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
In an aqueous solution of ammonia, water can donate a proton and acts as an acid with respect to the solute NH3:
Ammonia (base1) reacts with water (acid2), to give the conjugate base (base2) of the acid water.
Water acts as a proton acceptor, or base, in an aqueous solution of nitrous acid.
Base Acid Conjugate Conjugate
Acid Base2
Base Acid Conjugate Conjugate
Acid Base2

9. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
The conjugate base of the acid HNO2 is nitrate ion.
The conjugate acid of water is the hydrated proton, H3O+ (hydronium ion).
Figure 1: Possible structures for the hydronium ion.

10. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
Amphiprotic species:
Species that have both acidic and basic properties.
Example: Dihydrogen phopshate
Base Acid Acid Base2
Acid Base Base Acid2

11. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
The simple amino acids are an important class of amphiprotic compounds that contain both a weak acid and a weak base functional group.
When dissolved in water, an amino acid, such as glycine, undergoes a kind of internal acid/base reaction to ptoduce a zwitterion.
A zwitterion is an ion that has both a positive and a negative charge.
Glycine zwitterion

12. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
Water is the classic example of an amphiprotic solvent, that is, a solvent that can act either as an acid or as a base, depending on the solute.
Other common amphiprotic solvents are methanol, ethanol, and anhydrous acetic acid.

13. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
Strengths of acids and bases
Figure 2 shows the dissociation reactions of a few common acids in water.
Figure 2: Dissociation reactions & relative strengths of some common acids & their conjugate base.

14. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
The common strong bases include NaOH, KOH, Ba(OH)2, and the quaternary ammonium hydroxide R4NOH (R = an alkyl group: CH3 or C2H5).
The common strong acids include HCl, HBr, HI, HClO4, HNO3, the first proton in H2SO4, and the organic sulfonic acid RSO3H.
Perchloric and hydrochloric acids are strong acids in water.
If anhydrous acid, a weaker proton acceptor that water, is substituted as the solvent, neither of these acids undergoes complete dissociation.

15. THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS
Equilibria below is established:
In a differentiating solvent, various acids dissociate to different degrees and have different strengths.
In a leveling solvent, several acids are completely dissociated and show the same strength.
Base Acid Acid Base 2

16. CHEMICAL EQUILIBRIUM
Many reactions used in analytical chemistry never result in complete conversion of reactants to products.
Instead, they proceed to a state of chemical equilibrium in which the ratio of concentrations of reactants and products is constant.
Equilibrium-constant expressions are algebraic equations that describe the concentration relationship among reactants and products at equilibrium.

17. CHEMICAL EQUILIBRIUM The Equilibrium State:
Consider the chemical reaction:
The rate of this reaction and the extent to which it proceeds to the right by monitoring the appearance of the orange-red color of the triiodide ion I3-.
The final position of a chemical equilibrium is independent of the route to the equilibrium state.

18. CHEMICAL EQUILIBRIUM
The equilibrium state is altered by applying stress (temperature, pressure, concentration of reactant or product) to the system.
These effects can be predicted qualitatively from the Le Chatelier’s principle.
The Le Chatelier’s principle:
It states that the position of an equilibrium always shifts in such a direction as to relieve a stress that is applied to the system.

19. CHEMICAL EQUILIBRIUM
An increase in temperature of a system alters the concentration relationship in the direction that tends to absorb heat, and an increase in pressure favors those participants that occupy a smaller total volume.
The mass-action effect is a shift in the position of an equilibrium caused by adding one of the reactants or products to a system.

20. CHEMICAL EQUILIBRIUM Equilibrium is a dynamic process.
Although chemical reactions appear to stop at equilibrium, in fact, the amounts of reactants and products are constant because the rates of the forward and reverse processes are exactly the same.
Chemical thermodynamics is a branch of chemistry that concerns the flow of heat & energy in chemical reactions.
The position of a chemical equilibrium is related to these energy changes.

21. CHEMICAL EQUILIBRIUM Equilibrium-constant expressions
The influence of concentration or pressure on the position of a chemical equilibrium is conveniently described in quantitative terms by means of an equilibrium-constant expression.
They allow us to predict the direction and completeness of chemical reactions.
However, yields no information concerning the rate of reaction.

22. CHEMICAL EQUILIBRIUM
Consider a generalized equation for a chemical equilibrium: wW + xX ↔ yY + zZ
The equilibrium-constant expression:
If a reactant or product is a pure liquid, a pure solid, or the solvent present in excess, no term for this species appears in the equilibrium-constant expression.
Molar concentration or partial pressure in atmospheres if they are gas phase reactants or products.

23. The activities of species Y, Z, W, X
CHEMICAL EQUILIBRIUM
A temperature dependent equilibrium constant:
Table 1 summarizes the types of chemical equilibria and equilibrium constant that are of importance in analytical chemistry.
The activities of species Y, Z, W, X

24. CHEMICAL EQUILIBRIUM
Table 1: Equilibria & Equilibrium constants important in analytical chemistry

25. CHEMICAL EQUILIBRIUM Applying the ion-product constant for water
Aqueous solutions contain small concentrations of hydronium and hydroxide ions as a consquence of the dissolution reaction.
An equilibrium constant for this reaction can be formulated:

26. Ion-product constant for water
CHEMICAL EQUILIBRIUM
When compared with the concentration of hydrogen and hydroxide ions:
At 25oC, Kw = x
Table 2:
Variation
of KW with
temperature
Ion-product constant for water

27. CHEMICAL EQUILIBRIUM Example 1:
Calculate the hydronium and hydroxide ion concentrations of pure water at 25oC and 100oC.

28. CHEMICAL EQUILIBRIUM Using solubility-product constants:
Sparingly soluble salts are essentially completely dissociated in saturated aqueous solution.
When an excess of barium iodate is equilibrated with water, the dissociation process is adequately described by:

29. Solubility-product constant or Solubility product
CHEMICAL EQUILIBRIUM
Solubility-product constant or Solubility product

30. CHEMICAL EQUILIBRIUM Example 2:
What mass (in grams) of Ba(IO3)2 (487 g/mol) can be dissolved in 500 ml of water at 25oC?

31. CHEMICAL EQUILIBRIUM
The effect of a common ion in the solubility of a precipitate:
The common-ion effect is a mass-action effect predicted from Le Chatelier principle.
The solubility of an ionic precipitate decreases when a soluble compound containing one of the ions of the precipitate is added to the solution.

32. CHEMICAL EQUILIBRIUM Example 3:
Calculate the molar solubility of Ba(NO3)2 in a solution that is M in Ba(NO3)2.

33. LE CHATELIER’S PRINCIPLE
When a change is imposed on a system at equilibrium the position of the equilibrium shifts in a direction that tends to reduce the effect of that change.
Henri Le Chatelier
Studied mining engineering.
Interested in glass and ceramics.
It is like the “undo” button on your computer!

34. LE CHATELIER’S PRINCIPLE
Factors that Affect Equilibrium:
Concentration
Temperature
Pressure (For gaseous systems only!)
The presence of a catalyst
(i) Effect of a Change in Concentration

35. LE CHATELIER’S PRINCIPLE
Add more reactant  Shift to products
Remove reactants  Shift to reactants
Add more product  Shift to reactants
Remove products  Shift to products
The reaction quotient for an equilibrium system is the same as the equilibrium expression, but the concentrations are NOT at equilibrium!

36. LE CHATELIER’S PRINCIPLE
Q = [NO2]2
[N2O4]
Keq at
equilibrium
Changes in concentration are best understood in terms of what would happen to “Q” if the concentrations were changed.
If Q< K then there are too many reactants, the reaction will shift in the forward direction (the products).
If Q>K then there are too many products, the reaction will shift to the reactants.
N2O4(g)  2NO2(g)

37. LE CHATELIER’S PRINCIPLE
(ii) Effect of a Change in Pressure (Volume)
The system is initially at equilibrium.
The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.

38. LE CHATELIER’S PRINCIPLE
Decreasing volume:

39. LE CHATELIER’S PRINCIPLE
It only affects equilibrium systems with unequal moles of gaseous reactants and products.
Increase Pressure
Stress of pressure is reduced by reducing the number of gas molecules in the container.
There are 4 molecules of reactants vs. 2 molecules of products.
Thus, the reaction shifts to the product ammonia.
N2(g) + 3H2(g) = 2NH3(g)

40. LE CHATELIER’S PRINCIPLE
Increasing the volume:
The system shifts in the direction that increases its pressure.
Decrease Pressure:
Stress of decreased pressure is reduced by increasing the number of gas molecules in the container.
There are two product gas molecules vs. one reactant gas molecule.
Thus, the reaction shifts to the products.
PCl5(g) = PCl3(g) + Cl2(g)

41. LE CHATELIER’S PRINCIPLE
(iii) Effect of a Change in Temperature
The value of K changes with temperature (We can use this to predict the direction of this change).
Exothermic reaction – produces heat (heat is a product)
Adding energy shifts the equilibrium to the left (away from the heat term).
Endothermic reaction – absorbs energy (heat is a reactant)
Adding energy shifts the equilibrium to the right (away from the heat term).

42. LE CHATELIER’S PRINCIPLE
The meaning of K:
K>1  the equilibrium position is far to the right
K<1  the equilibrium position is far to the left
(iv) Presence of a Catalyst
A catalyst lowers the activation energy and increases the reaction rate.
It will lower the forward and reverse reaction rates.

43. LE CHATELIER’S PRINCIPLE
Therefore, a catalyst has NO EFFECT on a system at equilibrium!
It just gets you to equilibrium faster!
Presence of an Inert Substance
An inert substance is a substance that is not- reactive with any species in the equilibrium system.
These will not affect the equilibrium system.
If the substance does react with a species at equilibrium, then there will be a shift!

44. LE CHATELIER’S PRINCIPLE
Example 4
Given:
S8(g) O2(g)  8 SO3(g) kcals
What will happen when:
Oxygen gas is added?
The reaction vessel is cooled?
? ? ?
Shifts to products 
? ? ?
Shifts to Products – to replace heat

45. LE CHATELIER’S PRINCIPLE
The size of the container is increased?
Sulfur trioxide is removed?
A catalyst is added to make it faster?
V increases, Pressure decreases,
shifts to more particles – to reactants !
? ? ?
Shift to products to replace it !
? ? ?
? ? ?
No change !

46. EXAMPLE 1
Since OH- and H3O+ are formed only from the dissolution of water, their concentration must be equal:
Then,
At 25C,
= 1.00 x 10-7 M

47. EXAMPLE 1
At 100C,
= 7.0 x 10-7 M

48. EXAMPLE 2
The solubility-product constant for Ba(IO3)2 is 1.57 x 10-9 (Appendix 2).
Equilibrium between the solid and its ions in solution is described:
The equilibrium reveals that 1 mol of Ba2+ is formed for each mole of Ba(IO3)2 that dissolves.

49. EXAMPLE 2 Molar solubility of Ba(IO3)2 = [Ba2+]
Since two moles of iodate are produced for each mole of barium ion, the iodate concentration is twice the barium ion concentration:
The equilibrium constant expression:
= 1.57 x 10-9

50. EXAMPLE 2
= 7.32 x 10-4 M
Since 1 mol of Ba2+ is produced for every mole of Ba(IO3)2,
Solubility = 7.32 x 10-4 M
The number of milimoles of Ba(IO3)2 dissolved in 500 ml of solution,

51. EXAMPLE 2 The mass of Ba(IO3)2 in 500 ml is given by: Mass Ba(IO3)2 =

52. EXAMPLE 3 Molar solubility of Ba(IO3)2 = 0.5 [IO3-]
There are two sources of barium ions: Ba(NO3)2 and Ba(IO3)2.
The contribution from the nitrate is M, and that from the iodate is equal to the molar solubility, or 0.5 [IO3-].

53. EXAMPLE 3 The solubility-product expression, Assumption:
Simplified equation:
=2.80 x 10-4 M

54. EXAMPLE 3 Solubility of Ba(IO3)2 =0.5 [IO3-] =0.5 x 2.80 x 10-4
= 1.40 x 10-4 M