1. 18.2 Calculations involving acids and bases
TOPIC 18 ACIDS AND BASES
18.2
Calculations involving acids and bases

2. ESSENTIAL IDEA
The equilibrium law can be applied to acid-base reactions. Numerical problems can be simplified by making assumptions about the relative concentrations of the species involved. The use of logarithms is also significant here.
NATURE OF SCIENCE (1.9)
Obtaining evidence for scientific theories – application of the equilibrium law allows strengths of acids and bases to be determined and related to their molecular structure.

3. INTERNATIONAL-MINDEDNESS
Mathematics is a universal language. The mathematical nature of this topic helps chemists speaking different languages to communicate more objectively.

4. GUIDANCE
Know that the value for Kw depends upon temperature.

5. Water as an Equilibrium System
Water has the ability to act as either a Bronsted Lowry acid or base.
Autoionization – spontaneous formation of low concentrations of [H+] and OH-] ions by proton transfer from one molecule to another.

6. The Ionization of Water
The reaction for the ionization of water is:
H2O(l) ↔ H+(aq) + OH-(aq)
So Kw = [H+][OH-]
Kw is the ionic product constant of water and does not include water in the expression because water it is a liquid.

7. In pure water, [H+] = [OH-]
at 25°C, Kw = [H+][OH-] = 1.00 x 10-14
to solve for [H+], [H+] = √Kw
So at 25°C, [H+]=√(1.00 x 10-14)= 1.00x10-7
pH=-log[H+]=-log(1.00x10-7)= 7.00

8. Learning outcome
Deduce [H+] and [OH-] for water at different temperatures given Kw values.

9. Kw is Temperature Dependent
The dissociation of water is an endothermic reaction because it requires the breaking of bonds.
heat + H2O(l) ↔ H+(aq) + OH-(aq)
An increase in temp shifts equilibrium to the right which increases ion concentrations.
A decrease in temp shifts equilibrium to the left which decreases ion concentrations.

10. Water is always neutral because [H+] = [OH-]
The pH of water is 7.00 only at 25°C.
At higher temperatures, the ion concentrations are higher so pH’s are lower. Ex: @50°C pH is 6.63
At lower temperatures, the ion concentrations are lower so pH’s are higher. Ex: @0°C pH is 7.47
The pH’s can fluctuate above and below 7.00, but water is neutral because [H+] = [OH-]

11. To solve for [H+] of water at any temperature, just take the square root of the given Kw.
The [OH-] is then equal to the [H+].

12. UNDERSTANDING/KEY IDEA 18.2.A
The expression for the dissociation constant of a weak acid is Ka and a weak base is Kb.

13. Weak Acids in water HA(aq) + H2O(l) ↔ H3O+ + A- Ka = [H3O+][A-] [HA]or
HA(aq) ↔ H+ + A-
Ka = [H+][A-]
H+ and H3O+ are used interchangeably.

14. Ka Ka = acid dissociation or ionization constant
The larger the Ka, the more the weak acid has dissociated so we can compare strengths of weak acids by their Ka’s.

15. Weak Bases in water B(aq) + H2O(l) ↔ BH+ + OH- Kb = [BH+][OH-] [B]
You must ALWAYS include water in the base equation because you have to show that water is the source of H+.

16. Kb Kb = base dissociation or ionization constant
The larger the Kb, the more the weak base has dissociated so we can compare strengths of weak bases by their Kb’s.

17. UNDERSTANDING/KEY IDEA 18.2.B
The relationship between Ka and pKa is (pKa = -logKa) and between Kb and pKb is (pKb = - logKb).

18. pKa and pKb
Just as pH is a compact way to measure the H+ concentration, pKa and pKb are compact ways to measure Ka and Kb.
Ka and Kb are often very small numbers and involve negative exponents so the pKa and pKb values are easier to deal with.
pKa = -logKa and pKb = -logKb
Ka = 10(–pKa) and Kb = 10(-pKb)

19. The pKa and pKb scale is logarithmic so it compresses a very wide range of Ka’s and Kb’s into a much smaller scale of numbers.
This means that if you increase the pKa or pKb by one unit, the Ka or Kb decreases by 10 times and if you decrease pKa or pKb by one unit, Ka or Kb increases by 10 times.

20. pKa and pKb are usually positive and have no units.
The relationship between pKa and Ka and pKb and Kb are inverse.
That means as Ka increases, pKa decreases.
The smaller the Ka, the larger the pKa, the weaker the acid.
The larger the Kb, the smaller the pKb, the stronger the base.

21. UNDERSTANDING/KEY IDEA 18.2.C
For a conjugate acid base pair, Ka x Kb = Kw

22. More derivations The general acid equation:
HA(aq) ↔ H+ + A Ka = [H+][A-]
[HA]
The general conj base of the acid equation:
A- + H2O(l) ↔ HA + OH Kb = [HA][OH-]
[A-]

23. When you multiply the 2 equations:
Ka x Kb = [H+][A-] x [HA][OH-]
[HA] [A-]
The [HA]’s and [A-]’s cancel, leaving:
Ka x Kb = [H+][OH-] = Kw

24. Take the logarithms of both sides:
pKa + pKb = pKw
We know that at 25°C, Kw = 1.00x10-14
so pKw = 14.00
Therefore,
pKa + pKb = 14.00

25. APPLICATION/SKILLS
Be able to discuss the relative strengths of acids and bases using values of Ka, pKa, Kb, and pKb.

26. HA + H2O ↔ H3O+ + A- HA is the acid with A- its conjugate base.
H2O is the base with H3O+ as its conjugate acid.
There is a competition for the proton (H+) between the bases. The bases in this example are A- and water.
If water is a stronger proton acceptor than A-, then the acid is strong because it dissociates readily and the equilibrium lies to the right.
If A- is the stronger proton acceptor, the acid will be weak because most will stay in the HA form and the equilibrium will lie to the left.

27. A strong acid has a weak conj base.
Its conjugate base is weaker than water so water wins the competition for the H+ ions.
A weak acid has a strong conj base.
Its conjugate base is stronger than water. The water is not very successful in pulling the H+’s from the weak acid.

28. The higher the Ka and Kb, the stronger the acid or base because the equilibrium lies farther to the right and it dissociates more.
The higher the Ka and Kb, the lower the pKa and pKb.

29. APPLICATION/SKILLS
Be able to solve problems involving [H+], [OH-], pH, pOH, Ka, pKa, Kb, and pKb.

30. GUIDANCE
When making approximations in weak acid or weak base problems, always state the assumption. Quadratic equations will not be assessed.

31. RELATIONSHIPS OF pH and pOH
pH = -log [H+]
pOH = -log [OH-]
pH + pOH = 14
[H+] = 2nd log (-pH)
[OH-] = 2nd log (-pOH)
Sig Fig Rule for pH – the number of decimal places in the pH figure is equal to the number of sig figs in the original [H+].

32. pH’s of Strong Acids and Bases
Since the 6 strong acids and the 7 strong bases dissociate or ionize completely, you can calculate the pH directly from the given concentration.
2mol/dm3 HCl ↔ 2mol/dm3 H+ + 2 mol/dm3 Cl-
1mol/dm3 Ba(OH)2 ↔ 1mol/dm3 Ba mol/dm3 OH-

33. pH’s of Weak Acids and Bases
To calculate the pH of weak acids and bases, you must use equilibrium calculations.

34. LET’S RE-VISIT ICE TABLES
Only molar concentrations are used in ICE tables. ICE stands for “Initial” concentrations, “Change” in concentrations, and “Equilibrium” concentrations.

35. To solve for pH for weak acids and bases, you will use ICE tables.

36. Example 1 (Basic weak acid)
Calculate the [H+] and percent dissociation of a mol/dm3 solution of HF (Ka = 7.2x10-4).
What are the major species?
HF and H2O
Set up your ICE table.
HF ↔ H+ + F- I
C -x x x
E x x x
Assume 1.0-x = 1.0 because Ka is so small.

37. Set up your Ka expression and solve for x.
Ka = [H+][F-] = 7.2x10-4 = x x = [H+]=.027
[HF]
Calculate the % dissociation.
% dissoc = x = x 100% = 2.7%
[conc]ini

38. Weak Acid Equilibrium Constants
Sample problem . A certain weak acid dissociates in water as follows: HA + H2O  H3O+ + A- If the initial concentration of HA is 1.5 M and the equilibrium concentration of H3O+ is M. Calculate Ka for this acid Solution Ka = [H3O+ ] [A-] [HA] I C E Substituting [HA] 1.5 -x 1.5-x Ka = (0.0014)2 = 1.31 x 10-6 [A-] 0 +x x [H3O+ ] 0 +x x x = x =

39. Weak Base Equilibria
Weak bases, like weak acids, are partially ionized. The degree to which ionization occurs depends on the value of the base dissociation constant
General form: B + H2O  BH+ + OH-
Kb = [BH+][OH-]
[B]
Example
NH3 + H2O  NH OH-
Kb = [NH4+][ OH-]
[NH3]

40. Weak Base Equilibrium Constants
Sample problem . A certain weak base dissociates in water as follows:
B + H2O  BH+ + OH-
If the initial concentration of B is 1.2 M and the equilibrium concentration of OH- is M. Calculate Kb for this base
Solution
Kb = [BH+ ] [OH-]
[B]
I C E Substituting
[B] x x Kb = (0.0011) = x 10-6
[OH-] x x
[BH+ ] x x
x =
1.2-x =

41. Weak Acid Equilibria Concentration Problems
Problem 1. A certain weak acid dissociates in water as follows: HA + H2O  H3O+ + A-
The Ka for this acid is 2.0 x Calculate the [HA] [A-], [H3O+ ] and pH of a 2.0 M solution
Solution
Ka = [H3O+ ] [A-] = 2.0 x 10-6
[HA]
I C E Substituting
[HA] x x Ka = x = 2.0 x 10-6
[A-] x x x
[H3O+ ] x x If x <<< 2.0 it can be dropped from the denominator
The x2 = (2.0 x10-6)(2.0) = 4.0 x x = 2.0 x 10-3
[A-] = [H3O+ ] = 2.0 x [HA] = = 1.998
pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7

42. Weak Acid Equilibria Concentration Problems
Problem 2. Acetic acid is a weak acid that dissociates in water as follows: CH3COOH + H2O  H3O+ + CH3COO-
The Ka for this acid is 1.8 x Calculate the [CH3COOH],[CH3COO-] [H3O+ ] and pH of a M solution
Solution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH]
I C E Substituting
[CH3COOH] x x Ka = x = 1.8 x 10-5
[CH3COO- ] x x x
[H3O+] x x If x <<< it can be dropped from the denominator
The x2 = (1.8 x10-5)(0.100) = 1.8 x x = 1.3 x 10-3
[CH3COO--] = [H3O+ ] = 1.3 x [CH3COOH ] = =
pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88

43. Weak Base Equilibria
Example1. Ammonia dissociates in water according to the following equilibrium
NH3 + H2O  NH OH-
Kb = [NH4+][ OH-] = x 10-5
[NH3]
Calculate the concentration of [NH4+][ OH-] [NH3 ]and the pH of a 2.0M solution.
I C E Substituting
[NH3] x x Kb = x = 1.8x 10-5
[OH-] x x x
[NH4+] x x If x <<< 2.0 it can be from the dropped
denominator
The x2 = (1.8 x10-5)(2.0) = 3.6 x x = 6.0 x 10-3
[OH-] = [NH4+] = 6.0 x [NH3] = = 1.994
pOH = - log [OH-] =-log (6.0 x10-3) = 2.22
pH = 14-pOH = = 11.78

44. Review test yourself on page 338 in the HL IB Chemistry textbook.

45. Citations
International Baccalaureate Organization. Chemistry Guide, First assessment Updated Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.