1. Acid - Base Equilibria Part III: Ka and Kb Calculations
Jespersen Chap. 17 Sec 4 & 5
Dr. C. Yau
Fall 2013

2. Calculating Ka and pKa from pH
Example 17.1 p. 785
Lactic acid (HC3H5O3), which is present in sour milk, also gives sauerkraut its tartness. It is a monoprotic acid.
In a M soln of lactic acid, the pH is 2.44 at 25oC. Calc the Ka and pKa for lactic acid at that temperature.
This calls for an ICE table!

3. Calculating Kb and pKb from % Ionization
Example 17.2 p. 786
Methylamine, CH3NH2, is a weak base and one of several substances that give herring brine its pungent odor.
In M CH3NH2 only 6.4% of the base is ionized. what are Kb and pKb of methylamine?
What does "6.5% ionized" mean?

4. In 0. 100 M CH3NH2 only 6. 4% of the base is ionized
In M CH3NH2 only 6.4% of the base is ionized. what are Kb and pKb of methylamine? CH3NH2 + H2O CH3NH3+ + OH– x +x +x ( ) Now, simply calculate Kb and pKb. I C E

5. Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2
Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is M in nicotinic acid has a pH of 3.39 at 25 °C. What are the acid-ionization constant, Ka, and pKa for this acid at 25 °C? What is the degree of ionization of nicotinic acid in this solution?
Let HNic = nicotinic acid and Nic– = anion.
HNic(aq) + H2O(l) Nic–(aq) + H3O+(aq)
-x x x
(0.012 –x) x x
What does pH = 3.39 tell us?
[H+] = ?
Ka = ?
pKa = ?
Degree of ionization = ?
Do Pract Exer 17, 18, 19 p. 787 I C E
3.4%

6. Calculating Equilibrium Concentrations from Ka (or Kb) and Initial Concentrations
Almost any problem where you are given Ka or Kb falls into one of 3 categories:
Only solute is weak acid
Only solute is weak base
Two solutes, one is weak acid, other is conjugate base (buffer problem-section 17.7)

7. HA + H2O  H3O+ + A-
HA + H2O  H3O+ + A-

8. Determining the pH Of Aqueous Weak Acid Solutions
Previous examples (17.1 & 17.2) were ones where you had to calculate Ka and Kb. Here you are GIVEN Ka and Kb and have to calculate pH.
Dominant equilibrium is Ka reaction
Write the net ionic equation.
Look up the Ka value for the acid.
Set up ICE table.
Solve for x.
Calculate pH from the hydronium concentration at equilibrium.
ChemFAQ: How do I rearrange the expression for Ka in order to solve for the hydrogen ion concentration? 8

9. Calculate the value of [H+] and the pH for this soln.
Example 17.3 p. 790
A student planned an experiment that would use 0.10 M propionic acid, HC3H5O2.
Calculate the value of [H+] and the pH for this soln.
For propionic acid, Ka = 1.34x10-5
Rule is only for Ka and Kb where coefficient of reactant is always one.
Simplification General Rule:
X is negligible if [HA] >>100xKa
For bases If [B] >> 100xKb
Why 100xKa? See explanation on p. 793

10. Example 17.3 p. 790
A student planned an experiment that would use 0.10 M propionic acid, HC3H5O2. Calculate the value of [H+] and the pH for this soln.
For propionic acid, Ka = 1.34x10-5
Do Pract Exer 20, 21 p. 791

11. Now let’s see how this works with a weak base.
Example 17.4 p. 791
A solution of hydrazine, N2H4 has a conc of 0.25 M. what is the pH of the soln, and what is the percentage ionization of the hydrazine? Hydrazine has Kb = 1.3x10-6
Do Pract Exer 22, 23, 24 p. 793
Check: 0.25  100 x1.3x10-6
What does this mean??
To save lecture time, you should finish this yourself before checking your answer.
Ans. 0.23% ionized.

12. Simplifications in Acid-Base Equilibrium Calc’s
Ex.Morphine, C17H19NO3, is administered medically to relieve pain. It is a naturally occurring base, or alkaloid. What is the pH of a M solution of morphine at 25 °C? The base-ionization constant, Kb, is 1.6x10–6 at 25 °C. Let Mor = base morphine and HMor+ = conjugate acid. Mor(aq) + H2O(l)  OH–(aq) + HMor+(aq)

13. (cont’d) I 0.075 ~0 C E – x + x + x 0.075 – x x x
[Mor] (M)
[HMor+] (M)
[OH–] (M) I
0.075 ~0 C E
– x
+ x
+ x
0.075 – x x x
Assume x << 0.075,
then x  0.075
Plugging in for Kb and x gives
[Mor]=0.075>>100xKb

14. (cont’d) = 3.5 x 104 M x  [OH] = 3.5 x 104 M
check assumption: = 0.075
pOH = log (3.5 x 104) = 3.46
pH = – pOH
= – 3.46 = 10.54
Make sure you re-do this problem yourself without looking at the solution.