1. Ka – The Acid Dissociation Constant

2. HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq)
Strong Acids and Bases
Recall: strong acids and bases completely dissociate (break apart into ions) when dissolved in water
Example:
HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq)
is really H+ + Cl- + Na+ + OH-  H+ + OH- + Na+ + Cl-
Thus, it is easy to calculate the pH or pOH of these solutions.

3. Weak Acids and Bases Ka: the acid dissociation constant
a measure of how much an acid breaks apart into its ions
it is similar to the equilibrium constant Keq

4. CH3COOH + H2O  H3O+ + CH3COO-
Example #1
CH3COOH + H2O  H3O+ + CH3COO-
Ka = [H3O+][CH3COO-]
[CH3COOH]
This Ka value now tells us the tendency of CH3COOH to ionize in water.

5. Ka Values Large Ka value = stronger acid (more ionization)
Example: Ka of HF = 6.8 x 10-4
Small Ka value = weaker acid (less ionization)
Example: Ka of HCN = 6.2 x 10-10

6. What do we use Ka values for?
Since weak acids only partially dissociate, the calculation of pH using –log [H+] is not accurate.
Thus, Ka is used to determine the pH of weak acids.
You can also use the pH of weak acids to calculate Ka.

7. Determining Ka – Example #1
Determine the Ka for a 0.10 M solution of HCOOH if [H3O+] = 4.2 x 10-3 M.

8. Determining Ka – Whiteboards
Find the Ka of a 2.00 M HClO2 solution if [H3O+] = 0.14 M.
In a 0.50 M solution of a weak acid HX, the [H3O+] = 8.0 x 10-2 M. Calculate the acid dissociation constant.
Which acid is the stronger acid?
1) 0.011
2) 1.52 x 10-2

9. Calculating Ka from pH – Example #2
A 0.10 M solution of lactic acid (CH3CHOHCOOH) has a pH of Calculate the Ka of lactic acid.

10. Calculating Ka from pH – Whiteboards
A student prepared a M solution of HNO2 and measured the pH to be Calculate the Ka for HNO2.
A M solution of niacin (C5H4NCOOH) has a pH of What is the acid dissociation constant for niacin?
Which acid is the weaker acid? 1.
x 10-5

11. Calculating pH from Ka – Example #3
Determine the pH of a 0.30 M solution of CH3COOH, given Ka = 1.8 x 10-5.

12. Calculating pH from Ka – Example #3
CH3COOH H+
CH3COO-
Initial Molarity
0.30 M
Change in Molarity -x +x
Equilibrium
0.30 – x x
Ka = [H+][CH3COO-] = (x)(x) = 1.8 x 10-5
[CH3COOH] x
The x is so small in comparison to the Molarity of CH3COOH to begin with, we assume Molarity – x is negligible and just drop the x from the formula.
(x)(x) = 1.8 x 10-5
0.30

13. Calculating pH from Ka – Whiteboards
Calculate the pH of a 2.92 M HCN solution if the Ka = 6.2 x
Determine the pH of a M hypochlorous acid solution if the acid dissociation constant is 3.5 x 10-8.
Which acid is the weaker acid?
1) 4.37
2) 4.24
3) #1