1. AP Chem
Today:
Strong vs. Weak Acid Calculations
Acid/Base Equilibrium

2. Finding the pH of a Strong Acid
Strong acids
Ionize completely: HA  H+ + A-
So, the [strong acid] is the same as the [H+]

3. Finding the pH of a Strong Acid
Example – Find the pH of 0.10 M HNO3.
HNO3 is a strong acid, complete dissociates
HNO3  H+ + NO3-
[HNO3] = 0.10 M, therefore [H+] = 0.10 M
pH = - log [H+] = - log [0.10] = 1.00

4. Finding the pH of a Strong Acid
What if you have a very dilute amount of acid in water?
Example – Find the pH of 1.0 x M HBr.
pH = - log [H+] = - log [1.0 x10-10] = 10….
But an acid can’t have a pH greater than 7!

5. Find the pH of 1.0 x M HBr
HBr is an acid, water could be an acid
[HBr] = 1.0 x M, [H+] = 1.0 x M
[H+] in water = 1.0 x 10-7 M
So WATER is actually a better acid
pH = - log [H+] = - log [1.0 x x10-10] =

6. Finding the pH of a Weak Acid
Don’t ionize completely (only partially), so they make an EQUILIBRIUM HA ↔ H+ + A
They’ll have an equilibrium acid dissociation constant value, Ka, associated with them

7. A Couple Things about Ka
The relative strength of acids compared to each other can be found by comparing the Ka values for the acids
Strong acids have a Ka of infinity (very large)
Weak acids have known Ka values (will be given or can calculate)
You can also find the Kb of its conjugate base
Ka x Kb = 1.00 x 10-14
(same as [H+][OH-] = 1.00 x )

8. Finding the pH of a Weak Acid
Find the pH of 1.0 M HF. (Ka = 7.2 x 10-4)
HF (aq) + H2O (l)  F- (aq) + H3O+ (aq) I
C - x x x
E 1.0 – x x x
7.2 x 10-4 = [x][x]
[1.0-x]
x = 2.7 x 10-2 = [H3O+]
pH = - log [0.027] = 1.57

9. A Little Different Weak Acid calculation
Find the molarity of acetic acid if the pH of the solution is 4.15.
HC2H3O2 (aq) + H2O (l)  C2H3O2- (aq) + H3O+ (aq)
I ???
C - x x x
E ??? – x x x
1.8 x 10-5 =[7.1 x 10-5][7.1 x 10-5]
[??? – 7.1 x 10-5]
Solve for ???
=3.49 x 10-4 M
pH = 4.15
So [H3O+] = = 7.1 x 10-5 M = x