1. Strong Acid calculations Weak Acid calculations
Acids and Bases
Strong Acid calculations
Weak Acid calculations

2. Finding the pH of a Strong Acid
Strong acids
Ionize completely
HA  H+ + A-
Do it only one at a time
So??
The concentration of the strong acid is EXACTLY the same as the [H+]

3. Finding the pH of a Strong Acid
Example – Find the pH of 1.0 M HCl.
What contributes to H+?
HCl is an acid, water could be an acid
[HCl] = 1.0 M  [H+] = 1.0 M
[H+] in water = 1.0 x 10-7 M
So HCl is the only significant acid
pH = - log [H+] = - log [1.0] = 0.00

4. Finding the pH of a Strong Acid
Example – Find the pH of 0.10 M HNO3.
What contributes to H+?
HNO3 is an acid, water could be an acid
[HNO3] = 0.10 M  [H+] = 0.10 M
[H+] in water = 1.0 x 10-7 M
So HNO3 is the only significant acid
pH = - log [H+] = - log [0.10] = 1.00

5. Finding the pH of a Strong Acid
Example – Find the pH of 1.0 x M HBr.
What contributes to H+?
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 10-7] = 7.00

6. Finding the pH of a Weak Acid
Don’t ionize completely (only partially)
So they make an EQUILIBRIUM
Write the equation with water using the Bronsted-Lowry definition
They’ll have a Ka value 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
Weak acids have known Ka values (look them up)
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 10-14

8. Finding the pH of a Weak Acid
Example – find the pH of 1.0 M HF.
Look up the Ka for a weak acid.
HF (aq) + H2O (l)  F- (aq) + H3O+ (aq) I
C - x x x
E 1.0 – x x x

9. Finding the pH of a Weak Acid
7.4 x 10-4 = [x][x]
[1.0-x] very small
x = 2.7 x  5% rule?
What’s x?
So [H3O+] = 2.7 x 10-2
pH = - log [0.027] = 1.57

10. A Little Different Weak Acid calculation
Find the molarity of acetic acid is 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

11. A Little Different Weak Acid calculation
pH = 4.15
So [H3O+] = = 7.1 x 10-5 M
So x = 7.1 x 10-5 M
1.8 x 10-5 = [7.1 x 10-5] [7.1 x 10-5]
[??? – x]
Solve for ???
2.8 x 10-4 M

12. Last thing they could do
Give you the initial concentration of an unknown acid and its pH, find Ka
M HQ, pH of 5.31

13. Mixtures of Acids
When two or more acids are mixed together, you must calculate them in order from best “strongest” to weakest
Based on Ka

14. Mixtures of Acids If both are strong….
Basically finding the new molarity of H+ and taking the pH
Ex – calculate the pH of a mixture if 50.0 mL of 1.00 M HCl and mL of 3.00 M HNO3 are mixed.

15. Mixtures of Acids
Calculate the pH of a solution that contains 1.00 M HCN (Ka=6.2 x 10-10) and 5.00 M HNO2 (Ka=4.0 x 10-4).
Also calculate the [CN-] in this solution.

16. Mixtures of Acids Start with the better acid
HNO2 (aq) + H2O (l)  NO2- (aq) + H3O+ (aq) I
C x x x
E – x x x
4.0 x 10-4 = [x][x]  x = = [H3O+]
[5.00-x]

17. Mixtures of Acids
Then do the next acid (BUT YOU ALREADY HAVE A CONCENTRATION OF H3O+ FROM THE FIRST ACID)
HCN (aq) + H2O (l)  CN- (aq) + H3O+ (aq) I
C x x x
E – x x x
6.2 x = [ x][x]  x = 1.4 x 10-8 = [CN-] [1.00-x]
[H3O+] =  pH = - log [0.045] = 1.35

18. Polyprotic Acids Acids that furnish more than one H+
Ex: H2SO4, H2CO3, H3PO4
*****It is always one proton at a time****
H2CO3  H+ + HCO Ka1
HCO3-  H CO32- Ka2

19. Polyprotic Acids Typically: Ka1 > Ka2 > Ka3
Usually by at least a factor of 100 x
Why?
The conjugate base of a weak acid is a pretty “good” base – so it would rather act like a base than an acid in most cases

20. Polyprotic Acid Calculation
Find the pH of 5.0 M H3PO4 and the equilibrium [ ] of each conjugate base.
What are we trying to find?
[H2PO4-]
[HPO42-]
[PO43-]
[H3O+]  pH

21. Polyprotic Acid Calculation
Start with the first H+ (Ka1)
H3PO4 (aq) + H2O (l)  H2PO4- (aq) + H3O+ (aq) I
C x x x
E – x x x
7.5 x 10-3 = [x][x]  x = = [H3O+] = [H2PO4-]
[5.0-x]

22. Polyprotic Acid Calculation
Then do the next hydrogen (Ka2)– use the values for the things you just calculated in this one too!!
H2PO4- (aq) + H2O (l)  HPO42- (aq) + H3O+ (aq) I
C x x x
E – x x x
6.2 x 10-8 = [ x][x]  x = 6.2 x 10-8 = [HPO42-] [0.038-x]
[H3O+] now = x 10-8 = 0.038

23. Polyprotic Acid Calculation
Then do the last hydrogen (Ka3) – use the values for the things you just calculated in this one too!!
HPO42- (aq) + H2O (l)  PO43- (aq) + H3O+ (aq)
I x
C x x x
E 6.2 x 10-8 – x x x
4.8 x = [ x][x]  x = 7.8 x = [PO43-] [6.2 x x]
[H3O+] still M  pH = 1.42