1. Calculating the pH of Acids and Bases
Pg. 127

2. Strong Acids & Bases Dissociate completely in water
Also known as strong electrolytes
Electrolytes conduct electricity in aqueous solutions
The more ions dissociated…the more electricity conducted

3. Strong Acids and Strong Bases
HCl
HNO3
HClO4
H2SO4
All alkali metal hydroxides

4. Weak Acids & Bases Do not completely dissociate in water
The less dissociated they are…the weaker electrolytes they are
Weak bases are often difficult to recognize…look for the presence of –NH2, the amine group

5. HCl(aq)  H+(aq) + Cl-(aq)
pH of Strong Acids
Write the dissociation of HCl
HCl(aq)  H+(aq) + Cl-(aq)
Note the one-way arrow!
Since there is complete dissociation, the concentration of the acid will also be the concentration of the H+ ion.
Just plug into the pH formula (pH = -log[H+])

6. pH of Strong Acids example
If the concentration of HCl is 12 M, what is the pH of the solution?
pH = -log [H+]
pH = -log [12]
pH = -1.08

7. pH of Strong Acids (Multiprotic)
What is the pH of concentrated sulfuric acid, 18M?
This is a diprotic acid…that is, it has two dissociable hydrogen ions
Only one H+ dissociates at a time.
H2SO4(aq)  H+(aq) + HSO41-(aq)
HSO41-(aq)  H+(aq) + SO42-(aq)

8. pH of Strong Acids (multiprotic)
The first dissociation is complete, while the second is not. The second reaches equilibrium
Thus, the first dissociation really determines the pH of a strong, multiprotic acid.
pH = -log [18]
pH =

9. NaOH(aq)  Na+(aq) + OH-(aq)
pH of Strong Bases
Write the dissociation of NaOH
NaOH(aq)  Na+(aq) + OH-(aq)
Note the one-way arrow!
Again, since full dissociation, the concentration of the base is also the concentration of the [OH-] ion

10. pH of Strong Bases
Determine the pH of a 6 M solution of NaOH (remember pH + pOH = 14)
pOH = -log [OH-]
pOH = -log [6]
pOH =
pH + pOH = 14
pH = 14 – (-0.778)
pH = 14.8

11. pH of Weak Acids Note the two-way arrow!
Write the dissociation of the weak acid HC2H3O2
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
Note the two-way arrow!
Problem: Determine the pH of a 17.4M solution of concentrated HC2H3O2.

12. pH of Weak Acids Since it’s weak, it will reach equilibrium.
Since it will reach equilibrium, it has an equilibrium constant.
The equilibrium constant of a weak acid is called a Ka.
Write the Ka expression for the dissociation of acetic acid.

13. pH of Weak Acids Ka = [H+][C2H3O2-] [HC2H3O2]
The Ka value for acetic acid is
1.76 x 10-5 M
Make a chart for the dissociation

14. pH of Weak Acids
[HC2H3O2]
[H+]
[C2H3O2-] i
17.4  eq

15. pH of Weak Acids
[HC2H3O2]
[H+]
[C2H3O2-] i
17.4  -x +x eq

16. pH of Weak Acids
[HC2H3O2]
[H+]
[C2H3O2-] i
17.4  -x +x eq x x

17. pH of Weak Acids Plug into the Ka expression 1.76 x 10-5 = [x][x]
Neglect x
x =

18. pH of Weak Acids
x = = [H+]
pH = -log [0.0175]
pH = 1.76

19. pH of Weak Bases Note the two-way arrow!
Write the dissociation of the weak base NH3
NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)
Note the two-way arrow!
When dissociating a weak base, react it with water to justify the acceptance of the H+

20. pH of Weak Bases Since it’s weak, it will reach equilibrium.
Since it will reach equilibrium, it has an equilibrium constant.
The equilibrium constant of a weak base is called a Kb.
Write the Kb expression for the dissociation of ammonia.

21. pH of Weak Bases Kb = [NH4+][OH-] [NH3]
The Kb value for ammonia is 1.75 x 10-5 M
Make a chart for the dissociation of a 15.3M solution of NH3 in water.

22. pH of Weak Bases
[NH3]
[NH4+]
[OH-] i
15.3  eq

23. pH of Weak Bases
[NH3]
[NH4+]
[OH-] i
15.3  -x +x eq

24. pH of Weak Bases
[NH3]
[NH4+]
[OH-] i
15.3  -x +x eq
15.3 – x x

25. pH of Weak Bases Plug into the Kb expression 1.75 x 10-5 = [x][x]
Neglect x
x =

26. pH of Weak Bases x = 0.0164 = [OH-] pOH = -log [0.0164] pOH = 1.79

27. pH of Multiprotic Weak Acids
Write the dissociations of tartaric acid, H2C4H4O6 (found in cream of tartar)
H2C4H4O6(aq)  H+(aq) + HC4H4O61- (aq)
HC4H4O61-(aq)  H+(aq) + C4H4O62- (aq)

28. pH of Multiprotic Weak Acids
Determine the pH of a 100 mL solution that contains 5.00g of H2C4H4O6. Ka1 is 9.20 x 10-4 and Ka2 is 4.31 x 10-5.
First determine the initial molarity of the tartaric acid.
5g x 1mol x = 0.333M
150.1g L
Make a chart for the 1st dissociation.

29. pH of Multiprotic Weak Acids
[H2C4H4O6]
[H+]
[HC4H4O6-] i
0.333  eq

30. pH of Multiprotic Weak Acids
[H2C4H4O6]
[H+]
[HC4H4O6-] i
0.333  -x +x eq

31. pH of Multiprotic Weak Acids
[H2C4H4O6]
[H+]
[HC4H4O6-] i
0.333  -x +x eq x x

32. pH of Multiprotic Weak Acids
Write the Ka1 expression
Ka1 = [H+][HC4H4O61-]
[H2C4H4O6]
Plug your values into the expression.
9.20 x 10-4 = [x][x]
[ x]

33. pH of Multiprotic Weak Acids
x = = [H+]
More hydrogen ion will dissociate in the next dissociation…this is just the amount from the first dissociation.
Make a chart for the 2nd dissociation.

34. pH of Multiprotic Weak Acids
[HC4H4O61-]
[H+]
[C4H4O62-] i  eq

35. pH of Multiprotic Weak Acids
[HC4H4O61-]
[H+]
[C4H4O62-] i
0.0175  eq

36. pH of Multiprotic Weak Acids
[HC4H4O61-]
[H+]
[C4H4O62-] i
0.0175  -x +x eq

37. pH of Multiprotic Weak Acids
[HC4H4O61-]
[H+]
[C4H4O62-] i
0.0175  -x +x eq
– x x x

38. pH of Multiprotic Weak Acids
Write the Ka2 expression
Ka2 = [H+][C4H4O62-]
[HC4H4O61-]
Plug your values into the expression.
Neglect x
4.31 x 10-5 = [ x][x]
[ x]
Neglect x

39. pH of Multiprotic Weak Acids
x = 4.31 x 10-5 = [C4H4O62-]
So the total amount of hydrogen ion is represented by x 10-5…
[H+] =
pH = 1.76