1. UNIT IV pH

2. Ionization of Water
Pure distilled water still has a small conductivity. Why? 2H2O(l) + 59kJ H3O+(aq) + OH-(aq)

3. 2H2O(l) + 59kJ H3O+(aq) + OH-(aq)
Ionization of Water
2H2O(l) + 59kJ H3O+(aq) + OH-(aq)
In neutral water… [H3O+] = [OH-]
In acidic solutions… [H3O+] > [OH-]
In basic solutions… [OH-] > [H3O+]
Keq = [H3O+] [OH]
or...
Kw = [H3O+] [OH-]

4. 2H2O(l) + 59kJ H3O+(aq) + OH-(aq)
Ionization of Water
2H2O(l) + 59kJ H3O+(aq) + OH-(aq)
Since reaction is endothermic:  
At higher temperatures ______________are favoured and Kw is _____________er.
 At lower temperatures ______________ are favoured and Kw is _____________er.

5. Ionization of Water Always: [H3O+] [OH-] = Kw
At 250C only: [H3O+] [OH-] = 1.00 x 10-14

6. [H3O+] & [OH-] in Neutral Water
 At 25oC: (NOTE: Assume T = 25oC unless otherwise noted)  
[H3O+] [OH-] = 1.00 x 10-14
and [H3O+] = [OH-] if water is neutral
Substitute [H3O+] for [OH-]:
 [H3O+] [H3O+] = 1.00 x 10-14
 [H3O+]2 = 1.00 x 10-14
 √[H3O+] = √1.00 x = x M
Also [OH-] = [H3O+] = 1.00 x M

7. [H3O+] & [OH-] in Neutral Water
Example Given: Kw at 600C = 9.55 x Calculate [H3O+] & [OH-] at 600C.

8. [H3O+] & [OH-] in Acids and Bases
2H2O(l) H3O+(aq) + OH-(aq)
Add acid, H3O+ increases, so equilibrium shifts LEFT and [OH-] decreases.
Add base, [OH-] increases, so the equilibrium shifts LEFT and [H3O+] decreases.

9. [H3O+] & [OH-] in Acids and Bases
Ex. Find the [OH-] in M HCl. Ex. Find [H3O+] in M NaOH.

10. [H3O+] & [OH-] in Acids and Bases
Ex. Find [H3O+] in M Ba(OH)2 .
Ex. Calculate [OH-] in M HNO3 at 600C.
Kw at 600C = 9.55 x 10-14
Hebden Textbook Page 127 Questions #28-30

11. pH Shorthand method of showing acidity (or basicity/alkalinity)
pH= -log[H3O+]
If [H3O+] = 1.0 x 10-7
pH = -log (1.0 x 10-7 ) = 7

12. Converting [H3O+] to pH Ex. Find the pH of 0.030 M HCl.
**In pH or pOH notation, only the digits after the decimal are significant digits. The digits before the decimal come from the “non-significant” power of 10 in the original [H3O+].

13. Converting [H3O+] to pH
Ex. Find the pH of M NaOH at 250 C. Ex. Find the pH of neutral water at 250 C.

14. pH Scale In neutral water pH = 7.0 In acid solution pH < 7.0
In basic solution pH > 7.0

15. Converting pH to [H3O+]
[H3O+] = antilog (-pH) Ex.) If pH = , find [H3O+]. Ex.) If pH = calculate [H3O+].

16. Logarithmic Nature of pH
A change of 1 pH unit  a factor of 10 in [H3O+] (or acidity).
How many times more acidic is pH 3 than pH 7?
Natural rainwater pH ~ 6
Extremely acidic acid rain pH ~ 3
So, acid rain is 1000 times more acidic than natural rain water!

17. pOH = -log [OH-] [OH-] = antilog (-pOH)

18. pOH Ex. Calculate the pOH of 0.0020M KOH.
Ex. Find the pH of the same solution.
Notice: pH + pOH = 14.00

19. Relation of pH to pOH where pKw = -log Kw (definition of pKw) Since...
[H3O+] [OH-] = Kw
-log[H3O+ ] + -log [OH- ] = -log Kw
pH pOH = pKw
where pKw = -log Kw
(definition of pKw)
TRUE AT ALL TEMPERATURES!

20. Relation of pH to pOH Specifically at 250C: Kw = 1.00 x 10-14
pKw = -log (1.00 x 10-14)
pKw =
pH + pOH =
TRUE AT 25°C!

21. pH and pOH Ex. Find the pH of 5.00 x 10-4 M LiOH (250C).
Ex. Find the pOH of M HBr (250C).
See pOH scale & pH scale on page 140 of Hebden Textbook.

22. pH and pOH
When not at 250C: Ex. At 600C Kw = 9.55 x Find the pH of neutral water at 600C.

23. pH and pOH Is pH always 7.00 in neutral water?_________
At higher temperatures:
2H2O + heat  H3O+ + OH-
[H3O+] > 1.0 x so pH < 7
[OH-] > 1.0 x so pOH < 7

24. Summary of pH and pOH
These are very important? Make sure you study these!
In neutral water pH = pOH at any temperature
pH & pOH = 7.00 at 250C only
At lower temps pH and pOH are > 7
At higher temps pH and pOH are < 7
At any temperature: pH + pOH = pKw
At 250C: pH + pOH =
Hebden Textbook Pages Questions #49-53, 55-57

25. Summary of pH and pOH [H3O+] [OH-] pH pOH pH + pOH = 14
Kw = [H3O+][OH-]
[H3O+] [OH-]
pH =-log[H3O+] pOH = -log[OH-]
[H3O+] = antilog(-pH) [OH-] = antilog(-pH)
pH pOH
pH + pOH = 14