1. People forget how fast you did a job but they remember how well you did it.

3. Acids and Bases
Lesson 4

4. Anhydrides

5. Problem 1 Find the OH- if the [H+] is 2.3 x 10 -3 M. Remember
Kw= [H+] x [OH-]
1x M2 = 2.3 x M x [OH-]
Answer
4.35 x M

6. Problem 2 Find the [H+] if the [OH-] is 7.31 x 10 -3M. Remember
Kw= [H+] x [OH-]
1x M2 = 7.31 x M x [H+]
Answer
1.37 x M

7. Problem 3 Find the pH if the [H+] is 8.91 x 10 -5 M. Remember
pH = - log of [H+]
(-)Log of 8.91 x M
Answer
4.05

8. Arrhenius Came up with theory of ionization.
Stated that acids are compounds containing hydrogen that ionize to yield H+ in aqueous solution and bases are compounds that ionize yield OH- in aqueous solution.

9. Ionization
When ions form from a covalent compound.

10. Lewis acid
Substance that can accept a pair of electrons to form a covalent bond.
Acids are electron - pair acceptors.

11. Lewis base
Substance that can donate a pair of electrons to form covalent bond.
Bases are electron - pair donors.

12. Naming acids: Rule 1 If anion ends with ide.
The acid is named with the prefix hydro, the root name of the anion, and the suffix ic.
ide  Hydro (base word) ic
Ex: HCl - hydrochloric acid.

13. Naming acids: Rule 2 If the anion ends with ite.
The acid named is formed from the root name of the anion with a suffix of ous or rous, which replace ite
Example: H2SO3 - sulfurous acid . Anion was sulfite SO3-2.
Ite  (base word) ous

14. Naming acids: Rule 3 If the anion ends with ate.
The acid named is formed from the root name of the anion with a suffix of ic or ric, which replace ate.
Example: H2SO4 Sulfuric acid. Anion was sulfate SO4-2
ate (base word) ic

15. Strong acid plus a weak base
Will yield an aqueous salt and water.
The ions in the salt are dissociated.

16. Problem 4 Find the pH if the [OH-] is 3.71 x 10 -8 M. Remember
pH = - log of [H+] , but you have the [OH-] .
Find the [H+] first.

17. Work 1 x 10 -14 M2 = 3.71 x 10 -8 M x [H+] [H+] = 2.70 x 10-7 M
Take the –log of 2.70 x 10-7
Answer =
6.57

18. Problem 5 Find the pH if the [OH-] is 9.18 x 10 -5 M. Remember
pH = - log of [H+] , but you have the [OH-] .
Find the [H+] first.

19. Work 1 x 10 -14 M2 = 9.18 x 10 -5 M x [H+] [H+] = 1.09 x 10-10 M
Take the –log of 1.09 x 10-10
Answer =
9.96

20. Problem 6
Find the molarity of the sodium hydroxide solution if you have 57 g of sodium hydroxide dissolved in 800 ml of water.

21. Work
57 g NaOH 1 mol NaOH =
.8 L g NaOH
Answer
1.78 M