1. Roselyn Dooley, Tyler Schmidt, Kyle Doubleday and Deondré Robinson
Acids and Bases
Roselyn Dooley, Tyler Schmidt, Kyle Doubleday and Deondré Robinson

2. Properties of Acids -Sour taste -React with active metals
-Turns litmus paper Red
-React with bases to produce salt and water
-Conduct electricity
on pH scale

3. Properties of Bases -Bitter taste -Slippery feel
-Turns litmus paper blue
-Reacts with acids to produce salt and water
-Conduct electricity
on pH scale

4. Binary and Tertiary acids
Binary acid- an acid that contains only two different elements: hydrogen and one of the more electronegative elements.
Tertiary acid- an acid that contains hydrogen oxygen and one more element.

5. Theories of Acids and Bases
-Arrhenius
Acid- A substance that dissociates to produce hydrogen ions in water
Base- A substance that dissociates to produce hydroxide ions in water
-Bronsted-Lowry
Acid- Any substance that can donate H+ ions. (A proton donor)
Base- Any substance that can accept H+ ions. (A proton acceptor)

6. Theories of Acids and Bases Cont.
-Lewis
Acid- Any substance that can accept a pair of nonbonding electrons. (electron pair acceptor)
Base- Any substance that can donate a pair of nonbonding electrons. (electron pair donor)

7. Naming Acids Rule #1 Rule # 2
If the negative ion in the acid ends in "ide" you name the acid "Hyrdo (stem) ic acid
Ex: HCl (Chloride) would yield Hydrochloric acid
Rule # 2
If the negative ion in the acid ends in "ite", you name the acid "(stem)ous acid"
Ex: HNO2 (Nitrite) would yield Nitrous Acid
*Use only if there is an oxygen in the chemical formula

8. Naming Acids Cont. Rule # 3
If the negative ion ends in "ate", you name the acid "(stem)ic" acid.
Ex: HIO4 (Periodate) would yield Periodic acid
Note: The stem of Sulfur is Sulfur
Also the Stem of Phosphor is phosphor

9. Name that Acid!
HSCN
HClO2
HClO3
HBr
H2SO3
H3P

10. Answers Thiocyanic acid Chlorous acid Chloric acid Hydrobromic acid
Sulfurous acid
Hydrophosphoric acis

11. pH Scale

12. Six Strong Acids HCl HBr HI HNO3 H2SO4 HClO4
Everything else is considered a weak acid

13. Writing Acid-base reactions in aqueous solutions
General Formulas
Strong Acids
Weak Acids
HA H+1 + A-1
HCl H+1 + Cl-1
HA +H2O H3O+1 +A-1
HF + H2O H3O+1 + F-1

14. Now you try 1. HBr 2. H2SO4 3. HCN 4. HC2H3O2
Write the acid base reactions in aqueous solutions
1. HBr
2. H2SO4
3. HCN
4. HC2H3O2

15. Answers HBr H+1 + Br-1 H2SO4 H+1 + SO4-2 HCN + H2O H3O+1 + CN-1
HC2H3O2 + H2O H3O+1 + C2H3O2-1

16. Neutralization Reactions between Acids and Bases
Neutralization- The reaction of hydronium ions and hydroxide ions to form water molecules
Example equation-
HCl(aq) + NaOH(aq) NaCl(aq)+ H2O(l)
Note: This is basically just a double displacement reaction

17. Now you try HClO4(aq)+ NaOH(aq) HBr(aq) + Ba(OH)2 (aq)
***You might need your pink sheet

18. Answers HClO4(aq)+ NaOH(aq) NaClO4(aq)+ H2O(l)
2HBr (aq)+ Ba(OH)2 (aq) BaBr2 (aq) + 2H2O(l)

19. Calculate Hydronium and Hydroxide
[H+]=10-pH
ex. [H+]=10-4
[H+]=1 x 10-4 M
pH= 2.2
pH=3.6
pH=8.8
pOH=9
[OH-]=10-pOH
ex. [OH-]=10-7.5
[OH-]=3.16 x 10-8 M
pOH=7.8
pOH=9.3
pOH=5.6
pH=3

20. Hydronium and Hydroxide Answers
[H+]= M
[H+]= 2.51 x 10-4 M
[H+]= 1.58 x 10-9 M
[H+]= 1 x 10-9 M
[OH-]= 1.58 x 10-8 M
[OH-]= 5.01 x M
[OH-]= 2.51 x 10-6 M
[OH-]= .001 M

21. Calculate pH and pOH pH=-log[H+] pOH=-log[OH-]
ex. pH=-log[2.33 x 10-9 M]
pH=8.63
[H+]=7.24 x 10-5 M
[H+]=6.32 x 10-2 M
[OH-]=2.26 x 10-8 M
[H+]=4.54 x 10-3 M
pOH=-log[OH-]
ex. pOH=-log[7.65 x 10-3 M]
pOH=2.12
[OH-]=5.58 x 10-4 M
[OH-]=3.67 x 10-8 M
[OH-]=2.77 x 10-2 M
[H+]=4.49 x 10-7 M

22. pH and pOH Answers pH= 4.14 pH= 1.2 pH= 6.35 pH= 2.34 pOH= 3.25

23. Titration
The controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration.
Once the two solutions are chemically equivalent, the solution has reached the equivalence point.

24. Titration
Essentially, you add an acid of known molarity (concentration) to a base of unknown molarity in measured amounts to find the unknown, or vice versa.
Once at the equivalence point, the unknown concentration can be calculated using known concentration and volumes.

25. Walkthrough Problem 500 mL of .5 M HF titrates with 635 mL NaOH.
1) Balance the equation:
HF + NaOH NaF + H2O
2) Choose method: M1V1 = M2V2 or conversions
3) (.5M) (.500 L) = (x M) (.635 L)
4) .5 M (.500 L)
.635 L
5) .4 M NaOH

26. Titration Calculations
Ex 1) 25 mL of .3M HCl reaches an equivalence point with 75 mL of NaOH. What is the molarity of the NaOH?

27. Titration Calculations
1) Balance the equation:
HCl + NaOH NaCl + H2O
2) Because the mole ratios are equal (1:1), we can use the formula M1V1 = M2V2
3) (.3M) (.025 L) = (x M) (.075 L)
4) .3 M (.025 L)
.075 L
5) .1 M NaOH
= x M

28. Titration Calculations
Ex 2) 550 mL of H2SO4 of unknown concentration reaches an equivalence point with 775 mL of 2.0 M NaOH. What is the concentration of the H2SO4.

29. Titration Calculations
1) Balance the equation:
H2SO4 + 2NaOH Na2SO4 + 2H2O
2) Because the mole ratios are not equal (1:2), we must use conversion factors.
3) 2.0 M NaOH = (x) mols NaOH
.775 L NaOH
4) 1.55 mols NaOH x 1 mol H2SO4 = .755 mols H2SO4
2 mols NaOH
5) .755 mols H2SO4 = 1.4 M H2SO4
.550 L H2SO4

30. Questions for us

31. Sources http://www.elmhurst.edu/~chm/vchembook/184ph.html
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