1. Acids and Bases Chapter 14/15
Brendan Thomas
Justin Simmons
Sam Choi
Alina Wieprecht

2. Properties Acids Bases Sour Taste Conduct electricity
React with many metals to produce H2
Turn litmus paper red
Start with H (HCl, etc)
Lemons, oranges, grapefruits
Bitter taste
Conduct electricity (dissociate in water to create ions)
Do NOT react with metals
Turn litmus paper blue
Often end in OH (hydroxide)
Ivory soap, dish detergent
Sometimes called alkalis
Smooth and slippery

3. Scientists Arrhenius Arrhenius Acid: Arrhenius Base:
A substance that dissociates to produce hydrogen ions (H+) in water
Arrhenius Base:
A substance that dissociates to produce hydroxide ions (OH-) in water
Keyword: Dissociation

4. Scientists (cont.) Bronsted- Lowry Bronsted-Lowry Acid:
This theory assumes that H+ is a proton with no electrons.
Bronsted-Lowry Acid:
Any substance that can donate H+ ions (a proton donor)
Bronsted-Lowry Base:
Any substance that can accept H+ ions (a proton acceptor)
Keyword: Donate/accept H+ ions

5. Scientists (cont.) Lewis Lewis Acid: Lewis Base:
Any substance such as the H+ ion that can accept a pair of nonbonding electrons
(electron pair acceptor)
Lewis Base:
Any substance such as the OH- ion that can donate a pair of nonbonding electron
(electron pair donor)
Keyword: donate/accept electron pairs

6. Naming Binary and Ternary Acids
Binary Acids
Acids that consist of two elements
1) H+
2) nonmetal
Use the stem of the name of the nonmetal
hydro(stem)ic acid
Examples
HI _________________
HBr _________________
Tertiary
Acids that consist of three elements
1) H+
2) O2-
3) nonmetal
If the negative ion ends in “ite”
(stem)ous acid
If the negative ion ends in “ate”
(stem)ic acid
Examples
H2CO3 ________________
H2SO3 ____________________
Hydroiodic acid
Carbonic acid
Hydrobromic acid
Sulfurous acid

7. Writing Acid-Base Reactions in Aqueous Solutions
Acid-base aqueous reactions deal with the dissociation of strong and weak acids and bases. +1 -
HClO H + ClO3
NaOH Na+1 + OH-1
HCl _____________
KOH _____________
H+1 + Cl-1
K+1 + OH-1

8. Writing Neutralization Reactions
The acid and base will neutralize each other and form a product of salt and water.
Na3PO H2O
H3PO NaOH  ________ + ___________
N2CO NaOH
2 NaOH + H2CO3 ________ + ___________

9. Calculating Hydronium and Hydroxide Ion Concentrations
Hydronium [H3O+]
Hydroxide [OH-]
Given pH is 4.35, calculate the hydronium ion concentration.
4.35 = - log[H3O+]
= log[H3O+]
10^(-4.35) = [H3O+]
[H3O+] = 4.47 x 10^(-5) M
Given pOH is 8.7, calculate the hydroxide ion concentration.
8.4 = -log[OH-]
- 8.4 = log[OH-]
10^(-8.4) = [OH-]
[OH-] = 3.9 x 10^(-9) M
Units: M

10. More Hydronium and Hydroxide Ion Calculations
Given pH is 7.3, calculate the hydronium ion concentration.
7.3 = - log[H3O+]
- 7.3 = log[H3O+]
10^(-7.3) = [H3O+]
[H3O+] = 5.01 x 10^(-8) M
Given pOH is 6.7, calculate the hydroxide ion concentration.
6.7 = -log[OH-]
- 6.7 = log[OH-]
10^(-6.7) = [OH-]
[OH-] = 2 x 10^(-7) M
Units: M

11. pH and pOH Calculations
Given [H3O+] is 2.4 x 10^(-3) M, calculate pH.
pH = - log [2.4 x 10^(-3) M]
pH = 2.6
Given [OH-] is 7.8 x 10^(-5) M, calculate pOH.
pOH = - log [7.8 x 10^(-5) M]
pOH = 4.1

12. More pH and pOH Calculations
Given [OH-] = 5.9 x 10^(-6) M, calculate pH.
pOH = -log [5.9 x 10^(-6) M]
pOH = 5.23
pH = 14 – 5.23 = 8.77
Given [H3O+] = 9 x 10^(-2) M, calculate pOH.
pH = -log [9 x 10^(-2)]
pH = 1.05
pOH = 14 – 1.05 = 12.95

13. Titration
Is the controlled addition and measurement of the amount of the solution of known concentration required to react completely with a measured amount of a solution of unknown concentration
So, in English: when you’re using an indicator, the solution will turn a certain color once you’ve reached the equivalence point (where the two liquids are present in chemically equivalent amounts)
*using an indicator only works with strong acids and bases

14. Calculate the Molarity of an unknown acid or base using titration data-2
Need 20 mL of 1.0 x 10 mol NaOH to reach the end point in the titration of 10 mL of HCl
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Calculating Moles of NaOH -2
1.0 x 10 mol NaOH x 1L x 20.0mL NaOH = 2.0 x mol NaOH used
1L NaOH mL
Mole Ratio -4 -4
2.0 x 10 mol NaOH x 1 mol HCl = 2.0 x 10 mol HCl
1 mol NaOH
Calculating Molarity of HCl Solution -4 -2
2.0 x 10 mol HCl x 1000mL = 2.0 x mol HCl = x 10 M HCl -2
10.0mL HCl L L HCl

15. Calculate the Molarity of an unknown acid or base using titration data-3
Need 35.0 mL of 6.0 x 10 mol KOH to reach an end point in titration of mL of HBr
HBr(aq) + KOH(aq) KBr(aq) + H2O(l)
Calculating Moles of KOH -3
6.0 x 10 mol KOH x 1L x 35.0mL = 2.1 x 10 mol KOH used -4
1L KOH mL
Mole Ratio -4 -4
2.1 x 10 mol KOH x 1 mol HCl = 2.1 x 10 mol HBr
1 mol KOH
Calculating Molarity of HBr Solution -4 -2
2.1 x 10 mol HBr x 1000mL = 1.4 x mol HBr = 1.4 x 10 M HBr -2
15.0mL HBr L L HBr

16. Picture Websites https://science7acidbase.wikispaces.com/Litmus+Paper