1. Bases:
Acids:

2. ACIDS HCl(aq) → H+(aq) + Cl-(aq) Properties of Acids:
Cause blue litmus paper to turn pink.
Taste sour.
Form H2(g) in front of active metals.
ACIDS
Produce H+ when dissolve in water.
Form CO2(g) in front of carbonates.
HCl(aq) → H+(aq) + Cl-(aq)
(Arrhenius Model)
(An acid donates H+,protons)

3. Cause phenolphthalein to turn red.
Properties of Bases:
Taste bitter
&
feel slippery.
(Soap)
Cause red litmus paper to turn blue.
Bases
Produce OH- when dissolve in water.
Cause phenolphthalein to turn red.
NaOH(s) → Na+(aq) + OH-(aq)
(Arrhenius Model)
(A base donates OH-)

4. Ion Product Constant for Water (Kw):
H2O(l)
H+(aq) OH-(aq)
neutral
base
acid x
(1.0 x 10-7)
(1.0 x 10-7)
1.0 x 10-14
Kw = [H+] [OH-]
Kw=1.0 x 10-14

5. Example:
If the H+ ion concentration of an aqueous solution is 1 x 10-5M. What is the OH- ion concentration? Is the solution acid, basic, or neutral?
[H+] =1 x 10-5M
Kw = [H+] [OH-]
Kw = 1.0 x 10-14
[OH-] =
[OH-] = ?
[OH-] = 1.0 x ÷ 1 x 10-5
[OH-] = 1.0 x (-5) =
1.0 x
[OH-] = 1 x M
(Makes sense?)
The solution is acidic
(Because [H+] > [OH-] )

6. The solution is neutral.
Kw (10-14) = [H+] [OH-]
Solution
pH:
acidic
10-3M
>
10-11M
basic
10-12M
<
10-2M
acidic
10-4M
>
10-10M
neutral
10-7M =
10-7M
Conclusion:
If [H+] > [OH-]
The solution is acidic.
If [H+] < [OH-]
The solution is basic.
If [H+] = [OH-]
The solution is neutral.

7. The pH Scale:

8. The pH indicates the acidity level of a solution.
pH Scale:
Acidic level
increases
Basic level
increases
Neutral solution

9. pH and pOH: pH: Is the negative logarithm of the [H+].
pH = - log [H+]
(log XP = p log X)
Example:
pH= - log [H+] 1
[H+] =1 x 10-6 M
= - log 10-6
= -(-6) log 10
= - (-6) = 6
pH= 6

10. The solution is neutral. If pH = 7
Kw (10-14) = [H+] [OH-]
Solution
pH:
>
acidic 3
10-3M
10-11M
basic 12
10-12M
<
10-2M
acidic 4
10-4M
>
10-10M
neutral 7 =
10-7M
10-7M
Conclusion:
If pH < 7
The solution is acidic.
The solution is basic.
If pH > 7
The solution is neutral.
If pH = 7

11. pH + pOH = 14 pH and pOH: pOH: Is the negative logarithm of the [OH].
pOH = - log [OH-]
[H+] [OH-] = Kw
10-7 x 10-7 = 10-14
pH = -log 10-7 = -(-7) = 7
pOH = -log 10-7 = -(-7) = 7
= 14
pH + pOH = 14

12. Example: Calculate the pH and pOH of a solution in which
[H+] = 1 x 10-4M.
pH= - log [H+]
[H+] =1 x 10-4 M
= -log 10-4
pH = ?
= - (-4) = 4
pOH = ?
pH= 4
(acidic)
pH + pOH = 14
4 + pOH = 14
pOH =
pOH = 10

13. 10-pH [H+] = If pH= 8 [H+] = 10-4 M pH= 4 [H+] = 10-8 M pH= 8 pOH = 6

14. Complete the following table:
[OH-] pH
pOH
acid
base
neutral
1x10-4 M
1x10-10 M x 4 10
1x10-6 M x
1x10-8 M 8 6
1x10-7 M
1x10-7 M 7 7 x
1x10-13 M
1x10-1 M x 13 1