1. Acids and Bases

2. Some Properties of Acids
Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”

3. Some Properties of Bases
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes
Feel soapy, slippery
React with acids to form salts and water
pH greater than 7
Turns red litmus paper to blue “Basic Blue”

4. Acid/Base definitions Definition 1: Arrhenius
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
4.3

5. Acid/Base Definitions
Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen atom that has lost it’s electron!

6. + Cl H O acid base conjugate acid conjugate base
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor + Cl H O
acid
base
conjugate acid
conjugate base
conjugate acid-base pairs

7. A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
conjugate acid
conjugate base
base
acid

8. ACID-BASE THEORIES
The Brønsted definition means NH3 is a BASE in water — and water is itself an ACID

9. Conjugate Pairs

10. Learning Check!
Label the acid, base, conjugate acid, and conjugate base in each reaction:
HCl + OH-    Cl- + H2O
H2O + H2SO4    HSO4- + H3O+

11. The ph scale is a way of expressing the strength of acids and bases
The ph scale is a way of expressing the strength of acids and bases. instead of using very small numbers, we just use the negative power of 10 on the molarity of the H+ (or OH-) ion. Under 7 = acid 7 = neutral Over 7 = base

12. (Remember that the [ ] mean Molarity)
Calculating the pH
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5 pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74

13. Try These! pH = - log [H+] pH = - log 0.15 pH = - (- 0.82) pH = 0.82
pH = - log 3 X 10-7
pH = - (- 6.52)
pH = 6.52
Find the pH of these:
A 0.15 M solution of Hydrochloric acid
2) A 3.00 X 10-7 M solution of Nitric acid

14. pH calculations – Solving for H+
If the pH of Coke is 3.12, [H+] = ??? Because pH = - log [H+] then - pH = log [H+] Take antilog (10x) of both sides and get 10-pH = [H+] [H+] = = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button

15. More About Water Equilibrium constant for water = Kw
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION
Equilibrium constant for water = Kw
Kw = [H3O+] [OH-] = x at 25 oC

16. More About Water and so [H3O+] = [OH-] = 1.00 x 10-7 M Autoionization
Kw = [H3O+] [OH-] = x at 25 oC
In a neutral solution [H3O+] = [OH-]
and so [H3O+] = [OH-] = 1.00 x 10-7 M

17. pOH Since acids and bases are opposites, pH and pOH are opposites
pOH does not really exist, but it is useful for changing bases to pH.
pOH looks at the perspective of a base
pOH = - log [OH-]
Since pH and pOH are on opposite ends,
pH + pOH = 14

18. [H3O+], [OH-] and pH
What is the pH of the M NaOH solution? [OH-] = (or 1.0 X 10-3 M) pOH = - log pOH = 3 ; pH + pOH = 14 pH + 3 = 14 pH = 11

19. Strong and Weak Acids/Bases
Generally divide acids and bases into STRONG or WEAK ones.
STRONG ACID: HNO3 (aq) + H2O (l)  H3O+ (aq) NO3- (aq)
HNO3 is about 100% dissociated in water.
Weak acids are much less than 100% ionized in water.
*One of the best known is acetic acid = CH3CO2H

20. Strong and Weak Acids/Bases
Strong Base: 100% dissociated in water.
NaOH (aq)  Na+ (aq) + OH- (aq)
Strong bases are the group I hydroxides
Calcium, strontium, and barium hydroxides are strong, but only soluble in water to 0.01 M
Weak base: less than 100% ionized in water
One of the best known weak bases is ammonia
NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq)

21. Strong Acid
Weak Acid
15.4

22. What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
Start
0.002 M
0.0 M
0.0 M
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
End
0.0 M
0.002 M
0.002 M
pH = -log [H+] = -log(0.002) = 2.7
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Start
0.018 M
0.0 M
0.0 M
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
End
0.0 M
0.018 M
0.036 M
pOH = -log (0.036) = ; pH + pOH = 14
pH = = 12.56
15.4

23. Equilibria Involving Weak Acids and Bases
Consider acetic acid, HC2H3O2 (HOAc)
HC2H3O2 + H2O ↔ H3O+ + C2H3O2 -
Acid Conj. base
(K is designated Ka for ACID)

24. Ionization Constants for Acids/Bases

25. Equilibrium Constants for Weak Acids
Weak acid has Ka < 1

26. Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the equilibrium concentration of HOAc, H3O+, OAc-, and the pH.
Ka= 1.8 x 105
Step 1. Define equilibrium concentration in ICE table.
[HOAc] [H3O+] [OAc-]
initial
change
equilib
-x +x +x
1.00-x x x

27. Equilibria Involving A Weak Acid
Step 2. Write Ka expression
This is a quadratic. Solve using quadratic formula.

28. Equilibria Involving A Weak Acid
Step 3. Solve Ka expression
First assume x is very small because Ka is so small.
Now we can more easily solve this approximate expression.

29. Equilibria Involving A Weak Acid
Step 3. Solve Ka approximate expression
x2 = 1.8 x 10-5
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37

30. Equilibrium Constants for Weak Bases
Weak base has Kb < 1

31. Equilibria Involving A Weak Base
You have M NH3. Calc. the pH.
NH3 + H2O ↔ NH OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3] [NH4+] [OH-]
initial
change
equilib
-x +x +x
x x x

32. Equilibria Involving A Weak Base
Step 2. Solve the equilibrium expression
Assume x is small, so
x2 = 1.8 x 10-5 x = 1.8x10-7
x= 4.2 x 10-4

33. Equilibria Involving A Weak Base
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14
pH = 14 – pOH
pH = 10.63

34. Types of Acid/Base Reactions: Summary

35. 15.4

36. For a monoprotic acid HA
Ionized acid concentration at equilibrium
Initial concentration of acid
x 100%
percent ionization =
For a monoprotic acid HA
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration
15.5

37. Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq) Ka
A- (aq) + H2O (l) OH- (aq) + HA (aq) Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka = Kw Kb
Kb = Kw Ka
15.7

38. Exercise
Calculate the pH of 1 moldm-3 ethanoic acid with Ka is 1.7x10-5
CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)

39. \ Calculate the pH of 1 moldm-3 ethanoic acid with Ka is 1.7x10-5
CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)
Before
Equi 1-x x x
1.7x10-5 = x.x
1-x ~ 1
So, x = 4.12 x 10-3
pH = -log (4.12 x 10-3 ) ; pH = 2.4 \

40. Exercise Calculate the pH of 0.1 mol dm-3 ethanoic acid.
CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)

41. Calculate the pH of 0.1 mol dm-3 ethanoic acid.
Using the same method, we get:
Ka = x.x
0.1-x ~ 0.1
So, 1.7 x 10-5 = x.x
0.1
x= 1.30 x 10-3
pH = -log 1.30x10-3 ; pH = 2.9

42. Exercise
What are the concentrations of H3O+ and OH- ions in 0.05 mol dm-2 HCl?
HCl (aq) + H2O  H3O+ (aq) + Cl - (aq)

43. Example
What are the concentrations of H3O+ and OH- ions in 0.05 mol dm-2 HCl?
HCl (aq) + H2O  H3O+ (aq) + Cl - (aq)
[H3O+] = [HCl]
= 0.05 mol dm-2 (complete dissociation)
Kw = [H3O+] [OH- ]
1.0 x = [0.05 mol dm-2 ] [OH- ]
[OH- ] = 2.0 x 10-13mol dm-3

44. Thank you..