3. BASES ACIDS electrolytes  electrolytes sour taste bitter taste
turn litmus red
turn litmus blue
react with metals to form H2 gas
slippery feel
ammonia, lye, antacid, baking soda
vinegar, soda, apples, citrus fruits

4. HCl + H2O  H3O+ + Cl– Arhenius Definition – +
In aqueous solutions Acids form hydronium ions (H3O+)
HCl + H2O  H3O+ + Cl– H Cl O – +
Produce Hydronium
Acid

5. NH3 + H2O  NH4+ + OH- Arhenius Definition – +
In aqueous solutions Bases form hydroxide ions (OH-)
NH3 + H2O  NH4+ + OH- H N O – +
Produce Hydroxide
Base

6. Exceptions?
This did not encompass all of the compounds that we knew were basic and acidic.
Oh what to do?

7. Bronsted-Lowry Definition
Acids
Bases
are proton (H+) donors.
are proton (H+) acceptors.
HCl(aq) + H2O(l)  Cl–(aq) + H3O+(aq)
acid
base
conjugate base
conjugate acid

8. H2O + HNO3  H3O+ + NO3– Example (Acid) Conjugate Base Base Acid
Conjugate Acid 
Hydronium
Conjugate Base

9. NH3 + H2O  NH4+ + OH- Example(Base) B A CA CB Hydroxide
Conjugate Acid

10. BOTH H2O…Acid or Base? So water is a base. So water is an acid?
Amphiprotic:
A chemical species that can act as EITHER, an acid or a base.

11. Practice Activity HF HBr HI H3O+
Give the conjugate base for each of the following:
HF HBr HI H3O+

12. Practice Activity
Partner Up!
Partner 1, write the following bases on the back of a cue card. (one acid per card (3 cards).
Partner 2, write the conjugate acid for the acid on the other side of the card.

13. Br –(aq) HSO4-(aq) CO32-(aq) HBr(aq) H2SO4(aq) HCO3-(aq)
Give the conjugate acid for each of the following:
Br –(aq) HSO4-(aq) CO32-(aq)
HBr(aq) H2SO4(aq) HCO3-(aq)

14. Practice Activity
Partner Up!
Partner 1, write the following acids on the back of a cue card. (one acid per card (3 cards).
Partner 2, write the conjugate base for the acid on the other side of the card.

15. SO42-(aq) H2SO4(aq) HCl(aq) HCO3-(aq) HSO4-(aq) H2SO4(aq) H2CO32-(aq)
Give the conjugate base for each of the following:
H2SO4(aq) HCl(aq) HCO3-(aq)
HSO4-(aq) H2SO4(aq) H2CO32-(aq)
SO42-(aq)

16. Section 6.2
pH and pOH
calculations

17. Auto-ionization of Water
Most water molecules do not ionize. Only 1 in water molecules ionize! The other remain H2O!
Square Brackets indicate concentration
H2O + H2O H3O+ + OH-
Kw = [H3O+][OH-] = 1.0  10-14

18. pouvoir hydrogène (Fr.)
pH Scale 14 7
INCREASING
ACIDITY
NEUTRAL
INCREASING
BASICITY
pH = -log[H3O+]
pouvoir hydrogène (Fr.)
“power of hydrogen”

19. pH of Common Substances
Can go beyond 0 and 14

20. Super Acids/Super Bases
A very concentrated (really hot) strong acid can have a pH below 0! (-0.5, -1)
A very concentrated (really hot) strong base can have a pH above 14! (15, 16)

21. Relationship Between Hydronium Concentration [H3O+(aq)] and pH
[H3O+(aq)] = 1.00 x10-1
What’s the relationship?
pH = 3
[H3O+(aq)] = 1.00 x10-3
pH = 5
[H3O+(aq)] = 1.00 x10-5
pH = 9
[H3O+(aq)] = 1.00 x10-9
pH = 13
[H3O+(aq)] = 1.00 x10-13
Relationship:
[H30+(aq)] is related to pH by powers of 10.

22. pH = -log[H+] pOH = -log[OH-] [H+] = 1x10-pH [OH-] = 1x10-pOH
Formula’s (pH “Box”)
Kw = [H+][OH-]
[H+] [OH-]
pH = -log[H+] pOH = -log[OH-]
[H+] = 1x10-pH [OH-] = 1x10-pOH
pH pOH
pH + pOH = 14

23. Mr. K’s pH/pOH Diamond of Awesome-ness
Whenever you deal with pH and pOH, all you need is!.....
pH/H3O+
pH + pOH = 14
pOH/OH-
Perfect Diamond Shape
-log[OH-(aq)]
pH =
-log[H3O+(aq)]
pOH =
[H3O+(aq)]=
1 x 10-pH
[OH-(aq)]=
1 x 10-pOH
=Kw
[H3O+(aq)] [OH-(aq)]= 1.00 x 10-14

24. -log[H3O+(aq)] pH = -log[4.7 x 10-11] -log( pH = pH = + 4.7 x 10 11 -^ - 11
pH =

25. 1 x 10 ^ - 10.33 [H3O+(aq)]= 1 x 10-pH [H3O+(aq)]= 1 x 10-10.33

26. pH + pOH = 14 [OH-(aq)]= 1 x 10-pOH pOH = 14 - pH pOH = 14 – 5.3
hydroxide
pH + pOH = 14
[OH-(aq)]=
1 x 10-pOH
pOH = 14 - pH
pOH = 14 – 5.3
[OH-(aq)]=
1 x
pOH = 8.7 1 x 10 ^ -
8.7
[OH-(aq)]=

27. Example: What is the pH of 0.050mol/L HNO3? 1 HNO3(aq) + 1 H2O(l) ↔ 1
H3O+(aq) + 1
NO3-(aq)
C = mol/L
pH =
-log[H3O+(aq)]
pH =
-log[0.050]
pH =

28. Example: pH + pOH = 14 pH = 14 - pOH pH = 14 – 9.6 = 4.4 1 HBr(aq) + 1
What is the amount concentration of HBr in a solution that has a pOH of 9.6? 1
HBr(aq) + 1
H2O(l) ↔ 1
H3O+(aq) + 1
Br-(aq)
C = ?
[H3O+(aq)]=
1 x 10-pH
[H3O+(aq)]=
1 x
pH + pOH = 14
[H3O+(aq)]=
pH = 14 - pOH
pH = 14 – 9.6 =
4.4

29. Example: (N/G) (1/1) 1 HBr(aq) + 1 H2O(l) ↔ 1 H3O+(aq) + 1 Br-(aq)
What is the amount concentration of HBr in a solution that has a pOH of 9.6? N G 1
HBr(aq) + 1
H2O(l) ↔ 1
H3O+(aq) + 1
Br-(aq)
C = ?
C =
C =
(N/G)
(1/1)
[HBr(aq)] =

30. Why is the concentration of HBr the same as the hydronium ion concentration?
Look at the dissociation/ionization equation
HBr(aq)  H +(aq) + Br -(aq)
HBr(aq) + H2O(l)  H30+(aq) + Br-(aq)
There is a 1:1 relationship between HBr and the ions

32. Substances that change colour due to the acidity of a solution.
Acid – Base Indicators:

33. They are a weak acid – conjugate base pair that exist in two forms (two different colours) due to presence or lack of a single proton (Hydrogen atom) in the chemical formula.

34. Because of the complex nature of the chemical formula of each indicator.
Abbreviations are usually used to make using indicators less complex.
Ex: HLt – Lt- are the acid and conjugate base of litumus with Hlt being the red form and Lt- being the blue form

35. Example Reactions Placing red litmus paper in a base:
HLt(aq) + NaOH (aq)  H2O (l) + Na+ (aq) + Lt- (aq)
Placing blue litmus in an acid:
HCl (aq) + Lt- (aq)  HLt (aq) + Cl- (aq)

36. Universal Indicators ***most indicators DO NOT DO THIS***
An indicator substance that changes a variety of different colours to indicate a more precise acidity of the solution being tested.
***most indicators DO NOT DO THIS***
**Usually only do two colours**

37. Uses of Indicators pH = 2.8 – 3.2 0-4.8 2.8-8.0 7 14 0-3.2
Mark the end point of a titration (chp. 8)
to estimate the pH of a solution.
***We can use a series of indicators to get a fairly precise pH instead of using the more expensive pH meter.****
pH = 2.8 – 3.2
0-4.8 7 14
0-3.2
Indicator Table (Pg. 10)

38. Example (You Try) Phenolphthalein 0-8.2 7.6-8.2 7.6-14
Bromothymol Blue
7.6-14
pH =
8.0-14
Phenol Red

39. Indicator Practice Activity

41. Defining Acids Alternately, according to Bronsted – Lowry theory
According to the modified Arrhenius theory
Acids are substances that react with water (ionize in water) to produce hydronium ions.
Alternately, according to Bronsted – Lowry theory
Acids are proton donors that become basic (conjugate bases) once they donate their proton.

42. Example CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
This reaction can be explained using either definition and requires the acid react with water.
H2O(l)
CH3COOH(aq)
H3O+(aq)
CH3COO-(aq)

43. Example Bronsted-Lowry Arhenius NaHSO4(aq) Na+(aq) HSO4-(aq) SO42-(aq)
How can we explain using either the modified arrhenius theory or the Bronsted-Lowry theory that a solution of NaHSO4 will turn blue litmus paper red?
Acidic
Arhenius
Bronsted-Lowry
NaHSO4(aq)
Na+(aq)
HSO4-(aq)
SO42-(aq)
HSO4-(aq) +
H2O(l) →
H3O+(aq) +

44. Example Bronsted-Lowry Na+(aq) HSO4-(aq) SO42-(aq) HSO4-(aq) + H2O(l)
How can we explain using either the modified arrhenius theory or the Bronsted-Lowry theory that a solution of NaHSO4 will turn blue litmus paper red?
Acidic
Bronsted-Lowry
Na+(aq)
HSO4-(aq)
SO42-(aq)
HSO4-(aq) +
H2O(l) →
H3O+(aq) +
Acid
Conjugate Base

45. Bases
So far bases have been metal hydroxides that can be explained by simple dissociation to produce hydroxide ions according to the Arrhenius theory.
Ex: Ca(OH)2(aq)  Ca2+(aq) + 2OH-(aq)

46. Modified Arhenius Theory
The original Arrhenius theory doesn’t account for the basic nature of ammonia or baking soda.
The modified Arrhenius theory, that bases ionize in water (react with water) to produce hydroxide ions. Helps to explain why such substances are in fact basic.

47. Example Na2CO3(s)  2Na+(aq) + CO32-(aq) Then,
CO32-(aq) + H2O(l) OH-(aq) + HCO3-(aq)
The Bronsted – Lowry definition of a base as a proton acceptor can also help explain this reaction.

48. How did we know that the carbonate ion was going to produce a basic solution and that the sodium ion was a spectator?
Bases are proton acceptors and usually have a negative charge (water and ammonia are exceptions) and acids are proton donors. Therefore, the sodium ion cannot act as an acid or a base.
Na does not have H to give away and is + so it can accept an H.

49. A Special Case
Nonmetal oxides in water will form acidic solutions. There is a two step process to explain how this occurs.

50. Ex:
CO2(g) + H2O(l)  H2CO3(aq)
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)

51. The overall reaction could be combined into one equation:
CO2(g) + 2H2O(l)  H3O+(aq) + HCO3-(aq)

52. Neutralization Reactions
An acid – base neutralization is a double replacement reaction that produces water (HOH) and a salt (an ionic compound)
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

53. According to the modified Arrhenius definition acids produce hydronium and bases produce hydroxide in solution. Therefore, the reaction can be written as:
H3O+(aq) + OH-(aq)  2 H2O(l)
Neutralization can be defined as the reaction of hydronium and hydroxide to produce water.

54. How to get to that equation?
The first equation is called the molecular equation. It shows everything present in a typical double replacement which you should be familiar with.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

55. Next we need to write out the total ionic equation
Next we need to write out the total ionic equation. For this equation separate all soluble compounds into ions and strong acids into hydronium and the conjugate base (anion). Insoluble compounds and weak acids will remain the same.
H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) 
2 H2O(l) + Cl-(aq) + Na+(aq)

56. Now we cancel out spectator ions, ions that don’t change or react in the equation. This leaves us with the net ionic equation.
H3O+(aq) + OH-(aq)  2 H2O(l)

59. Strong Acids: Weak Acids:
React completely (more than 99%) with water to form hydronium ions. The more hydronium ions the greater the acidic properties such as conductivity and low pH.
Weak Acids:
React incompletely (less than 50%) with water to form hydronium ions. Lower concentration of hydronium ions leads to less acidic properties. They have a higher pH and are poor conductors of electricity.

60. Strong Bases:
Soluble ionic hydroxides that dissociate 100% in water to produce hydroxide ions.
Weak Bases:
Reacts partially with water (less than 50%) to produce fewer hydroxide ions.

61. Examples:
Explain the weak base properties of baking soda (sodium bicarbonate/sodium hydrogen carbonate).
Solid sodium acetate is dissolved in water. The final solution is tested and found to have a pH of about 8. Explain this evidence by writing balanced chemical equations.

62. Polyprotic Substances:
Polyprotic Acids:
Weak acids with multiple protons to donate and whose percent reaction with water decreases after each step.
Polyprotic Bases:
Weak bases that can accept multiple protons and whose percent reaction with water decreases after each step.
BicarbonateCaCO3

63. <50% <1% <0.00% H3PO4(aq) H+(aq) H2PO4-(aq) H+(aq) HPO42-(aq)

64. Polyprotic Base
YOU DRAW THIS ONE

66. Indicator Table (Pg. 10)
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67. Indicator Table (Pg. 10)
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