1. Why study chemical reactions in water?
3/4 of the earth surface covered by water
Life, as we know it, evolved in water
Reactions occur quickly in water
Water is relatively cheap (cents per gallon)
Reactions in water can be grouped into categories:
Reactions of acids and bases
Reactions that form precipitates
Reactions involving oxidation and reduction

2. Acids and Bases

3. Properties of Acids Change the color of acid / base indicators
Aqueous solution have sour taste
Change the color of acid / base indicators
Litmus paper for example turns red
React with active metals to release H2 gas
2HCl + Mg  MgCl2 + H2

4. Properties of Acids HCL + NaOH NaCl + H2O
React with bases to produce salts and H2O (neutralization)
HCL + NaOH NaCl + H2O
Conduct electricity - when put in water they form ions, they are considered electrolytes

5. Acid Nomenclature Binary Acid - contain H and 1 other element.
Prefix “hydro”
Root “name of 2nd element”
Suffix “ic”
Ex. Hydrochloric Acid HCl
Hydroiodic Acid - HI

6. Oxyacid
H, O and a 3rd element (mostly non-metal) Root - name the oxy ion, replace suffix
Suffix “ic” for “ate” “ous” for “ite” ions Carbonic Acid H2CO3
Sulfuric Acid H2SO H+ + SO sulfate
Sulfurous Acid H2SO H+ + SO32- sulfite

7. Common household substances that contain acids and bases
Common household substances that contain acids and bases. Vinegar is a dilute solution of acetic acid. Drain cleaners contain strong bases such as sodium hydroxide.

8. Common Industrial Acids
H2 SO4 - sulfuric Acid - most produced chemical in the world. Metallurgy, fertilizer, paper, paint, dyes
HNO3 nitric acid- suffocating odor, stains, protein yellow - explosives, rubber, plastics
H3PO4 - phosphoric acid - fertilizer, flavors beverages, cleaning agent
HCl - hydrochloric acid - found in stomach, cleaning agent, food processing, dilute forms called “muriatic acid” in stores.
CH3COOH - Acetic Acid- foul smelling, vinegar - 4% - 6% acetic acid : plastics, food additives.

9. Bases Aqueous solutions that taste bitter
Change color of acid/ base indicators
Litmus paper turns blue in the presence of a base
Dilute aqueous solutions feel slippery

10. Bases React with acids to produce salts and water (neutralization)
HCL + NaOH NaCl + H2O
Conduct electricity – when put in water they form ions, they are considered electrolytes

11. Strong Acid ionizes completely in water - strong electrolyte.
Remember covalent molecules ionize!
Complete Ionization equation example:
HBr + H2O H3O+ + Br-
Ex. 6 strong acids that will always ionize completely
3 strong oxyacids: H2SO4, HClO4, HNO3 ,
3 strong binary acids: HCl, HBr, HI

12. Weak Acid partially ionizes - weak electrolyte.
Incomplete ionization equation example
HCN + H2O  H3O CN –
*Remember the arrow for weak ionization equations mean that the molecule does not form ions as readily and will stay in the molecular form more often
Ex. H3PO4, HF, CH3COOH, H2CO3, H2S

13. Strong Base strong electrolyte - completely dissociate.
H2O
NaOH  Na+ + OH-
Ex. NaOH , KOH, LiOH
(Strong bases are generally metallic ions in the first and second column on the periodic table that combine with OH- )

14. Weak Base
Weak electrolyte- partially ionizes; Most of it stays in its molecular form
Notice this weak base ionizes!
Ex. NH3 + H2O  NH OH-
The arrow is showing the direction of the equation

15. Acid/Base Strength Strong Acids HNO3 HClO4 H2SO4 HCl HBr HI
Strong Bases
NaOH
KOH
LiOH
Ba(OH)2
Ca(OH)2 (slightly soluble)
Sr(OH)2(slightly soluble)

17. Models of Acids and Bases
Arrhenius Concept: Acids produce H+ ions in solution, bases produce OH ions in solution
HCl  H+ + Cl-
NaOH  Na OH –
Brønsted-Lowry: Acids are H+ donors, bases are proton acceptors.
HCl + H2O  Cl + H3O+
acid base
NH3 + H2O  NH OH –
base acid
Lewis Concept: Acids are electron pair accepters, bases are electron pair donors (we will not need to learn about this concept!)

18. Arrhenius Definition
Based on the idea that aqueous solutions of acids or bases produce ions
Acid – increases the H3O+ or H+ concentration in water
HCl  H+ + Cl-
H2SO4  2H+ + SO42-
Base – increases the OH- concentration in water
NaOH  Na+ + OH-
NH4OH  NH OH-

19. Bronsted - Lowry Acids and Bases
Based on whether a substance is a proton acceptor or donor in non-aqueous solutions.
Bronsted-Lowry is a way to study proton transfer!!
Acid – H+ donor Base – H+ acceptor
HCl + NH3  NH Cl -
HCl donated a proton (H +) to NH3.
Proton donor - acid
The NH3 accepted a proton from HCl
proton acceptor- base

20. Bronsted - Lowry Acids and Bases
Bronsted-Lowry is a way to study proton transfer!!
Acid – H+ donor Base – H+ acceptor
For Example:
HCl + NH3  NH Cl-
H2SO4 + 2H2O  2 H3O+ + SO42-

21. Amphoteric HCl + H2O  H3O + + Cl - proton acceptor (water)
can react as either an acid or base
HCl + H2O  H3O Cl -
proton acceptor (water)
H2O + NH  NH OH -
proton donor (water)

22. Conjugate Acids/Bases
Conjugate Acid – formed when BL base gains a proton
Conjugate Base – formed when BL acid looses a proton

23. Conjugate Acids/Bases
HCl + NH3  NH Cl-
H2SO4 +2H2O  2 H3O+ + SO42-
Acid Base Conj. Acid Conj. Base
Acid Base Conj. Acid Conj. Base

24. Conjugates Strength
The stronger the acid, the weaker its conjugate base; the stronger the base, the weaker its conjugate acid.
Proton transfers favor the production of weaker acids and weaker base. Therefore,
CH3COOH + Water  H3O ++ CH3COO -
(weak acid) (weak base) (stronger acid) (stronger base)
Reactants are favored!!

25. Monoprotic Acids
Ionization – when ions are formed from solute molecules by the action of the solvent.
Monoprotic donates 1 hydrogen
For example:
H2O (l) + HCl (s)  H3O+ (aq) + Cl- (aq)
Hydronium ion = H+

26. Polyprotic Acids/Bases
Some acids have more than one ionizable hydrogen and are called polyprotic: diprotic (2 H+), triprotic (3 H+).
For example:
2 H2O (l) + H2SO4 (s)  H3O+ (aq) + SO42- (aq)
Two moles of hydronium ions

27. Polyprotic Acids Ionization is in several distinct steps:
e.g., H2CO3: carbonic acid
H2CO3 + H2O  H3O+ + HCO3-
HCO3- + H2O  H3O+ + CO32-
Transfer of 2nd (or 3rd) proton are more difficult than 1st.

28. Self-ionization of water
a Very weak electrolyte.
Autoionization of water: 
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
hydronium hydroxide
When protons (H+) are produced in water, they bind to the lone pair e- of water to produce H3O+

29. [H+] is analogous to [H3O+] !
Acid/Base Equilibria
At 25 oC pure water, has a pH = 7 [H3O+] = [OH-] = 1 x 10-7
We use the [ ] to indicate concentration (which is measured in moles/liter or Molarity!)
Kw = ionization constant for water
Kw = 1.0 X @ 25 oC
Note: your book presents the autoionization of water on the reaction:
H2O  H+ + OH-;
[H+] is analogous to [H3O+] !

30. Acids and Bases H2O = HOH = H+ + OH- Acids Bases HCl NaOH HNO3 KOH
HF NH4OH
Remember water can be written like HOH because you are combining a hydrogen ion from the acid and a hydroxide ion from the base to produce water!

31. The pH Scale To Find pH or pOH pH = log[H3O +] pOH = log[OH -]
pX = -log [ion]
X= H+ or OH-
To Find [H3O +] or [OH -]
[H3O +]= 10 -pH
[OH -] = 10 -pOH
[ion] = 10 –pX
X=H+ or OH-
pH in water ranges from 0 to 14.
Kw = 1.00  1014 = [H3O+] [OH]
pH + pOH =
As pH rises, pOH falls (sum = 14.00)

32. The pH Scale pH is measured in exponential values
A pH change from 6 to 5 is 10 times more [H3O+]
A pH change from 6 to 4 is 100 times more [H3O+]
A pH change from 6 to 3 is 1000 times more [H3O+]
Reminder: as exponential values get larger the value gets smaller
Ex:
1 x = 0.001
1 x =

33. pH Scale
pH – stands for power of hydrogen, related to the concentration of H3O + ions in solutions.
The more H3O+ ions, the lower the pH.
pH dictates if something is an acid or a base
Ex. pH 3 = acid pH 9=base

34. The pH scale and pH values of some common substances.

35. pH scale pH 7 = H3O+ ions and OH- ions are equal
pH 0 = many H3O+ ions and few OH- ions
pH 14= many OH- ions and few H3O+ ions
Good
conductor
Non-
conductor
Good
conductor

37. HIn  H+ + In- For phenolphthalein: pH 0 to 8
HIn  H In- For phenolphthalein: pH 0 to 8.2 = colorless; then pink, then pH 10 = red
Add H+ then shift to left
Add OH- then shift to right

38. Phenolphthalein indicator simple diagram
When the H+ leaves the Indicator it changes shape, which causes it to change color b/c it will refract a different wavelength of light
When you add an OH- they attract the H+
OH-  In +
In-
H+ + OH- H+
Pink in the presence of a base H+
H and OH will eventually combine to form water
Clear in the presence of an acid
This is reversible by adding an acid
What makes the slight pink color? Occurs when There are just slightly more OH- ions than H+ ions at the indicators end point so as soon as all of the H+ ions are used up the pink color will show up.

39. pH Calculation Practice
pX = -log [ion]; [ion] = 10-pX 14
1x10-14 13
1x10-12 11
1x10-10 9
1x10-8 7
1x10-7 5
1x10-6 3
1x10-4 A
1x10-1 1 2
1x10-2
AorB
[H3O+] pH

40. pH Practice pX = -log [ion]; [ion] = 10-pX B 1x10-14 14 1x10-13 13 1211 10
1x10-10
1x10-9 9 8
1x10-8 N
1x10-7 7 A
1x10-5 5 6
1x10-6
1x10-3 3 4
1x10-4
1x10-1 1 2
1x10-2
AorB
[H3O+] pH

41. More pH Practice pX = -log [ion]; [ion] = 10-pX bananas 7.8 eggs 2.0
stomach acid
8.5
seawater
milk of mag.
3.1
apples N
4.0x10-8
7.4
blood
AorB
[H3O+] pH
Item
3.2x10-11

42. More pH Practice pH = -log [H3O+]; [H3O+] = 10-pH bananas 7.8 eggs 2.0
stomach acid
8.5
seawater
milk of mag.
3.1
apples N
4.0x10-8
7.4
blood
AorB
[H3O+] pH
Item
7.9x10-4 A
10.5
3.2x10-11 B
3.2x10-9 B
1.0x10-2 A
1.6x10-8 B
4.6 A

43. pX = -log [ion]; [ion] = 10-pX
More pH Practice
pX = -log [ion]; [ion] = 10-pX
AorB
[OH-]
pOH
Item

44. pX = -log [ion]; [ion] = 10-pX
More pH Practice
pX = -log [ion]; [ion] = 10-pX
(for A or B, think pH)
A or B
[OH-]
pOH
Item OJ
10.4
3.98 x 10-11
A (pH 3.6)
B (pH 9.9)
Toothpaste
4.1
7.9 x 10-5
Rainwater
7.5
3.2 x 10-8
A (pH 6.5)
Great Salt Lake
3.9
.00012
B (pH 10.1)

45. Only one number to left of decimal
Kw = [H3O+][OH-] = 1 x 10-14
[H3O+]
1 x 10-4
[OH-] Kw
1 x 10-10
1 x 10-14
1 x 10-9
1 x 10-5
1 x 10-14
10.08 x 10-15
4.8 x 10-4
2.1 x 10-11
round
10 x 10-15
Only one number to left of decimal
1 x 10-14
9.9 x 10-15
6.2 x 10-9
1.6 x 10-6
1 x 10-14

46. Kw and pOH Practice
pX= -log [ion]; pH + pOH = 14; Kw = [H3O+][OH-] = 1 x 10-14
2.5x10-5
4.6
bananas
1.6x10-8
7.8
eggs
1.0x10-2
2.0
stomach acid
3.2x10-9
8.5
seawater
3.2x10-11
10.5
milk of mag.
7.9x10-4
3.1
apples
4.0x10-8
7.4
blood
[OH-]
[H3O+] pH
Item
pOH

47. Kw and pOH Practice
pX= -log [ion]; pH + pOH = 14; Kw = [H3O+][OH-] = 1 x 10-14
4.0x10-10
2.5x10-5
4.6
bananas
6.25x10-7
1.6x10-8
7.8
eggs
1.0x10-12
1.0x10-2
2.0
stomach acid
3.1x10-6
3.2x10-9
8.5
seawater
3.1x10-4
3.2x10-11
10.5
milk of mag.
1.27x10-11
7.9x10-4
3.1
apples
2.5x10-7
4.0x10-8
7.4
blood
[OH-]
[H3O+] pH
Item
pOH

48. pX = -log [ion]; [ion] = 10-pX
pH + pOH = 14
pX = -log [ion]; [ion] = 10-pX
[H3O+]
Concentration M pH
pOH
[OH-]
A/B/Neutral
3.2 x M
2.5 x 10-3 M
4.2 M
2.7
1.0 x 10-8 M

49. pX = -log [ion]; [ion] = 10-pX
pH + pOH = 14
pX = -log [ion]; [ion] = 10-pX
[H3O+]
Concentration M pH
pOH
[OH-]
A/B/Neutral
3.2 x M
9.49
4.51
3.1 x 10-5 M B
4 x M
11.4
2.6
2.5 x 10-3 M
6.3 x 10-5 M
4.2
9.8
1.6 x M A M
2.36
11.63
2.3 x M
5 x M
11.3
2.7
1.9 x 10-4 M
1.0 x 10-8 M 8 6
1.0 x 10-6 M

50. pH word problems If the [H3O+] ion concentration is
3.8 x 10-9 M what is the pOH?
2 step problem
Calculate the pH
pH= - log [3.8 x 10-9 ]
=8.4
2) Subtract the pH from 14 to get the pOH
= 5.6

51. pH word problem
If the pOH is 8.2 what is the [H3O+] ion concentration?
2 Step problem
Find the pH
= 5.8
2. Calculate the ion concentration
Inverse log (or antilog) of the –pH
10^-5.8 = 1.6x10-6

52. pH word problems
If the [H3O+] ion concentration is 4.8x10-8 M what is the [OH-] ion concentration
Find the pH, calculate the pOH then calculate the [OH-] ion concentration
-log (4.8x10-8 ) = pH 7.3
= pOH 6.7
10^ -6.7 = [OH-] 2.0x10-7

53. Molar solutions and Titrations
Molarity (M) = moles solute/liter of solvent
We can use molarity as a conversion unit
Equivalents (eq factor): # of H+ ions or OH- ions donates for an acid or a base
1 mole of HCl has 1 moles of H+
1 mole of H2SO4 has 2 moles of H+
For HCl to have the same # moles of H+ as 1 mole of H2SO4 , you need 2 moles HCl
If you have 10 moles of H2SO4 how many moles of HCl would you need to have the same # of H+ ions? 20
OH- are the same: With 4 moles of NaOH you would need 2 Mg(OH)2 to have the same # of OH-

54. Molar solutions and Titrations
Normality= # equivalents/liter of solution OR Molarity x Equivalents
USE Normality concentration to determine pH or pOH
px = -log [N conc.]

55. Molar solutions and Titrations
To titrate, the H+ ions from the acid will react with the OH- from the base to form water.
This is then a neutral solution.
To reach an equivalence point in titration (neutralization point) between an acid and a base, the number of H+ ions must = the number of OH- ions.

56. Molar solutions and Titrations
Titration – experimental technique that provides a sensitive means of determining the chemically equivalent amounts of acid and base
Titration Equation (algebraic) :
Ma x Eqa x Va = Mb x Eq b x Vb

57. Molar solutions and Titrations
Titration equation using a T-chart:
*Knowing more information about the base:
volume base mole base mole acid
liter base mole base volume acid
= Moles/liter = Molarity of Acid
Base molarity
Mole ratio
*Knowing more information about the acid:
volume acid mole acid mole base
liter acid mole acid volume base
= Moles/liter =Molarity of Base
Mole ratio
Acid molarity

58. Titrations
Equivalence point – point in reaction when equal # moles of acid and base have reacted.
Neutralization reaction equation:
HCl NaOH  NaCl H2O
1 mol
1 mol
1 mol
1 mol
acid
base
salt
water

59. End Point: The point in the reaction where the indicator starts changing color
The end point for Phenolphthalein is around a pH of 8
The end point for methyl orange indicator is around pH 4

60. Refer back to this slide to help you with the Acid Base pre-lab question about what indicator to use for each acid base titration based on the end point and the equivalence point

61. Titrations
We can now look at titrations more quantitatively. Possible combinations we will consider:
strong acid (SA) - strong base (SB)
weak acid (WA) - strong base (SB)
strong acid (SA) - weak base (WB)
Consider each case before/at/after equivalence point (EP).

62.
Notice the end point of the indicators and the equivalence point of the titrations vary based on the combination of acid and base

63. Titrations
A Titration Curve shows pH at various points in a titration experiment(before, at, and after equivalence point (EP)).
Can generate by:
experimentally measuring pH during a titration experiment
calculating pH at various points for an acid-base reaction

64. Titrations
The Titration curve of acids with bases (or visa versa) has four different environments to consider:
Before rxn (before titration begins)
Before the EP (at the very beginning of the titration)
At the EP (moles of acids = moles of base)
After the EP

65. SA/SB Titrations KOH (aq) + HBr (aq)  KBr (aq) + H2O (l)
Net reaction: OH- (aq) + H+ (aq)  H2O
@ Equivalence Point (EP):
What compounds are present at EP?
H2O, K+, and Br-
K+ and Br- are conjugates of strong base and acid; too weak to react with water!
EP, [OH-] = [H3O+]; pH=7

66. SA/SB Titrations The Titration Curve for a SA/SB will look like this.
But: Acid/Base “neutralization” reactions do not always result in “neutral” solutions!

67. SA/SB Titrations Before & After Equivalence Point (EP)
In this case, we need to consider what species are present and at what concentrations. At each stage – we need the [H3O+] or [OH-] concentration!
Example: KOH + HBr
Have 20.0 ml 2.0 M HBr; titrate with 1.0 M KOH

68. EP will occur at 0.04 L or 40 ml KOH
SA/SB Titrations
20.0 ml 2.0 M HBr; titrate 1.0 M KOH
Before rxn:
2.0 M HBr = 2.0 M H3O+ pH = ~ 0
L (2.0 M) = mol HBr present
MAVA = MBVB
(2.0 M)(0.02 L) = (1.0 M)(vol KOH)
EP will occur at 0.04 L or 40 ml KOH

69. SA/SB Titrations 20.0 ml 2.0 M HBr; titrate 1.0 M KOH
Before Equivalence Point (30.0 ml KOH):
30.0 ml KOH = L = mol KOH added
mol HBr have reacted
= mol HBr in 50.0 ml  0.20 M H3O+
pH = +0.70

70. SA/SB Titrations 20.0 ml 2.0 M HBr; titrate 1.0 M KOH
@ Equivalence Point (40.0 ml KOH):
40.0 ml KOH = L = mol KOH added
mol HBr have reacted
= mol HBr in 60.0 ml  [OH-] = [H3O+];
pH = 7

71. SA/SB Titrations 20.0 ml 2.0 M HBr; titrate 1.0 M KOH
After Equivalence Point (50.0 ml KOH):
50.0 ml KOH = L = mol KOH added
all HBr reacted; mol KOH leftover!
0.010 mol KOH in 70.0 ml  [OH-] = M OH-
pOH = 0.84; pH = 13.15

72. WA/SB Titrations F- + H2O  HF + OH-
This equilibrium will make the solution basic.
The EP of a WA/SB reaction is always basic.

73. SA/WB Titrations NH4+ + H2O  H3O+ + NH3
This equilibrium will make the solution acidic.
The EP of a SA/WB reaction is always acidic.

74. Titration Summary Strong Acid/ Strong Base

75. Titration Summary Weak Acid/ Strong Base

76. Titration Summary Strong Acid/ Weak Base

77. Titration Summary

78.
Click on Titration curve
Then, you pick the correct
Curve in the question

79. Polyprotic Acid Titrations
When polyprotic acids are titrated with strong bases, there are multiple equivalence points. The titration curve of a polyprotic acid shows an equivalence point for the each acid H+:

80. Molarity by dilution M1 x V1 = M2 x V2 Mc x Vc = Md x Vd
Acids are usually acquired from chemical supply houses in concentrated liquid form. These acids are diluted to the desired concentration by adding water.
Since moles of acid before dilution = moles of acid after dilution, and moles of acid = M x V, then:
M1 x V1 = M2 x V2
Mc x Vc = Md x Vd
1 = what you have or start with, initial (or c for concentrate)
2 = what you want to make / prepare, final (or d for dilution)

81. Molarity by dilution practice
How much concentrated 18 M sulfuric acid is needed to prepare 250 ml of a 6.0 M solution?
(18M) (V1) = (6M) (250ml)
V1= 83.3ml of 18M Sulfuric acid is needed
How much water do you add to bring your final volume to 250 ml?
= 166.7ml of water
(just bring to volume – you do NOT measure out the water)

82. Molarity by dilution and titration practice
You want to make up 500 ml of a 3 M solution of HCl from concentrated 12 M HCl. How much of the 12 M solution do you need? (Dilution equation)
12M x V1 = 3M x 50ml
V1= 125ml of 12M HCl
Now you take 10 mls of the solution that you made (hopefully 3 M), add indicator and titrate it with a 1 M NaOH solution. You find it takes 28 mls NaOH. What is the LAB Molarity results of the HCl? (Use the Titration equation to calculate the M of the acid – don’t put in 3 M for the acid - that is what you HOPE you will get - your theoretical amount)
Lab molarity of HCl = 2.8M
Finally, calculate your % error. The theoretical amount should be is 3 M HCl
6.7%