Properties
Acid-Base Theories
Acid-Base Reactions
Most assignments are from the Ch 13 handout.
1

Both conduct electricity (electrolytes) because they break apart to some degree in water.
Acids produce H + (proton) in water.
Bases produce OH (hydroxide) in water.
Samples:
Acids: vinegar(acetic acid), lactic acid in sour milk, citric acid,
Bases: ammonia, lye (NaOH), Milk of Magnesia
Mg(OH)
2
.
2

1. Sour taste. NEVER taste acids in lab situations.
2. Change color of indicators.
3. Some acids react with metals & release H
2 gas.
4. Acids react with bases to produce salt & water. When neutralization occurs, #1.-#3 disappear.
5. Conduct electric current.
3

• Binary Acids contain Hydrogen and another element:
• Hydro + root of 2 nd element + ic
– HF hydrofluoric acid
– HCl hydrochloric acid
– HBr hdrobromic acid
– HI hydroiodic …
– H
2
S hydrosulfuric …
4

• Contain H, O, and a 3 rd element. More are listed in your book.
5

• Sulfuric
• Nitric
• Phosphoric
• Hydrochloric
• Acetic
6

• Bitter taste (NEVER taste bases in labs).
• Change the color of indicators.
• Slippery feel (dilute bases, don’t touch concentrated bases)
• React with acids to produce salt & water
• Conduct electric current.
7

Acid + Base --> Salt + Water
HCl + NaOH --> NaCl + H
2
0
H
2
SO
4
+ Ca(OH)
2
--> CaSO
4
+ 2H
2
0
Assignment from Ch 13 A-B handout:
117/1-3 AND 123/1 (naming/formulas)
8

• Arrhenius Acid is a chemical compound that increases the concentration of hydrogen ions, H+, in aqueous solutions.
• Arrhenius Base is a chemical compound that increases the concentration of hydroxide ions,
OH-, in aqueous solutions.
• When put with water, these compounds dissociate (break apart) forming ions
9

• HNO
3
(l) + H
2
0 (l) --> NO
3
(aq) + H
3
0 + (aq)
• When put in water, HNO
3
, ionizes and the charged particles formed can conduct electricity.
• The amount of H
3
0 + (hydronium) produced is an indication of the acid’s strength.
10

Strong Acids ionize completely in water.
Strong Acids:
• HI
• HClO
4
• HBr
• HCl
• H
2
SO
4
• HClO
3
Weak Acide release few hydrogen ions in water.
Weak Acids:
• HSO
4
-
• H
3
PO
4
• HF
• CH
3
COOH
• H
2
CO
3
• H
2
S
• HCN
• HCO
3
-
11

12

Strong Bases ionize completely.
Strong Bases
• Ca(OH)
2
--> Ca 2+ + 2OH -
• Sr(OH)
2
• Ba(OH)
2
• NaOH
• KOH
• RbOH
• CsOH
Weal Bases ionize slightly.
Weak Bases
• NH
3
+ H
2
O NH
4
+
• C
6
H
5
NH
2
+ OH -
“ “ means the reaction is reversible
13

• Book: 437/33,34 (strengths & ionization)
Answer these on page 118(bottom) of handout.
• H/out: 117/5,6 (reactions) and 122/5,6ac
(neutralization reactions, balancing and mole ratio)
14

Bronsted-Lowry Acids donate protons (H + )
Molecules or ions can donate protons.
HCl + NH
3
 NH
4
+ + Cl -
Na+
-
Na+
15

The HCl is a Bronsted-Lowry Acid. It donates a proton to water
Water can act as a Bronsted-Lowry Acid also as in the following reaction:
H
2
O (l) + NH
3
OH + NH 4+
16

accept protons . In the equation below, ammonia is the base, because it accepts the proton to become an ammonium ion.
acid base
3

4
+
-
17

• Monoprotic acids can only donate one proton per molecule. Ex.: HCl, HNO
3
• Polyprotic acids can donate 2 or more protons per molecule. Ex.: H
2
SO
4
, H
3
PO
4
• For polyprotic acids the donations occur in stages, losing one H + at a time.
18

• Arrhenius and Bronsted-Lowery definitions have some limitations. Lewis classification is based on bonding and structure including substances without hydrogen. The Lewis classification is more complete than the other
2 methods.
19

is an atom, ion or molecule that accepts an electron pair to form a covalent bond.
Dot notation
Structural formula
– a bar represents what?
A pair of shared electrons.
20

is an atom, ion, or molecule that donates an electron pair to form a covalent.
21

• is the formation of one or more covalent bonds between an electron-pair donor and an electron-pair acceptor.
Pair of donated electrons
22

• 119/1-3 (ionization in stages)
• 120/4,5ab (BL a&b, L’LP)
23

Sample: Dilute HCl(aq) and KOH(aq) are mixed in chemically equivalent quantities.
a) Write the formula equation for the reaction.
HCl(aq) + KOH(aq) --> KCl(aq) + H
2
O(l) b) Write the overall ionic equation.
H
3
O + (aq) + Cl (aq) + K + (aq) + OH (aq) -->
K + (aq) + Cl (aq) + 2H
2
0(l) c) Write the net ionic equation.
H
3
O + (aq) + OH (aq) --> 2H
2
0(l)
24

Sample: Write the formula equation and net ionic equation for this reaction.
Formula equation for: Zn(s) + HCl(aq) -->
Zn(s) + 2HCl(aq) --> ZnCl
2
(aq) + H
2
(g)
Overall ionic equation:
Zn(s) + 2H
3
O + (aq) + 2Cl (aq) -->
Zn 2+ (aq) + 2Cl (aq) + H
2
(g) + 2H
2
0(l)
Net ionic equation:
Zn(s) + 2H
3
0 + (aq) --> Zn 2+ (aq) + H
2
(g) + 2H
2
0(l)
25

• Now we are going to use Bronsted-Lowry description to explore acid-base reactions.
• What was the Bronsted-Lowery theory?
• B-L acid donates protons
• B-L base accepts protons
• Proton: H + (a hydrogen nucleus)
26

• A conjugate base is the species that remains after a Bronsted-Lowery acid has given up a proton.
• A conjugate acid is the species that forms when a Bronsted-Lowery base gains a proton.
27

28

Using Bronsted-Lowry definitions to study Acid-
Base reactions, continued:
The species that remains after a Bronsted-Lowry acid has given up a proton is the conjugate base of that acid.
HF + H
2
O F + H
3
0 +
Acid conjugate base
29

The species that is formed when a Bronsted-
Lowry base gains a proton is the conjugate acid of that base.
HF(aq) + H
2
O(l) F (aq) + H
3
0 + (aq)
Base conjugate acid
30

HF(aq) + H
2
O(l) F (aq) + H
3
0 + (aq)
Acid Base conjugate conjugate base base
1 acid
1 base
Conjugate pairs:
2
(1) HF and F -
(2) H
2
0 and H
3
0 + acid acid
2
31

• 121/1,2 (conjugate A&B)
• 123/2,3 (same as above)
32

• On Page 1 of your handout for this chapter, you have a table which lists and compares the strengths of various acids and their conjugate bases. Get your Ch. 14 handout out now.
33

The stronger an acid is, the weaker its conjugate base will be.
The stronger a base is, the weaker its conjugate acid will be.
From these concepts, we can predict the outcome of a reaction.
34

35

36

37

38

)
Sample problem on next page:
39

Sample: Identify the proton donor or acid and the proton acceptor or base. Label each acidbase conjugate pair.
CH
3
COOH + H
2
0 H
3
0 + + CH
3
COO acid base conjugate conjugate acid base
40

Write the formula equation, the overall ionic equation, and the net ionic equation for a neutralization reaction that would form
RbClO
4
.
Formula equation:
RbOH(aq) + HClO
4
(aq) --> RbClO
4
(aq) + H
2
0(l)
41

Overall Ionic equation:
Rb + (aq) + OH (aq) + H
3
0 + (aq) + ClO
4
(aq) -->
Rb + (aq) + ClO
4
(aq) + 2H
2
0(l)
Net ionic equation:
H
3
0 + (aq) + OH (aq) --> 2H
2
0(l)
42

These can act as either an acid or a base.
Water acts as a base in this reaction:
H
2
SO
4
(aq) + H
2
0(l) --> H
3
0 + (aq) + HSO
4
(aq) acid
1 base
2 acid
2
But, water acts as an acid here: base
1
NH
3
(g) + H
2
0(l) NH
4
+ (aq) + OH (aq)
Base
1 acid
2 acid
1 base
2
43

• 123/5 (amphoteric/stages)
• 124/6ac (conduction, strength)
44

Acids Bases
Arrhenius concentration of: [H + ] [OH ]
Bronsted-Lowry H + donor H + acceptor
Lewis, e- pair: acceptor donor
45

46

pH is an indication of the hydronium ion concentration present in a solution.
[H
3
0 + ] is the symbol for concentration of hydronium ion in moles per liter, mol/L or M pOH is an indication of the hydroxide ion concentration present in a solution.
[OH-] is the symbol for concentration of hydroxide ion in mol/L or M
47

H
2
0(l) + H
2
0(l) H
3
0 + (aq) + OH (aq)
In the above reaction, two water molecules produce a hydronium ion and a hydroxide ion by transfer of a proton. Water is self Ionizing.
At 25 o C, the concentrations of H
3
0 + and OH are each only 1.0x10
-7 mol/L of water.
48

is a constant, k w
K w
= [H
3
, the ionization constant of water .
0 + ] [OH ] = 1.0x10
-7 (1.0x10
-7 ) =1.0x10
-14
This occurs at 25 o C. If the temperature changes, the ion product, K changes.
When both [H
3 w
0 + ] and[OH ] are 1.0x10
-7 , the solution is neutral.
If [H
3
0 + ] is greater than 1.0x10
-7 , the solution is
Acidic. (10 -6 or 10 -4 would be greater)
If [OH ] is greater than 1.0x10
-7 , the solution is
Basic.
49

K w
= [H
3
0 + ] [OH ] = 1.0x10
-7 (1.0x10
-7 ) =1.0x10
-14
Let’s say that the [H
3
0 + ] is 1.0x 10 -6 and you are asked to find the [OH ].
K w
= [H
3
0 + ] [OH ] --> [OH ] = k w =
1.0x10
-14
[H
3
0 + ] 1.0x10
-6
-14 – (-6) = -14 + 6 = -8 so: [OH ] = 10 -8 mol/Liter
More practice: 10 -14 /10 -2 = 10 -12 and 10 -14 /10 -9 = 10 -5
50

51

3
+
-
Your own scientific calculator is a MUST here!!!
Find these keys: 2 nd , either EE or EXP, and change sign (-) or (+/-) on your calculator.
Let’s practice putting in numbers in sci. not.
1x10 -7 : Press keys in this sequence:
1 2 nd EE (-) 7 on your display you see something similar to this: 1E -7
To get 2 x10-4, press: 2 2 nd EE (-) 4 display: 2 E -4
52

The [H
3
0 + ] is 2.34 x 10 -5 M in a solution. Calculate the [OH ] of the solution.
[OH ] = K w =
1.0x10
-14
[H
3
0 + ] 2.34 x 10 -5
Key sequence:
1 2 nd EE (-) 14
:
2.34 2 nd EE (-) 5 enter
Display: 4.27 E -10 which means: 4.27 x 10 -10 M
53

Calculate hydronium and hydroxide ion concentrations in a solution that is 1x10 -4 M HCl.
HCl is a strong acid that ionizes completely. So the concentration of H
3
0 + is 1x10 -4 M.
Find [OH-]: [OH ] = K w =
1.0x10
-14
[H
3
0 + ] 1x10 -4
Answer: [OH ] = 10 -10 M
54

pH of a solution is the negative of the common logarithm of the hydronium ion concentration.
pH = - log [H
3
0 + ]
A common logarithm of a number is “the power to which
10 must be raised to equal the number.”
55

The logarithm of 1.0x10
-7 is - 7.0
The pH = - log [H
3
0 + ] = - log (1.0x10
-7 ) = 7.0
pOH is the negative log of [OH-].
pOH = -log [OH-]. In a neutral solution where
[OH-] is 1.0x10
-7 , the pOH = -log [OH-] = -log 1.0x10
-7 = 7.0
56

K w
= [H
3
0 + ] [OH ] = 1.0x10
-7 (1.0x10
-7 ) =1.0x10
-14
From above: pH + pOH = 14.0
pOH can also be found by: pOH = 14.0 – pH = 14.0 – 7.0 = 7.0
57

a. 1x10 -3 M HCl. HCl is a strong acid that ionizes completely. So the concentration of H
3
0 + is 1x10 -3 M. pH = - log [H
3
0 + ] = - log (1.0x10
-3 ) = 3.0
b. 1x10 -4 M NaOH. NaOH is a strong base that ionizes completely. The concentration of OH is 1x10 -4 M
[H
3
0 + ] = K w
= 10 -14
[OH ] 10 -4
= 10 -10 pH = -log(10 -10 ) = 10.0
58

Find the pH of a solution where [H
3 pH = -log [H
3
0 + ] = -log 2.8 x 10 -5
0 + ] is 2.8 x 10 -5 M?
= 4.55 or 4.6
Key sequence: (-) log 2.8 2 nd EE (-) 5 =
Find the pH of a 4.7 x 10 -2 M NaOH solution.
Find the concentration of [H
3
[H
3
0 + ] = K w
= 10 -14
[OH ] 4.7x10
-2
0 + ] first:
[H
3
0 + ] = 2.1x10
-13 pH = -log [H
3
0 + ] = - log 2.1x10
-13 pH = 12.7
59

• Asgn: H/O: 125/4a-d and 129/1-3. Hint for
3a: H
2
SO
4 will produce double [H
3
0 + ]
60

3
+
-
pH = -log [H
3
0 + ] log [H
3
0 + ] = -pH
[H
3
0 + ] = antilog (-pH)
[H
3
0 + ] = 10 -pH
If pH = 7, [H
3
0 + ] = 1x10 -7
If pH = 2, [H
3
0 + ] = 1x10 -2
61

3
+
pH = -log [H
3
0 + ] log [H
3
0 + ] = -pH
[H
3
0 + ] =antilog(-pH) = antilog (-7.52)
= 1x10 -7.52 = 3.0 x 10 -8 M H
3
0 +
Raise 10 to the -7.52 power using this
Flow chart:
[2 nd ] [log] [+/- ] [7.52] = 3.10x10
-8 M
62

2. If the pH = 12.0 then [H
3
0 + ] = 1x10 -12 M
3. The pH of an aqueous solution is measured as
1.50. Calculate the [H
3
0 + ] and [OH ] .
[H
3
0 + ] = [2 nd ] [Log] [+/-] [1.5]
[H
3
0 + ] = 1x10 -1.5 = 3.16x10
-2 M
[OH ] = ___Kw___ = 1x10 -14 = 3.16 x 10 -13 M
3.16x10
-2 3.16x10
-2

The pH of an aqueous solution is 3.67.
Determine [H
3
0 + ] .
[H
3
0 + ] = antilog (-pH) = antilog (-3.67) …
OR
[2 nd ] [log] [+/-] [3.67] = 2.14 x10 -4 M
Asgn from book: 437/30-32
(labeling A+B),
42,61-63
(calc pH) and 439/64-68 a-c only
(calc pH + pOH)

Molarity – one mole of solute dissolved in enough solvent (water) to make exactly one liter of solution.
Molality – one mole of solute dissolved in exactly
1,000 grams of solvent.
Normality – one gram equivalent weight (gew) of solute dissolved in enough solvent to make exactly one liter of solution.
These are on your handout titled: “ph/Acid/Base
Equations”
65

M = grams of solute given
GFW of solute (L of solvent)
Grams of solute needed= M(GFW of solute)(Liters Solvent)
L of solvent needed = g solute/GFW solute (M)
66

You have 3.50L of solution that contains 90.0g of
NaCl. What is the molarity of the solution?
g NaCl x 1mol NaCl = mol NaCl, the solute g NaCl mol of Solute = molarity of solution
L of solution
90.0g NaCl x 1 mol NaCl = 1.54 mol NaCl
58.44 g NaCl
1.54 mol NaCl = 0.440 M NaCl
3.50 L
67

1. What is the molarity of a solution composed of 5.85g KI, dissolved in enough water to make 0.125 L of solution?
5.85g KI x 1 mol KI = 0.0352 mol KI
166g KI
0.0352 mol KI = 0.282 M KI
0.125 L
68

m = g of solute given X 1000
GFW of solute X g of solvent g of solute needed = m(GFW)(g solvent)
1000 g solvent needed = g solute(1000)
GFW solute x m
Molality = molarity if water is the solvent
(aqueous solutions)
69

Normality, N – one gram equivalent weight of solute dissolved in enough solvent to make exactly one liter of solution.
N = _________g solute_____
GEW solute x L solvent g solute = N X GEW solute X L of solvent
L solvent = ____g of solute___
GEW of solute X N
Normality to Molarity
M = N(valence of cation)(subscript of cation)
GEW = _________GFW of solute_______ charge X subscript of solute cation
GEW – gram equivalent weight
70

One equivalent of an acid is the quantity, in grams, which donates one mole of protons.
1 equiv HCl = 1molHCl x 36.5g = 36.5g HCL
1mol H
3
0 + mol mol H
3
0 +
1equiv H
2
SO
4
= 1mol H
2
SO
4 x 98g = 49 g H
2
SO
4
2mol H
3
0 + mol mol H
3
0 +
A diprotic acid (H
2
SO
4
) has 2 atoms of ionizable hydrogen per molecule.
71

One equivalent of a base is the quantity, in grams, which accepts one mole of protons, or supplies one mole of
OH- ions.
1equiv KOH = 1 mol KOH x 56g = 56g KOH
1 mol OH mol mol OH -
1equiv Ca(OH)
2
=1 mol Ca(OH)
2
2 mol OH x 74g = 37 g Ca(OH)
2 mol mol OH -
1 Mole of Ca(OH)
2 supplies 2 equiv of OH therefore ½ mole of Ca(OH)
2
ions, is 1 equiv of Ca(OH)
2
72

Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration.
More simply: it is using a known concentration of a solution to determine the concentration of a solution of unknown concentration.
73

Picture retrieved from: cikguwong.blogspot.com
When a base is added to an acid the solution will become neutral and this will be shown by an indicator changing color.
This will occur when equal numbers of
H
3
0 + and OH are present.
74

Equivalence point
The point at which two solutions used in a titration are present in chemically equivalent amounts.
The figures below shows typical pH curves for various acidbase titrations. The equivalence points and end points are different for the various combinations of strong and weak acids and bases.
End point
The point in a titration at which an indicator changes color.
Retrieved from: chemguide.co.uk
75

In this section we are going to look at: indicators, pH meters, and titrations.
You saw 2 indicators demonstrated recently to determine how acidic or basic several solutions were.
Indicators used were litmus paper and pH paper
(Hydrion).
Acid-base indicators are sensitive to pH of acids and bases. They will change color as a result of the ions present.
76

In acidic solutions, an indicator will be one color
(litmus turns red ) and in basic solutions, an indicator will be another color (litmus turns blue ).
77

Methyl red, Bromthymol blue, Methyl orange,
Phenolphthalein, Phenol red are indicator samples.
These will ionize in solution and, depending upon their acid or base strength will change color over a range of pH values until the end point is reached.
The range over which an indicator changes color is called its transition interval.
78

Litmus gives a very broad reading – a solution is either acidic or basic.
Indicators are more specific in reading the pH of an acid or base.
But the most accurate method of measuring pH is with a pH meter. A pH meter determines the pH of a solution by electrically measuring the voltage between the two electrodes placed in a solution.
79

Determination of the acidity/alkalinity of salt solutions produced by neutralization reactions.
• The relative pH of a neutralized solution can be determined utilizing the following scale showing A/B strengths.
pH 1-3 strong acid with a weak base pH 3-5 strong acid with a moderate base pH 5-7 moderate acid with a weak base pH 7(neutral)equal strength acid/base reacton pH 7-9 moderate base with a weak acid pH 9-12 strong base with a moderate acid pH 12-14 strong base with a weak acid
See your Ch 13 handout for more information.
80

• Standard solution – the solution that contains the precisely known concentration of a solute.
• Primary standard – highly purified solid compound used to check the concentration of the known solution in a titration.
• Knowing the molarity and volume of a known solution used in a titration, the molarity of a given volume of a solution with unknown concentration can be found.
81

Your book has a more complete explanation.
1. Fill one buret with an acid. Record volume.
2. Fill other buret with standard solution base.
Record volume.
3. Indicator (Phenolphthalein) will be in a flask.
4. Add a given amount of A to the flask.
5. Begin adding B to the flask until the pink color of the indicator begins to form. Swirl the contents constantly.
6. As the pink color begins to remain for longer periods of time, you are nearing the end point.
7. When the pink color remains after 30 seconds of swirling, the equivalence point is reached.
8. Record the exact volume of the base put in the flask.
Retrieved from web.ysu.edu.
82

Ma x Va = Mb x Vb
Ma – molarity of acid
Va – volume of acid
Mb – molarity of base
Vb – volume of base
Na x Va = Nb x Vb
Na – normality of acid
Va – volume of acid
Nb – normality of base
Vb – volume of base
See Ch 13 handout for more info.
83

What is the molarity of a 1.65 L solution of Al(OH)
3 totally neutralizes 58.0 g of H
3
PO
4 if it in a 6.00 L solution?
Find M of the acid: H=3x1=3, P=1x31=31, O=4x16=64 for total of: 98g/mol
58g x 1mol = 0.592 mol of acid
98 g
M acid
= 0.592 mol = 0.0986M
6.00 L
Mb = MaVa = 0.0986M(6.00L) = 3.59 x 10 -1 M Al(OH)
3
Vb 1.65L
84

Determine the volume of a 0.85 N sulfuric acid needed to neutralize 625 g of Sn(OH)
4 in 8.5 L of water. Notice: the base provides 4 equiv of OH .
Find Nb of : Sn=119, O=4(16)=64, H=4(1)=4 for a total of 187g/4 equiv/mol=46.8g/equiv.
Nb = 625g x 4 equiv x mol = 1.57 Nb
8.5L mol 187g
Va = NbVb = 1.57N (8.5L) = 15.7 L acid
Na 0.85N
85

Finish Ch13 (14&15) handout pages 125-30. This is due on Friday.
Complete Acid-Base Reactions Worksheet
86