Acids and Bases
Acids and Bases
pH and pOH
Weak Acids and Bases
Neutralization Reactions
Titrations
Buffers
In Your Textbook
Acids and Bases: pp. 576–581 and 594-598
pH and pOH: pp. 582 - 593
Weak Acids and Bases: pp. 600 – 604
Neutralization Reactions: pp. 613 – 614
Titrations: pp. 615 – 627
Buffers: pp. 628 -630
Assignment 1: Acids and Bases
Characteristic
Taste
Conductivity
Feels
Reactions
Definitions:
Arrhenius
Bronsted-Lowry
Examples
From
Home
Acids
Bases
Assignment 2: pH
pH is defined as - log [H3O]
The pH scale goes from 0 to 14.
Sketch and label the [H3O+] and the [OH-] on a pH scale.
Use the filled in chart to answer the following questions:
What happens to the [OH] going right across the chart?
The hydroxide ion concentration increases left to right.
What happens to the [H+] going right across the chart?
The hydronium ion concentration decreases left to right.
What is true when the [H+] is 1 x 10-7 ?
The solution is neutral and the [OH-] is also 1 x 10-7 M
Which is more acidic, [H+] = 1 x 10-4 or [H+] = 1 x 10-2?
[H3O+] = 1 x 10-2 M
What is the pH if the [H+] is 1 x 10-8
__________
Is this an acid or a base?
What is the pH if the [H+] is 1 x 10-4
__________
Is this an acid or a base?
What is the pH if the [H+] is 0.00010
__________
Would this taste bitter or sour?
What is the pH if the [H+] is 0.10
__________
Does this solution have more H+ or OH-?
If the pH is 5, what is the [H+]?
Does this solution have more H+ or more OH-?
__________
If the pH is 10, what is the [H+]?
Does this solution have more H+ or more OH-?
__________
Which is more acidic, lemon juice or grapefruit juice?
A lot of things that we eat are (acids or bases)?
__________
What characteristic makes them good for that application?
List some examples:
Cleaning agents tend to be (acids or bases)?
What characteristic makes them good for that?
List some examples
__________
Assignment 3: More pH and pOH
What is the most acidic substance on your “pH Values for Common Items” handout?
Hydrochloric acid at pH = 0.1
What is the most basic?
Soda lye (sodium hydroxide) at pH = 14
What is the [H+] for baking soda?
Is that a base or an acid?
A base
1 x 10-8
What is the pOH for baking soda?
1 x 10-14 = [OH-] [1 x 10-8]
1 x 10-6
Use your calculator to solve:
What is the [H+] of lemon juice?
- 2.3 = log[H3O+]
[H3O+] = 5.0 x 10-3 Molar
What is the [H+] of tomato juice?
- 4.2 = log [H3O+]
[H3O+] = 6.3 x 10-5 Molar
What is the pH of an HCl solution that is 0.036 Molar?
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.36]
0
0
-0.36
+0.36
+0.36
0
[0.36]
[0.36]
pH = - log [0.36]
= 0.44
If the pH is 8.1, what is the [H+]?
[7.9 x 10-9]
If the pH is 8.1, what is the pOH?
5.9
If the pH is 8.1, what is the [OH-]?
[1.3 x 10-6]
Calculate the pH of each solution:
[H3O+] = 1 x 10-5 M
5
[H3O+] = 4.4 x 10-11 M
10.4
[OH-1] = 2.2 x 10-7 M
7.3
pOH = 1.4
12.6
Assignment 4: Neutralization Reactions
Write the equation for these neutralization reactions:
HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l)
H2SO4 (aq) + 2 KOH (aq)  K2SO4 (aq) + 2 HOH (l)
2 HBr (aq) + Mg(OH)2 (aq)  MgBr2 (aq) + 2 HOH (l)
2 HNO3 (aq) + Ca(OH)2 (aq)  Ca(NO3)2 (aq) + 2 HOH (l)
Name the acid in each reaction.
Name the salt formed in each.
hydrochloric acid
sodium chloride
sulfiuric acid
potassium sulfate
hydrobromic acid
magnesium bromide
nitric acid
calcium nitrate
If in the 1st reaction shown above, 25.0 mL of 0.50 Molar HCl is used to neutralize 17.0
mL of NaOH, what is the molarity of the NaOH?
HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l)
25.0mL
17.0mL
0.50M
?M
0.50M = x moles HCl
0.0250 L
0.0125 moles HCl x
x = 0.0125 moles HCl
1 mole NaOH
1 mole HCl
M = 0.0125 moles NaOH
0.017 L solution
What is the standard in this titration?
= 0.0125 moles NaOH
= 0.74 M
the HCl solution
If in the 1st reaction shown above, 23.5 mL of 0.55 Molar NaOH is used to neutralize
33.0 mL of HCl, what is the molarity of the HCl?
HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l)
33.0mL
23.5mL
?M
0.55 M
0.55M = x moles NaOH
0.0235 L
0.0129 moles NaOH x
x = 0.0129 moles NaOH
1 mole HCl = 0.0129 moles HCl
1 mole NaOH
M = 0.0129 moles HCl = 0.43 M
0.0330 L solution
What is the pH of the acid solution?
0.37
Write the dissociation for the dissolving of the acid.
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.43]
0
0
-0.43
+0.43
+0.43
0
[0.43]
[0.43]
pH = - log [0.43]
= 0.37
Assignment 5: Neutralization Reactions and pH
3
What is the pH if the [H+] is 0.001
11
What is the pH if the [OH-] is 0.001
1.9
What is the pH is the [H+] is 0.0125?
0.6
10-5
What is the pH of an HCl solution that is 0.25 Molar?
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.25]
0
0
-0.25
+0.25
+0.25
pH = - log [0.25]
0
[0.25]
[0.25]
What is the pH of an NaOH solution that is 0.25 Molar?
NaOH (s)  Na+ (aq) + OH- (aq)
pOH = - log [0.25]
[0.25]
0
0
pOH = 0.6
-0.25
+0.25
+0.25
pH = 14 – 0.6
0
[0.25]
[0.25]
If the pH is 5, what is the [H+]?
10-2
If the pH is 12, what is the [OH]?
13.4
6.3x10-4
If the pH is 3.2, what is the [H+]?
1.5x10-11
If the pH is 3.2, what is the [OH-]?
pOH = 14 – 3.2 = 10.8
What is the molarity of a hydrochloric acid solution that has 12.5g dissolved in 2.0 liters.
What is the pH of the solution? Write the dissociation for the dissolving of the acid.
12.5g x 1 mole HCl = 0.34 moles HCl
M = 0.34 moles = 0.17M
36.5 g HCl
2.0 L soln
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.17]
0
0
-0.17
+0.17
+0.17
0
[0.17]
[0.17]
pH = - log [0.17]
pH = 0.77
What is the molarity of a sodium hydroxide solution that has 12.5g dissolved in 2.0 liters.
What is the pH of the solution? Write the dissociation for the dissolving of the base.
12.5g x 1 mole NaOH = 0.312 moles NaOH
M = 0.312 moles = 0.16M
40.0 g NaOH
2.0 L soln
NaOH (s)  Na+ (aq) + OH- (aq)
[0.16]
0
0
-0.16
+0.16
+0.16
0
[0.16]
[0.16]
pOH = - log [0.16]
pOH = 0.80
pH = 14 – 0.80 = 13.2
Write the equation for these neutralization reactions:
Hydrochloric acid and calcium hydroxide
2 HCl (aq) + Ca(OH)2 (aq)  CaCl2 (aq) + 2 HOH (l)
Sulfuric acid and potassium hydroxide
H2SO4 (aq) + 2 KOH (aq)  K2SO4 (aq) + 2 HOH (l)
Hydrofluoric acid and sodium hydroxide
HF (aq) + NaOH (aq)  NaF (aq) + HOH (l)
Phosphoric acid and sodium hydroxide
H3PO4 (aq) + 3 NaOH (aq)  Na3PO4 (aq) + 3 HOH (l)
What is the concentration of a sodium hydroxide solution when 30.0mL of 0.50M
hydrochloric acid are needed to neutralize 50.0mL of the base?
HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l)
30.0mL
50.0mL
0.50M
??? M
0.50M = x moles HCl
0.0300 L
0.015 moles HCl x
x = 0.015 moles HCl
1 mole NaOH
1 mole HCl
M = 0.015 moles NaOH
0.0500 L solution
= 0.015 moles NaOH
= 0.30 M
What is the pH of the base
Write the dissociation for the base.
NaOH (s)  Na+ (aq) + OH- (aq)
[0.30]
0
0
-0.30
+0.30
+0.30
0
[0.30]
[0.30]
pOH = - log [0.30]
pOH = 0.52
pH = 14 – 0.52 = 13.4
An antacid tablet containing 0.50g of magnesium hydroxide neutralizes 450.0mL of
stomach acid. What is the molarity of the HCl solution?
2 HCl (aq) + Mg(OH)2 (aq)  2 NaCl (aq) + 2 HOH (l)
450.0mL
0.50 g
?M
0.50g Mg(OH)2 x
1mole MgOH)2
58.3g Mg(OH)2
8.58 x 10-3 moles Mg(OH)2x
= 8.58 x 10-3 moles Mg(OH)2
2 mole HCl
= 0.017 moles HCl
1 mole Mg(OH)2
M = 0.017 moles HCl = 0.038 M
0.450 L solution
What is the pH of the stomach acid?
Write the dissociation for the acid.
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.038]
0
0
-0.038
+0.038
+0.038
0
[0.038]
[0.038]
pH = - log [0.038]
pH = 1.4
Assignment 6: Neutralization Reactions and pH
1
What is the pH if the [H+] is 0.10
13
What is the pH if the [OH-] is 0.10
2.6
What is the pH is the [H+] is 2.5 x 10-3?
1.4
What is the pH of an HCl solution that is 0.036 Molar?
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.036]
0
0
-0.036
+0.036
+0.036
pH = - log [0.036]
0
[0.036]
[0.036]
pH = 1.4
_____ What is the pH of an NaOH solution that is 0.036 Molar?
NaOH (s)  Na+ (aq) + OH- (aq)
pOH = - log [0.036]
[0.036]
0
0
pOH = 1.44
-0.036
+0.036
+0.036
pH = 14 – 1.4 = 12.6
0
[0.036]
[0.036]
10-10
If the pH is 10, what is the [H+]?
10-13
If the pH is 1, what is the [OH]?
7.9x10-9
If the pH is 8.1, what is the [H+]?
1.3x10-6
If the pH is 8.1, what is the [OH-]?
pOH = 14 – 8.1 = 5.9
0.05
What is the molarity of an HCl solution that has a pH of 1.3?
0.16
What is the molarity of an NaOH solution that has a pH of 13.2
Write the equation for these neutralization reactions:
Nitric acid and potassium hydroxide
HNO3 (aq) + KOH (aq)  KNO3 (aq) + HOH (l)
Hydrochloric acid and calcium hydroxide
2 HCl (aq) + Ca(OH)2 (aq)  CaCl2 (aq) + 2 HOH (l)
Acetic acid and sodium hydroxide
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) + HOH (l)
If in the 1st reaction above, 42.7 mL of 0.498 Molar standard is used to neutralize 30.0mL
of the base, what is the molarity of the base?
HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l)
42.7mL
30.0mL
0.498M
??? M
0.498M = x moles HCl
0.0427 L
0.0213 moles HCl x
x = 0.0213 moles HCl
1 mole NaOH
1 mole HCl
M = 0.0213 moles NaOH
0.0300 L solution
= 0.0213 moles NaOH
= 0.71 M NaOH
What is the pH of the base?
_______________
NaOH (s)  Na+ (aq) + OH- (aq)
[0.71]
0
0
-0.71
+0.71
+0.71
0
[0.71]
[0.71]
pOH = - log [0.71]
pOH = 0.15
pH = 14 – 0.15 = 13.85
What is the concentration of acetic acid in vinegar when 32.5mL of 0.56M sodium
hydroxide is needed to neutralize 15.0mL of the vinegar?
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) + HOH (l)
?? M
.56M
15.0mL
32.5mL
0.56M NaOH =
x mole NaOH
0.0325L
0.0182 moles NaOH x
= 0.0182 moles NaOH
1 mole HCl
1 mole NaOH
M = 0.0182 moles HCl = 1.2 M
0.0150 L solution
= 0.0182 moles HCl
Assignment 7: Weak Acids and Bases
Acids
Strong
Bases
Weak
Strong
Weak
Ions
Present
Definition
Examples
Equilibrium
Constant
Hydrocyanic acid is a weak acid with a Ka of 4.9 x 10-10.
Write the reaction for the dissociation of the acid.
HCN (aq) + H2O (l)   H3O+ (aq) + CN - (aq)
Write the expression for the acid dissociation constant (Ka).
Ka = [H3O+] [CN-]
[HCN]
=
What is the [H3O+] concentration when 0.25 moles of HCN is reacted with enough water
to make 1 liter of solution.
HCN (aq) + H2O (l)   H3O+ (aq) + CN - (aq)
[0.25]
0
0
4.9x10-10 = (x)(x)
-x
+x
+x
[0.25]
0.25 - x
[x]
[x]
x = 1.1 x 10-5M = [H3O+]
The larger the Ka, the stronger the acid.
Read pp. 600-604 and look at sample problem 20-8
Complete problems 26 -33 on page 605.
26.
HCOOH (aq) + H2O (l)  H3O+ (aq) + HCOO- (aq)
[0.1]
0
0
-0.0042
+0.0042
+0.0042
0.0958
[0.0042]
[0.0042]
Ka = [H3O+] [HCOO-] =
[HCOOH]
27.
[0.0042] [0.0042] = 1.8 x 10-4
[0.0958]
HX (aq) + H2O (l)  H3O+ (aq)
[0.2]
0
-9.86x10-4
+9.86x10-4
0.1990
[9.86x10-4]
Ka = [H3O+] [X-]
[HX]
=
+
X- (aq)
0
+9.86x10-4
[9.86x10-4]
[9.86x10-4] [9.86x10-4] = 4.9 x 10-6
[0.1990]
pH = - log [H3O+]
= - log [9.86 x 10-4]
= 3.01
28.
A strong acid dissociates completely into H3O+ ions and anions. A weak acid
only dissociates partially. The acid with the highest Ka is the stronger acid (it
dissociates more).
29.
pH = - log [H3O+]
-1.80 = log [H3O+]
[H3O+] = 0.0158M
CH2ClCOOH (aq) + H2O (l)  H3O+ (aq) + CH2ClCOO- (aq)
[0.18]
0
0
-0.0158
+0.0158
+0.0158
[0.164]
[0.0158]
[0.0158]
Ka = [H3O+] [CH2ClCOO-]
[HCOOH]
=
[0.0158] [0.0158]
[0.164]
= 1.5 x 10-3
30.
Table 20.7 is a list of acids ranked according to “strength” which means how
much they dissociate. The weakest acid (at the bottom of the acid list) would
have the lowest Ka indicating that is doesn’t dissociate much.
31.
If an acid has a low Ka, the acid stays mostly undissociated (HX) and there is not
much dissociated (H+).
32.
a. HNO3 + H2O (l) H3O+ (aq) + NO3- (aq)
b. CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
c. NH3(aq) + H2O (l)   NH4 + (aq) + OH- (aq)
d. Mg(OH)2 (s)  Mg+2 (aq) + OH- (aq)
33.
“Strength” refers to whether the acid or base dissociates all the way.
“Concentration” refers to how much of the acid or base is dissolved in water. It
may or may not be all dissociated. At the same concentration, a strong acid has a
much higher concentration of H3O+ ions causing a lower pH. However, a weak
acid could have a high concentration and so have a lower pH than a very dilute
strong acid.
Assignment 8: pH problems with strong and weak acids/bases:
1. What is the pH of a 0.75M solution of hydrochloric acid?
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[0.75]
0
0
-0.75
+0.75
+0.75
pH = - log [0.75]
0
[0.75]
[0.75]
pH = 0.12
2. What is the pH of a 0.75M solution of sodium hydroxide?
NaOH (g)  Na+ (aq) + OH- (aq)
[0.75]
0
0
-0.75
+0.75
+0.75
pOH = - log [0.75]
0
[0.75]
[0.75]
pOH = 0.12
pH = 14 – 0.12
pH = 13.88
3. What is the concentration of a solution of hydrochloric acid if the pH is 3.5?
-3.5 = log [H3O+]
[H3O+] = 3.16 x 10-4 M
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
[3.16x10-4]
0
0
-3.16x10-4
+3.16x10-4 +3.16
0
[3.16x10-4]
[3.16x10-4]
so [HCl] = 3.16x10-4M
4. What is the concentration of a solution of sodium hydroxide if the pH is 8.5?
pOH = 14 – 8.5 = 5.5
NaOH (g)  Na+ (aq) + OH- (aq)
-5.5 = log [OH_]
[3.16x10-6]
0
0
[OH] =
-3.16x10-6
+3.16x10-6
+3.16x10-6
0
[3.16x10-6] [3.16x10-6]
-6
so [NaOH] = 3.16x10 M
5. What is the pH of a 0.50M solution of acetic acid? Ka = 1.8 x 10-5
HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)
[0.50]
0
0
-x
+x
+x
[0.50-x]
[x]
[x]
*the x is negligible
1.8 x 10-5 =
[x] [x]
[0.5]
= 0.0030 = x = [H3O+]
pH = - log [0.0030]
pH = 2.5
6. What is the pH of a 0.50M solution of carbonic acid? Ka = 4.3 x 10-7
H2CO3 (aq) + H2O (l) H3O+ (aq) + HCO3- (aq)
[0.50]
0
0
-x
+x
+x
[0.50-x]
[x]
[x]
*the x is negligible
4.3 x 10-7 =
[x] [x]
[0.5]
= 0.00046 = x = [H3O+]
pH = - log [0.00046]
pH = 3.3
7. Why are the pH’s not the same in the previous 2 problems?
The pH’s for the two acids are not the same because they dissociate to different
degrees. Acetic acid dissociates a little more than the carbonic acid (it has a
higher Ka) therefore the concentration of hydronium ions for acetic acid is higher
causing a lower pH.
8. What is the concentration of hydrocyanic acid if the pH is 4.6? Ka = 4.9 x 10-10
- 4.6 = log [H3O+]
[H3O+] = 2.5 x 10-5 M
HCN (aq) + H2O (l) H3O+ (aq) + CN- (aq)
[1.3]
0
0
-5
-5
- 2.5 x 10
+ 2.5 x 10
+ 2.5 x 10-5
[x]
[2.5 x 10-5]
[2.5 x 10-5]
Ka = [H3O+] [CN-]
[HCN]
4.9 x 10-10 = [2.5 x 10-5]2
[HCN]
[HCN] = 1.3M at equilibrium
initial would be:
1.3 – 2.5x10-5 = 1.3M
(the x is negligible)
Assignment 9: pH Problems with Acids (weak and strong):
1.
Calculate the molarity of an acetic acid solution that has [H3O+] of 9.0 x 10-4.
(Ka = 1.8 x 10-5)
2.
Calculate the pH of a 1.00 M solution of hydrocyanic solution. (Ka = 4.8 x 10-10)
3.
The [H3O+] in a 0.072 M solution of benzoic acid is 2.1 x 10-3 M. Compute the
Ka. (Benzoic acid is HC7H5O2)
4.
Calculate the concentration of the ions of a 0.100 M formic acid (HCOOH)
solution. (Ka = 1.7 x 10-4)
What is the pH of the solution?
5.
Calculate the concentration of the ions of a 0.100 M HCl solution.
What is the pH of the solution?
6.
Explain any difference between problem 4 and problem 5.
pH Problems with Bases (weak and strong):
7.
Write the reaction for the dissociation of potassium hydroxide in water.
What is the pH of a 0.400 potassium hydroxide solution?
Explain how you know the [OH-].
8.
Write the reaction for the dissociation of ammonia in water. Water is part of the
reaction.
What is the pH of a 0.400 M ammonia solution? (Kb = 1.8 x 10-5)
Explain the difference between #7 and #8.
9.
If the pH of a solution of calcium hydroxide is 9.5, what is the concentration of
the solution?
How many grams of calcium hydroxide would be dissolved in 50.0 mL of
solution?
10.
The pH of an ammonia solution is 9.5. What is the concentration of the solution?
11.
A 0.500 M solution of hydrochloric acid is titrated against a solution of ammonia
of unknown concentration. If 7.45 mL of the acid is required to neutralize 10.65
mL of the ammonia, what is the concentration of the ammonia solution?
What is the pH of the ammonia solution?
Assignment 10: Buffers:
1. Determine which of the following would be a buffer system and explain how a
buffer works.
a. HCl/KCl
(b) would be a buffer system as it is a weak base
b. NH3/NH4NO3
with the salt of that base. It sets up an equilibrium
that can react with both and acid or a base.
2. Write the equation for the dissociation of weak acid HCN and label the conjugate
acid and the conjugate base.
HCN (aq) + H2O (l)   H3O+ (aq) + CN- (aq)
conj. acid
conj. base
3. Write the equation for the dissociation of weak acid HClO and label the conjugate
acid and the conjugate base. What do you think the name of this acid is?
HClO (aq) + H2O (l)   H3O+ (aq) + ClO- (aq)
conj. acid
conj. base
The acid is names hypochlorous acid (it is hydrogen hypochlorite…the
“ite” gives the “ous” ending.)
4. Write the equation for the dissociation of acetic acid and label the conjugate acid
and the conjugate base.
HC2H3O2 (aq) + H2O (l)   H3O+ (aq) + C2H3O2 - (aq)
conj. acid
conj. base
Write the reaction that occurs when base (OH-) is added:
OH- (aq) + HC2H3O2 (aq)  C2H3O2 – (aq) + HOH (l)
(the conjugate acid reacts with the OH- ions)
Write the reaction that occurs when acid (H3O+) is added:
H3O+ (aq) + C2H3O2 – (aq)  H2O (l) + HC2H3O2 (aq)
(the conjugate base reacts with the H3O+ ions.)
5. Calculate the pH of a buffer solution with 0.15M NH3 / 0.35M NH4Cl. The Kb of
ammonia is 1.8 x 10-5.
NH4Cl (s)  NH4+ (aq) + Cl- (aq)
[0.35]
0
0
- 0.35
+ 0.35
+ 0.35
0
[0.35]
[0.35]
NH3 (g) + HOH (l)   NH4+ (aq) + OH- (aq)
[0.15]
[0.35]
0
-x
+x
+x
0.15-x
0.35+x
x
1.8 x 10-5 = (x) (0.35) = 7.7 x 10-6 M
0.15
*the NH4+ is from the salt
above.
*The “x” for NH3 and NH4+
is negligible
pOH = - log [7.7 x 10-6] = 5.1
pH = 14 – 5.1 = 8.9
6. Calculate the pH of a buffer solution prepared by adding 20.5g of HC2H3O2 and
17.8g of NaC2H3O2 to enough water to make 500.0mL of solution.
17.8g NaC2H3O2 x 1 mole = 0.217 mol
M = 0.217 mol = [0.434]
82g
0.5000L
20.5g HC2H3O2 x
1 mole = 0.342 mol
60g
M = 0.234 mol = [0.683]
0.5000L
NaC2H3O2 (s)  Na+ (aq) + C2H3O2- (aq)
[0.434]
0
0
- 0.434
+ 0.434
+ 0.434
0
[0.434]
[0.434]
HC2H3O2 (aq) + HOH (l)   H3O+ (aq) + C2H3O2- (aq)
[0.683]
0
[0.434]
* C2H3O2- is from
-x
+x
+x
the salt above.
0.683-x
x
0.434+x
*The “x” for HC2H3O2 and
C2H3O2- is negligible.
-5
-5
1.8 x 10 = (x) (0.434) = 2.8 x 10 M
0.683
pH = - log [2.8 x 10-5] = 4.55
7. Calculate the pH of 1.00L of a buffer that is 1.0M NaC2H3O2 and 1.00M
HC2H3O2.
NaC2H3O2 (s)  Na+ (aq) + C2H3O2- (aq)
[1.0]
0
0
- 1.0
+ 1.0
+ 1.0
0
[1.0]
[1.0]
HC2H3O2 (aq) + HOH (l)   H3O+ (aq) + C2H3O2- (aq)
[1.0]
0
[1.0]
* C2H3O2- is from
-x
+x
+x
the salt above.
1.0-x
x
1.0+x
*The “x” for HC2H3O2 and
C2H3O2- is negligible.
-5
-5
1.8 x 10 = (x) (1.0) = 1.8 x 10 M
1.0
pH = - log [1.8 x 10-5] = 4.7
What is the pH after the addition of 0.08 mol NaOH?
NaOH (s)  Na+ (aq) + OH – (aq)
[0.08]
0
0
-0.08
+0.08
+0.08
0
[0.08]
[0.08]
The added 0.08M OH- completely reacts with the conj acid (HC2H3O2)
HC2H3O2 (aq) + OH- (aq)  C2H3O2- (aq) HOH (l)
[1.0]
[0.08]
[1.0]
-0.08
-0.08
+0.08
[0.92]
0
[1.08]
from original
buffer soln.
so now:
HC2H3O2 (aq) + HOH (l)   H3O+ (aq) + C2H3O2- (aq)
[0.92]
0
[1.08]
-x
+x
+x
0.92-x
x
1.08+x
* x’s are
negligible.
1.8 x 10-5 = (x) (1.08) = 1.53 x 10-5 M
0.98
pH = - log[1.53 x 10-5] = 4.8
Impressive, isn’t it?
What is the pH after the addition of 0.12 mol HCl?
H2O (l) + HCl (s)  H+ (aq) + Cl – (aq)
[0.12]
0
0
-0.12
+0.12
+0.12
0
[0.12]
[0.12]
The added 0.12M H3O+- completely reacts with the conj base (C2H3O2-)
C2H3O2- (aq) + H3O+ (aq)  HC2H3O2- (aq) H2O (l)
[1.0]
[0.12]
[1.0]
-0.12
-0.12
+0.12
[0.88]
0
[1.12]
bold [ ]
from original
buffer soln.
so now:
HC2H3O2 (aq) + HOH (l)   H3O+ (aq) + C2H3O2- (aq)
[1.12]
0
[0.88]
-x
+x
+x
1.12-x
x
0.88+x
1.8 x 10-5 = (x) (0.88) = 2.29 x 10-5 M
1.12
pH = - log[2.29 x 10-5] = 4.6
* x’s are
negligible.
Acids and Bases Review
You should be able to:
Describe acids and bases as to physical and chemical properties.
Name acids (review your notes from 1st semester)
Write the reaction for the dissociation of acids and bases in water.
Describe the pH scale, how it is derived, why it goes from 0-14, and what that has to do
with water.
Calculate pH of strong acids and bases. (and pOH)
Given the pH (or the pOH), calculate the concentration of the acid or base.
Describe the difference between a strong acid or base and a weak acid or base.
Write the reaction for the dissociation of weak acids and bases in water.
Calculate the pH of a weak acid or base using the dissociation constant.
Given the pH (or pOH) of a weak acid or base, calculate the concentration using the
dissociation constant.
Write the reaction for a neutralization reaction between an acid and a base.
Describe how a titration works and what it is used for.
Use titration data to calculate concentrations of unknown acid or bases and then calculate
the pH of that unknown.
Describe a buffer solution, what it does, and how it works.
Calculate a change in pH when acid or base is added to the buffer.
Discuss acid rain, its causes, problems, and possible solutions.
Review Problems:
1. A solution of hydrochloric acid is 1.5 x 10-2 M.
a. What is the equation for the dissociation of the acid in water?
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
b. What is the hydronium ion concentration [H3O+]?
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
1.5x10-2
0
0
+
-1.5x10-2
1.5x10-2
+1.5x10-2
0
[1.5x10-2]
[1.5x10-2]
c. What is the pH of the solution?
pH = - log[1.5x10-2]
= 1.8
d. What is the hydroxide ion concentration [OH-]?
1 x 10-14 = [OH-] [1.5x10-2]
[OH-] = 6.7 x 10-13
e. Does this solution:
turn litmus blue? no
turn phenolphthalene pink? no
taste sour? yes
2. A solution of sodium hydroxide is 1.4 x 10-4 M.
a. Write the equation for the dissociation of the base in water.
NaOH (s)  Na+ (aq) + OH- (aq)
b. What is the hydronium ion concentration ?
NaOH (s)  Na+ (aq) + OH- (aq)
1.4x10-4
0
0
-4
-4
-1.4x10
+1.4x10
+1.4x10-4
0
[1.4x10-4]
[1.4x10-4]
1 x 10-14 = [H3O+] [1.4x10-4]
[H3O+] = 7.1 x 10-11
c. What is the pH of the solution?
pH = - log[7.1 x 10-11]
pH = 10.1
d. Does this solution:
taste bitter? yes
react with metals no
react with bases to form salt water? No, it is a base.
e. Does the solution in problem #1 or problem #2 have more hydronium ions? 1
Which has more hydroxide ions? 2
Which is more acidic? 1
3. A solution of acetic acid is 1.5 x 10 -2M.
a. What is the equation for the dissociation of the acid in water?
HC2H3O2 (aq) + H2O (l)   H3O+ (aq) + C2H3O2- (aq)
b. What is the expression for the ionization constant (Ka)?
Ka = [H3O+] [C2H3O2-] =
[HC2H3O2]
c. What is the hydronium ion concentration [H3O]? (Ka = 1.8 x 10-5)
HC2H3O2 (aq) + H2O (l)   H3O+ (aq) + C2H3O2- (aq)
1.5x10-2
0
0
-x
+x
+x
1.5x10-2 - x
[x]
[x]
1.8 x 10-5 = (x) (x) = 5.2 x 10-4 M
1.5x10-2
d. What is the pH of the solution?
pH = - log [5.2 x 10-4]
pH = 3.3
e. Why is the pH of the acetic acid different from the hydrochloric acid in
problem #1 while they have the same concentration.
Because they have different dissociation constants. The acetic acid has a
higher constant, meaning that it dissociates more making more hydronium
ions. That makes the pH lower (more acidic).
f. What is the hydroxide ion concentration?
1 x 10-14 = [5.2 x 10-4] [OH-]
[OH-] = 1.9 x 10-4 M
g. Does this solution:
conduct electricity? yes
have more hydroxide or hydronium ions? Hydronium ions
feel slippery? no
h. If the Ka for carbonic acid is 4.4 x 10-7, does a 0.50M solution of carbonic acid
or a 0.50M solution of acidic acid have a lower pH (more acidic)?
At the same concentration, the weak acid with the larger Ka would have more
hydronium ions and so a lower pH. Therefore, the acetic acid would be more
acidic.
i. Which is a stronger acid, the carbonic acid or the acetic acid?
They are both weak, but the acetic is stronger.
4. A solution of lithium hydroxide of unknown concentration is titrated with a
0.50M solution of hydrochloric acid. 50.0mL of the acid was required to
neutralize 62.0mL of the base.
a. What is the neutralization reaction?
LiOH (aq) + HCl (aq)  LiCl (aq) + HOH (l)
62.0mL
50.0mL
0.50M
b. What is the molarity of the sodium hydroxide? (what is the normality?)
0.50M HCl =
x mole HCl
0.0500L
0.025 moles HCl x
1 mole LiOH
1 mole HCl
M = 0.025 moles LiOH
0.0620 L solution
or:
= 0.025 moles HCl
= 0.025 moles LiCl
= 0.40 M which is 0.40N because LiOH
gives 1 OH- per mole.
Na Va = Nb Vb
(0.50N)(50.0mL) = Nb (62.0mL)
Nb = 0.40N = 0.40M
c. What is the hydroxide ion concentration [OH-] of the lithium hydroxide
solution?
LiOH (s)  Li+ (aq) + OH- (aq)
[0.40]
0
0
-0.40
+0.40
+0.40
0
[0.40]
[0.40]
d. What is the hydronium ion concentration [H3O+] of the lithium hydroxide
solution?
1 x10-14 = [H3O+] [0.40]
[H3O+] = 2.5 x 10-14 M
e. What is the pH of the lithium hydroxide?
pH = - log [2.5 x 10-14]
pH = 13.6
5. A solution of calcium hydroxide with a concentration of 0.50M is titrated
with a solution of hydrocyanic acid with unknown concentration. 50.0mL
of the acid was required to neutralize 62.0mL of the base.
a. What is the neutralization reaction?
Ca(OH)2 (aq) + 2 HCN (aq)  Ca(CN)2 (aq) + 2 HOH (l)
[0.50]
?M
62.0mL
50.0mL
b.
What is the molarity of the hydrocyanic acid? (what is the normality?)
0.50M Ca(OH)2 = x mole Ca(OH)2
= 0.031 moles Ca(OH)2
0.0620L
0.031 moles Ca(OH)2 x
2 mole HCN
1 mole Ca(OH)2
= 0.062 moles HCN
M = 0.062 moles HCN = 1.24 M which is 1.24 N because HCN
0.0500 L solution
gives 1 H3O+ per mole.
or:
Nb Vb = Na Va
(1.0N)(62.0mL) = Na (50.0mL)
Na = 1.24N = 1.24M
*the 0.50M Ca(OH)2 is 1.0N because is gives 2 OH- ions per mole of base.
c. Write the equation for the dissociation of the weak acid in water.
HCN (aq) + H2O (l)   H3O+(aq) + CN- (aq)
d. What is the pH of the acid? The Ka is 4.8 x 10-10.
HCN (aq) + H2O (l)   H3O+(aq) + CN- (aq)
[1.24]
0
0
-x
+x
+x
1.24-x
x
x
4.8 x 10-10 = _(x) (x) = 2.4 x 10-5 M
1.24
pH = - log [2.4 x 10-5]
pH = 4.6
6. A buffer solution is made with 0.10M HCN/ 0.10M NaCN. Ka is 4.8 x 10-10
a. Write the reaction for the dissociation of the weak acid in water.
HCN (aq) + H2O (l)   H3O+(aq) + CN- (aq)
b. What is the conjugate acid and what is the conjugate base?
HCN (aq) + H2O (l)   H3O+(aq) + CN- (aq)
Acid
Base
c. What is the pH of the buffer solution?
NaCN (s)  Na+ (aq) + CN- (aq)
[0.10]
0
0
-0.10
+0.10
+0.10
0
[0.10]
[0.10]
the CN- goes as initial below.
HCN (aq) + H2O (l)   H3O+(aq) + CN- (aq)
[0.10]
0
[0.10]
-x
+x
+x
0.10-x
x
0.10+x
x is negligible
4.8 x 10-10 = (x) (0.10)
0.10
x = 4.8 x 10-10 M
pH = - log [4.8 x 10-10]
pH = 9.3
d. If 0.05M NaOH was added, what would the new pH be?
NaOH (s)  Na+ (aq) + OH – (aq)
[0.05]
0
0
-0.05
+0.05
+0.05
0
[0.05]
[0.05]
The added 0.05M OH- completely reacts with the conj acid (HCN)
HCN (aq)
[0.10]
-0.05
[0.05]
so now:
+
OH- (aq)  CN- (aq) HOH (l)
[0.05]
[0.10]
-0.05
+0.05
0
[0.15]
from original
buffer soln.
HCN (aq) + HOH (l)   H3O+ (aq) + CN- (aq)
[0.05]
0
[0.15]
-x
+x
+x
0.05-x
x
0.15+x
4.8 x 10-10 = (x) (0.15) = 1.6 x 10-10 M
0.05
pH = - log[1.6 x 10-10] = 9.8
* x’s are
negligible.