Chapter 10 Acids and Bases
pH Scale
The pH of a solution
10.5
The pH Scale
§  Is used to indicate the acidity of a solution.
§  Has values that usually range from 0 to 14.
§  Is acidic when the values are less than 7.
§  Is neutral with a pH of 7.
§  Is basic when the values are greater than 7.
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pH of Everyday Substances
2
Learning Check
Identify each solution as
A. acidic
B. basic
N. neutral
1. ___ HCl with a pH = 1.5
2. ___ Pancreatic fluid [H3O+] = 1 x 10−8 M
3. ___ Sprite soft drink pH = 3.0
4. ___ pH = 7.0
5. ___ [OH−] = 3 x 10−10 M
6. ___ [H3O+ ] = 5 x 10−12
3
Solution
1. A
HCl with a pH = 1.5
2. B
Pancreatic fluid [H3O+] = 1 x 10−8 M
3. A
Sprite soft drink pH = 3.0
4. N
pH = 7.0
5. A
[OH-] = 3 x 10−10 M
6. B
[H3O+] = 5 x 10−12
4
Testing the pH of Solutions
The pH of solutions can be determined using
§  a) pH meter, b) pH paper, or
§  c) indicators that have specific colors at different pH
values.
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6
1
Calculating pH
Significant Figures in pH
Mathematically pH
§  Is the negative log of the hydronium ion concentration,
pH = - log [H3O+]
§  For a solution with [H3O+] = 1 x 10−4
pH = −log [1 x 10−4 ]
pH = - [-4.0]
pH = 4.0
Note: The number of decimal places in the pH equals
the significant figures in the coefficient of [H3O+].
When expressing log values, the number of decimal
places in the pH is equal to the number of significant
figures in the coefficient of [H3O+].
coefficient
[H3O+] = 1 x 10-4
[H3O+] = 8.0 x 10-6
[H3O+] = 2.4 x 10-8
decimal places
pH = 4.0
pH = 5.10
pH = 7.62
1 SF in 1 x 10-4 = 4.0
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Guide to Calculating pH
8
Sample pH Problem
Find the pH of a solution with a [H3O+] of 1.0 x 10−3:
STEP 1 Enter [H3O+]
Enter 1 x 10-3 (press 1 EE 3
The EE key gives an exponent of 10 and
change sign (+/- key or – key)
STEP 2 Press log key and change sign
- log (1 x 10−3) = -[-3]
STEP 3 Adjust figures after decimal point to equal the
significant figures in the coefficient to give
Two significant figures in 1.0 x 10−3 = 3.00
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Learning Check
10
Solution
What is the pH of coffee if the [H3O+] is 1 x 10−5 M?
What is the pH of coffee if the [H3O+] is 1 x 10−5M?
3) pH = 5.0
pH = -log [1 x 10−5] = -(-5.0) = 5.0
1) pH = 9.0
2) pH = 7.0
3) pH = 5.0
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2
Learning Check
Solution
A. The [H3O+] of tomato juice is 2 x 10−4 M.
What is the pH of the solution?
1) 4.0
2) 3.7
3) 10.3
A. 2) 3.7
pH = - log [ 2 x 10-4] = 3.7
2 (EE) 4 (+/-) log (+/-)
B. The [OH−] of a solution is 1.0 x 10−3 M.
What is the pH of the solution?
1) 3.00
2) 11.00
3) -11.00
B. 2) 11.00
Use the Kw to obtain [H3O+] = 1.0 x 10−11
pH = - log [1.0 x 10−11]
1.0 (EE) 11 (+/-) log (+/-)
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[H3O+], [OH-] and pH Values
14
Calculating [H3O+] from pH
§  The [H3O+] for pH value of 8.0 ( a whole number) is
calculated [H3O+] = 1 x 10−pH
TABLE 10.7
For pH = 8.0, the [H3O+] = 1 x 10−8
STEP 1 Enter the pH value, change sign
-8.0
STEP 2 Use 2nd and 10x keys or inverse and log keys
1-08
STEP 3 Adjust the significant figures
1 x 10-8
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16
Calculating [H3O+] from pH
Learning Check
§  Calculate the [H3O+] for pH = 3.80 is calculated as
follows:
A. What is the [H3O+] of a solution with a pH of
10.0?
1) 1 x 10−4 M
2) 1 x 1010 M
3) 1 x 10−10 M
B. What is the [H3O+] of a solution with a pH of
9.30?
1) 1.0 x 10−9.3 M
2) 3.0 x 10−9 M
3) 5.0 x 10-10 M
STEP 1 Enter the pH value, change sign
-3.80
STEP 2 Use 2nd and 10x keys or inverse and log keys
1.584893 -06
STEP 3 Adjust the significant figures
1.6 x 10-6
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3
Solution
Chapter 10 Acids and Bases
10.6
Reactions of Acids and Bases
A. What is the [H3O+] of a solution with a pH of 10.0?
3) 1 x 10-10 M
1 x 10-pH
B. What is the [H3O+] of a solution with a pH of
9.30?
3) 5.0 x 10-10 M
[H3O+] = -9.30 2nd key 10x key sci key
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20
Acids and Metals
Acids and Carbonates
Acids react with metals
§  Such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn .
§  To produce hydrogen gas and the salt of the metal.
Molecular equations:
2K(s) + 2HCl(aq)
2KCl(aq) + H2(g)
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Acids react
Net ionic equation:
2K(s) + 2H+(aq)
Zn(s) + 2H+(aq)
§  With carbonates and hydrogen carbonates to produce
carbon dioxide gas, a salt, and water.
2HCl(aq) + CaCO3(s)
CO2(g) + CaCl2(aq) + H2O(l)
HCl(aq) + NaHCO3(s)
CO2(g) + NaCl (aq) + H2O(l)
2K+(aq) + H2(g)
Zn2+ (aq) + H2(g)
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Learning Check
22
Solution
Write the products of the following reactions of acids:
Write the products of the following reactions of acids:
A. Zn(s) + 2 HCl (aq)
A. Zn(s) + 2 HCl(aq)
B. MgCO3(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
MgCl2(aq) + CO2(g) + H2O(l)
B. MgCO3 (s) + 2HCl(aq)
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4
Neutralization Reactions
Neutralization Equations
Neutralization is the reaction of
§  An acid such as HCl and a base such as NaOH.
HCl + H2O
H3O+ + Cl−
NaOH
Na+
+ OH−
§  The H3O+ from the acid and the OH− from the
base to form water.
H3O+ + OH−
2H2O
§  In the equation for neutralization, an acid and a
base produce a salt and water.
acid
base
salt
water
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
2HCl(aq) + Ca(OH)2(aq)
CaCl2(aq) + 2H2O(l)
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Balancing Neutralization
Reactions
26
Balancing Neutralization
Reactions
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
STEP 1 Write the base and acid formulas.
Mg(OH)2(aq) + HNO3(aq)
STEP 2 Balance OH- and H+.
Mg(OH)2(aq) + 2HNO3(aq)
STEP 3 Balance with H2O
Mg(OH)2(aq) + 2HNO3(aq)
salt +
2H2O(l)
STEP 4 Write the salt from remaining ions.
Mg(OH)2(aq) + 2HNO3(aq)
Mg(NO3)2(aq) + 2H2O(l)
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Learning Check
Solution
Select the correct group of coefficients for the following
A. 3) 3, 1, 1, 3
neutralization equations
A. HCl (aq) + Al(OH)3(aq)
1) 1, 3, 3, 1
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3HCl(aq + Al(OH)3(aq)
AlCl3(aq) + H2O(l)
2) 3, 1, 1, 1
AlCl3(aq) + 3H2O(l)
3) 3, 1, 1, 3
B. 2) 3, 2, 1, 6
B. Ba(OH)2(aq) + H3PO4(aq)
1) 3, 2, 2, 2
Ba3(PO4)2(s) + H2O(l)
2) 3, 2, 1, 6
3) 2, 3, 1, 6
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3Ba(OH)2 (aq) + 2H3PO4(aq)
Ba3(PO4)2(s) + 6H2O(l)
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5
Antacids
Learning Check
Antacids
Write the neutralization reactions for stomach acid
HCl and Mylanta.
§  are used to neutralize stomach acid (HCl).
TABLE 10.8
31
Solution
32
Acid-Base Titration
Titration
§  Is a laboratory
procedure used to
determine the
molarity of an acid.
§  Uses a base such
as NaOH to
neutralize a
measured volume
of an acid.
Write the neutralization reactions for stomach acid
HCl and Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2
3HCl(aq) + Al(OH)3(aq)
AlCl3(aq) + 3H2O(l)
2HCl(aq) + Mg(OH)2(aq)
MgCl2(aq) + 2H2O(l)
Base
(NaOH)
Acid
solution
33
Indicator
34
End Point of Titration
At the end point,
§  The indicator has a
permanent color.
§  The volume of the base
used to reach the end
point is measured.
§  The molarity of the acid
is calculated using the
neutralization equation
for the reaction.
An indicator
§  Is added to the acid
in the flask.
§  Causes the acid
solution to change
color when the acid
is neutralized.
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36
6
Calculating Molarity from A
Titration with A Base
Guide to Calculating Molarity
What is the molarity of an HCl solution if 18.5 mL of a
0.225 M NaOH are required to neutralize 10.0 mL HCl?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
STEP 1 Calculate the moles of base.
18.5 mL NaOH x 1 L
x 0.225 mole NaOH
1000 mL
1L
= 0.00416 mole NaOH
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38
Calculating Molarity (continued)
Learning Check
STEP 2 Calculate the moles of HCl.
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
0.00416 mole NaOH x 1 mole HCl
1 mole NaOH
= 0.00416 mole HCl
Calculate the mL of 2.00 M H2SO4 required to
neutralize 50.0 mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)
K2SO4(aq) + 2H2O(l)
1) 12.5 mL
2) 50.0 mL
STEP 3 Calculate the molarity of HCl.
10.0 mL HCl = 0.010 L HCl
0.00416 mole HCl = 0.416 M HCl
0.0100 L HCl
3) 200. mL
39
Solution
40
Learning Check
1) 12.5 mL
0.0500 L x 1.00 mole KOH x 1 mole H2SO4
1L
2 mole KOH
A 25.0 mL sample of phosphoric acid is neutralized by
42.6 mL of 1.45 M NaOH. What is the molarity of the
phosphoric acid solution?
x
3NaOH(aq) + H3PO4 (aq)
1L
x
2 mole H2SO4
1000 mL =
1L
Na3PO4(aq) + 3H2O(l)
1) 0.620 M
12.5 mL
2) 0.841 M
3) 0.185 M
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42
7
Solution
Chapter 10 Acids and Bases
10.7
Acid-Base Properties of Salt
Solutions
2) 0.841 M
0.0426 L x 1.45 mole NaOH x 1 mole H3PO4
1L
3 mole NaOH
= 0.0206 mole H3PO4
0.0206 mole H3PO4 = 0.841 mole/L = 0.841 M
0.0245 L
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Salts that Form Neutral Solutions
44
Salts that Form Basic Solutions
A salt solution containing the anion of a weak acid and
A salt solution containing the anion of a strong
the cation of a strong base
acid and the cation of a strong base
§  Is a basic solution.
§  Does not produce or attract H+ from water.
§  Has an anion of the weak acid that attracts H+ from
water.
§  Is a neutral solution.
§  Of KNO3, for example, is neutral because it
contains a cation from a strong base (KOH)
and an anion from a strong acid (HNO3).
§  Of, KHCO3, for example, is basic because its anion
from a weak acid (H2CO3) attracts H+ from water and
produces OH-.
HCO3- + H2O
anion of
weak acid
H2CO3 + OH-
Forms
Solution is basic
weak acid
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Salts that Form Acidic Solutions
46
Changes in pH for Salts in Water
A salt solution containing the ions of a strong acid and a
TABLE 10.9
weak base
§  Is an acidic solution.
§  Has an cation of the weak base that produces H+ in
water.
§  Of NH4Cl, for example, is acidic because its cation from
a weak base (NH4OH) releases H+ to form H3O+.
NH4+ + H2O
NH3
+ H3O+
cation of
weak base
Forms
weak base
Solution is acidic
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8
Learning Check
Solution
Predict whether a solution of each salt will be
1) acidic
2) basic
or
3) neutral
A. Li2S (aq)
B. Mg(NO3)2 (aq)
C. NH4Cl (aq)
Predict whether a solution of each salt will be
1) acidic
2) basic
or
3) neutral
A. Li2S (aq)
2) basic; S2- forms HS- and OHB. Mg(NO3)2 (aq)
3) neutral
C. NH4Cl (aq)
1) acidic; NH4+ produces H3O+
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Chapter 10 Acids and Bases
50
Buffers
Buffers
10.8
Buffers
§  Resist changes in pH from the addition of acid or
base.
§  In the body, absorb H3O+ or OH¯ from foods and
cellular processes to maintain pH.
§  Are important in the proper functioning of cells and
blood.
§  In blood maintain a pH close to 7.4. A change in
the pH of the blood affects the uptake of oxygen
and cellular processes.
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52
Buffers
Components of a Buffer
When an acid or base
Is added
The components of a buffer solution
§  Are acid-base conjugate pairs.
§  Can be a weak acid and a salt of its conjugate
base.
§  Typically has equal concentrations of a weak
acid and its salt.
§  Can also be a weak base and a salt of its
conjugate acid.
§  To water, the pH
changes drastically.
§  To a buffer solution,
the pH does not
change very much;
pH is maintained.
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9
Learning Check
Solution
A. HCl + KCl
B. H2CO3 + NaHCO3
No; HCl is a strong acid.
Yes; This has a weak acid
and its salt
C. H3PO4 + NaCl
No; NaCl does not contain a
conjugate base of H3PO4.
D. CH3COOH + CH3COOK
Yes; This is a weak acid and
its salt
Does each combination make a buffer solution or not?
Explain.
A. HCl and KCl
B. H2CO3 and NaHCO3
C. H3PO4 and NaCl
D. CH3COOH and CH3COOK
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56
Function of the Weak Acid in a
Buffer
Buffer Action
An acetic acid/acetate buffer contains the weak acid
acetic acid (CH3COOH) and the salt of its conjugate base
sodium acetate (CH3COONa)
§  Acid dissociation
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
§  The function of the weak acid in a buffer is to
neutralize added base.
§  The acetate ion produced becomes part of the
available acetate.
CH3COOH + OH−
CH3COO− + H2O
acetic acid
base
acetate ion
water
§  The salt provides a higher concentration of the
conjugate base CH3COO- than provided by the
dissociation of the weak acid by itself.
CH3COONa(aq)
CH3COO-(aq) + Na+ (aq)
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58
Function of the Conjugate Base
Summary of Buffer Action
§  The function of the acetate ion CH3COO− is to
neutralize added H3O+.
Buffer action occurs because
§  The weak acid in a buffer neutralizes base.
§  The conjugate base in the buffer neutralizes
acid.
§  The pH of the solution is maintained.
§  The acetic acid produced contributes to the
available weak acid.
CH3COO− + H3O+
acetate ion
acid
CH3COOH + H2O
acetic acid
water
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10
pH of a Buffer
Calculating Buffer pH
§  The [H3O+] in the Ka expression is used to determine
the pH of a buffer.
The weak acid H2PO4- in a blood buffer H2PO4-/HPO42has Ka = 6.2 x 10-8. What is the pH of the buffer if it is
0.20 M in both H2PO4- and HPO42-?
Weak acid + H2O
Ka =
[H3
H3O+ + Conjugate base
O+][conjugate
base]
[H3O+] = Ka x [H2PO4-]
[HPO42-]
[H3O+] = 6.2 x 10-8 x [0.20 M] = 6.2 x 10-8
[0.20 M]
pH = -log [6.2 x 10-8] = 7.17
[weak acid]
[H3O+] = Ka x
[weak acid]
[conjugate base]
pH = -log [H3O+]
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Learning Check
62
Solution
What is the pH of a H2CO3 buffer that is 0.20 M H2CO3
and 0.10 M HCO3- ? Ka (H2CO3) = 4.3 x 10-7
What is the pH of a H2CO3 buffer that is 0.20 M
H2CO3 and 0.10 M HCO3-? Ka(H2CO3) = 4.3 x 10-7
[H3O+] = Ka x [H2CO3]
[HCO3-]
[H3O+] = 4.3 x 10-7 x [0.20 M] = 8.6 x 10-7 M
[0.10 M]
pH = -log [8.6 x 10-7] = 6.07
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