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Acids and BasesSection 18.1: Calculations involving Acids and Bases
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
H2O (l) H+ (aq) + OH- (aq)
The Ion Product of Water
Kc =[H+][OH-]
[H2O][H2O] = constant
Kc[H2O] = Kw = [H+][OH-]
15.2
2 H2O (l ) H3O+ (aq) + OH- (aq)
Kc =[H3O +][OH-]
[H2O]2 [H2O] = constant
Kc[H2O] 2 = Kw = [H+][OH-]
The Ion Product of Water
The ion-product constant (Kw) is
the product of the molar concentrations of H+ and OH- ions
at a particular temperature.
At 250CKw = [H+][OH-] = 1.0 x 10-14
when [H+] = [OH-]
when [H+] > [OH-]
when [H+] < [OH-]
Solution Is
neutral
acidic
basic
15.2
The Ion Product of Water
The ion-product constant (Kw) is the product of the molar concentrations of H+ and
OH- ions at a particular temperature. At 250CKw = [H+][OH-] = 1.0 x 10-14
15.2
Note: The autoionization of water is an endothermic process.
What will happen to the Kw if the temperature increases?
• equil shifts to the right – more H+ and OH- produced…• Kw increases… pH decreases• water is still neutral… but pH is slightly less than 7 at
higher temps • ex: @ 50°C [H+] = [OH-] = 3.05 x 10-7 ;pH is 6.5
heat + H2O (l) H+ (aq) + OH- (aq)
What is the concentration of OH- ions in a solution whose hydrogen ion concentration is 1.00 x 10-4 M?
15.2
pH – A Measure of Acidity
pH = -log [H+]
[H+] = 10-pH
pH [H+]
15.3
pOH = -log [OH-]
[OH-] = 10-pOH
pH
[OH-]
[H+] [OH-]
15.3
[H+][OH-] = Kw = 1.0 x 10-14
-log [H+] – log [OH-] = 14.00
pH + pOH = 14.00
What is the pH of a 2 x 10-3 M HNO3 solution?
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
15.4
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Start
End
0.002 M
0.002 M 0.002 M0.0 M
0.0 M 0.0 M
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.615.4
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Weak Acids (HA) and Acid Ionization Constants
HA (aq) H+ (aq) + A- (aq)
Ka =[H+][A-][HA]
Ka is the acid ionization constant
if Ka
strength of weak acid
15.5
if Ka is small, acid is weak
What is the pH of a 0.5 M HF solution (at 250C)?
Ka =
Eqn:
Initial (M)
Change (M)
Equilibrium (M)
0.50 0.00
-x +x
0.00
+x
15.5
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]
= 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50 0.00
-x +x
0.50 - x
0.00
+x
x x
Ka =x2
0.50 - x= 7.1 x 10-4
Ka x2
0.50= 7.1 x 10-4
0.50 – x 0.50Ka << 1
x2 = 3.55 x 10-4 x = 0.019 M
[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72
[HF] = 0.50 – x = 0.48 M15.5
When can I use the approximation?
0.50 – x 0.50Ka << 1
When x is less than 5% of the value from which it is subtracted.
x = 0.0190.019 M0.50 M
x 100% = 3.8%Less than 5%
Approximation ok.
What is the pH of a 0.05 M HF solution (at 250C)?
Ka x2
0.05= 7.1 x 10-4 x = 0.006 M
0.006 M0.05 M
x 100% = 12%More than 5%
Approximation not ok.
Must solve for x exactly using quadratic equation or method of successive approximation. IB WILL NOT REQUIRE THIS!!! 15.5
NOTES: Solving weak acid ionization problems:
1. Identify the major species that can affect the pH.
• assume you can ignore the autoionization of water – this works in most cases (pH < 6) ,
• ignore [OH-] because it is determined by [H+].
2. Use ICE to express the equilibrium concentrations in terms of single unknown x.
3. Write Ka in terms of equilibrium concentrations.
15.5
Solving weak acid ionization problems cont:
4. Solve for x by the approximation method.
• approximation assumes that the acid is quite weak… so not much dissociates… (<5%) [If approximation is not valid, solve for x exactly… NOT REQD FOR IB.]
5. Use the value for x to calculate concentrations of all species and/or pH of the solution.
15.5
What is the pH of a 0.122 M monoprotic acid whose Ka is 2.5 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
Ka =
Ka
If Ka << 1; do approximation
15.5
Your Turn: Practice Problem # 1
A 0.0100 M solution of a weak acid has a pH of 5.00. What is the dissociation constant of the acid?
Answer: 1.00 x 10-8 M
Ka is often expressed as pKa
pKa = -log Ka
Ka = 10-pKa
15.3
acid strength
pKa
Your Turn: Practice Problem # 2
Benzoic acid has a pKa of 4.2. What is the pH of a 0.100 M solution of this acid?
Answer: 2.6
Your Turn: Practice Problem # 3a
a. What concentration of hydrofluoric acid is required to give a solution of pH 2.00?
Answer: 0.148 M
percent ionization = Ionized acid concentration at equilibrium
Initial concentration of acidx 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0
x 100% [HA]0 = initial concentration
15.5
Your Turn: Practice Problem # 3b
b. What percentage of the hydrofluoric acid is dissociated at this pH, if the dissociation constant of the acid is 6.76 x 10-4 M?
Answer: 6.76 %
B (aq) + H2O (l) BH+ (aq) + OH- (aq)
Weak Bases and Base Ionization Constants
Kb =[BH+][OH-]
[B]
Kb is the base ionization constant
Kb
weak basestrength
15.6
Solve weak base problems like weak acids except solve for [OH-] instead of [H+].
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Weak Bases and Base Ionization Constants
Kb =[NH4
+][OH-][NH3]
Kb is the base ionization constant
Kb
weak basestrength
15.6
Solve weak base problems like weak acids except solve for [OH-] instead of [H+].
Kb is often expressed as pKb
pKb = -log Kb
Kb = 10-pKb
15.3
basestrength
pKb
Conjugate acid pKa pKb Conjugate base
Stronger acid
(but still weak)
Weaker base
H3PO4 2.1 11.9 H2PO4–
HF 3.3 10.7 F–
CH3COOH 4.8 9.2 CH3COO–
H2CO3 6.4 7.6 HCO3–
NH4+ 9.3 4.7 NH3
Weaker acid Stronger base
(but still weak)
Your Turn: Practice Problem # 4
What is the pH of a 0.0500 M solution of ethylamine (pKb = 3.40)?
Answer: 11.60
15.7
Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq)
A- (aq) + H2O (l) OH- (aq) + HA (aq)
Ka
Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka = Kw
Kb
Kb = Kw
Ka
pKa+ pKb = 14.00
Homework
Read Section 18.1 pp. 217-220
Do Ex 18.1 p 220-221
#1-5, 6a,6c,6e, 7, 8, 9, 10