acids and bases section 18.1: calculations involving acids and bases copyright © the mcgraw-hill...
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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.
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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-]
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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
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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)
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What is the concentration of OH- ions in a solution whose hydrogen ion concentration is 1.00 x 10-4 M?
15.2
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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-]
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15.3
[H+][OH-] = Kw = 1.0 x 10-14
-log [H+] – log [OH-] = 14.00
pH + pOH = 14.00
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Ka is often expressed as pKa
pKa = -log Ka
Ka = 10-pKa
15.3
acid strength
pKa
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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
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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
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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
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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 %
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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+].
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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+].
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Kb is often expressed as pKb
pKb = -log Kb
Kb = 10-pKb
15.3
basestrength
pKb
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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)
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Your Turn: Practice Problem # 4
What is the pH of a 0.0500 M solution of ethylamine (pKb = 3.40)?
Answer: 11.60
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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
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Homework
Read Section 18.1 pp. 217-220
Do Ex 18.1 p 220-221
#1-5, 6a,6c,6e, 7, 8, 9, 10