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04/23/23 Chapter 17 1
CH160 General Chemistry IILecture Presentation
Applications of Acid-Base Equilibria
Chapter 17
Sections 1-4
04/23/23 Chapter 17 2
Common Ion Effect• Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + A-
• What affect will adding salt NaA to the solution have on the acid ionization and solution pH?
NaA Na+ + A-
04/23/23 Chapter 17 3
Common Ion Effect• Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + A-
A- has 2 sources: HA and NaA. Adding
NaA increases [A-].
04/23/23 Chapter 17 4
Common Ion Effect• Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + A-
A- is a “common ion”
04/23/23 Chapter 17 5
Example 1(1a on the Example Problem Handout)
• Calculate the percent ionization of the acid and the pH of the solution that contains 0.500M HC2H3O2 (Ka = 1.8 x 10-5) and 0.200 M NaC2H3O2.
• (ans.: pH = 4.34, 0.0090%)
04/23/23 Chapter 17 6
Calculations Using Ka • Basic Steps for Weak Acid Calculations Using KBasic Steps for Weak Acid Calculations Using Kaa
– Write balanced chemical equation and the expression for Ka
• Look up value for Ka
– For each chemical species involved in the equilibrium (except H2O), write:
• Initial concentration• Equilibrium concentration
– Let the change in the [H3O+] be the variable “x”
– Substitute the equilibrium concentrations into Ka and solve for x using either
• quadratic approach• simplified approach
– Calculate pH, equilibrium concentrations, % ionization, etc., as specified in the problem.
04/23/23 Chapter 17 7
0.5 M CH3COOH
pH = 2.5
0.5 M CH3COOH +
0.200 M CH3COONa
pH = 4.3
0.6%
0.009%
+ CH3COONa
(problem 8a)
(problem 1a)
04/23/23 Chapter 17 8
Common Ion Effect• What affect does NaA have on weak acid
HA ionization?:HA + H2O <=> H3O+ + A-
Which way does equilibrium shift? What do the results of problems
8a and 1a tell us? Does this agree with LeChatelier’s principle?
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Common Ion Effect• What affect does NaA have on weak acid
HA ionization?:HA + H2O <=> H3O+ + A-
In presence of NaA, HA ionization shifts left.
“common-ion effect”
04/23/23 Chapter 17 11
Buffer Solutions• What is a buffer solution?
– solution with ability to resist pH changes upon addition of small amounts of either acid or base
• Requirements– must contain an acid to neutralize added OH-
ions – must contain a base to neutralize added H3O+
ions – acidic and basic species in buffer must not
neutralize each other.
04/23/23 Chapter 17 12
Buffer Action
• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of acid
A- + H+ HA
04/23/23 Chapter 17 13
Buffer Action
• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of acid (small amount)
A- + H+ HA
[A-] decreases slightly
[HA] increases slightly
Added acid is neutralized.
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Buffer Action
• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of base
HA + OH- H2O + A-
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Buffer Action
• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of base
HA + OH- H2O + A-
[HA] decreases slightly
[A-] increases slightly
Added base is neutralized.
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Buffer Action
• Two important properties of buffer solutions:– Buffer capacity
• Amount of acid or base the buffer can react with before giving a significant pH change (1 pH unit)
• Determined by how much buffer acid and base are used to make buffer
– pH• Determined by Ka and relative amounts of buffer
acid and base present
04/23/23 Chapter 17 17
Calculation of Buffer pH
• Calculating pH for buffer containing both weak acid HA and salt NaA. The major equilibrium is:– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
04/23/23 Chapter 17 18
Calculation of Buffer pH
• Calculating pH for buffer containing both weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
A- has 2 sources: HA and NaA
(This seems familiar! Didn’t we
just do this?)
04/23/23 Chapter 17 19
Calculation of Buffer pH
• Calculating pH for buffer containing both weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
Neither CHA nor CNaA change much since only a very small amount of HA ionizes.
04/23/23 Chapter 17 20
Calculation of Buffer pH
• Calculating pH for buffer containing both weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
• Since % ionization is small: CA- [A-] and CHA [HA]
04/23/23 Chapter 17 21
Calculation of Buffer pH• Calculating pH for buffer containing both
weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
• Since % ionization is small: CA- [A-] and CHA [HA]
Solving for [H3O+] gives: [H3O+] = KaCHA/CA-
pH = -log[H3O+](This is just a common ion effect problem.)
04/23/23 Chapter 17 22
Calculation of Buffer pH
• We can also take the –log of our equation:
[H3O+] = KaCHA/CA-
-log[H3O+] = -log(KaCHA/CA-)
-log[H3O+] = -logKa - log CHA/CA-
pH = pKa - log CHA/CA-
04/23/23 Chapter 17 23
Example 2(2a on Example Problem Handout)
• Calculate the pH of a buffer solution that contains 0.25 M sodium acetate, NaC2H3O2, and 0.35 M acetic acid, HC2H3O2 (Ka = 1.8 x 10-5).
• (ans.: pH = 4.60)
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Calculating pH Changes in Buffers
• How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?
H3O+ + A- HA + H2O
or
OH- + HA A- + H2O
HA/NaA+ H3O+ or OH-
Buffer
Neutralization Rxn
Calculate new [HA] & [A-]
Calculate new [H3O+] from [H3O+]
= KaCHA/CA- pH
04/23/23 Chapter 17 25
Calculating pH Changes in Buffers
• How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?
H3O+ + A- HA + H2O
or
OH- + HA A- + H2O
HA/NaA+ H3O+ or OH-
Buffer
Neutralization Rxn
Calculate new [HA] & [A-]
This much is a stoichiometry problem. (Oh, oh! General
Chemistry I stuff here.).
04/23/23 Chapter 17 26
Calculating pH Changes in Buffers
• How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?
Calculate new [H3O+] from [H3O+]
= KaCHA/CA- pH
This part is an equilibrium calculation.
04/23/23 Chapter 17 27
Example 3(3a on Example Problem Handout)
• Calculate the pH of the solution formed and the change in pH observed when (a) 0.050 moles of HCl and (b) 0.025 moles of NaOH are added to 500 mL of the buffer in example (2a). (c) Calculate the change in pH that occurs when 0.050 moles HCl are added to 500 mL H2O.
(ans.: (a) pH = 4.27, pH = -0.33, (b) pH = 4.74, pH = +0.14, (c) pH = -6)
04/23/23 Chapter 17 28
Buffer Preparation
• What if I need to make a buffer solution of known pH? Select:– Buffer system.
• Often pKa of buffer acid is close to desired pH.
– Relative amounts of buffer acid and base.• Buffer capacity increases with concentrations.• Buffer effectiveness best with concentrations on
same order of magnitude.
04/23/23 Chapter 17 29
Example 4 (4 on Example Problem Handout)
• Starting with 1.0L of 0.100 M CH3COOH (Ka = 1.8 x 10-5), how many grams of sodium acetate, CH3COONa (FW = 82.034 g/mol), to give a buffer with a pH of 4.40? (Assume no volume change.)
• (ans.: 3.7 g)
04/23/23 Chapter 17 30
Quantitative Acid-Base Chemistry• How do we calculate the pH of a solution
formed by mixing an acid solution with a base solution?
• Consider addition of 0.1 M strong base, MOH, to 0.1 M strong acid, HX– pH changes can be observed from titration curve
• pH vs. volume standard
04/23/23 Chapter 17 31
Titration of 25 mL of 0.1 M HCl with 0.1 M NaOH
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50 55volume 0.1 M NaOH
pH
Rxn is complete = equivalence point. pH = 7.0 since only neutral NaX present.
Strong Acid-Strong Base Titration
HX + NaOH NaX + H2O
pH
mL NaOH
04/23/23 Chapter 17 32
Calculating pH in Acid-Base Reactions
• How do we calculate pH after adding a strong base to a strong acid?
HX + MOH MX + H2OHX+ MOH
Strong acid
Neutralization Rxn
Calculate [HX] or [MOH] left (ignore
neutral MX)
Calculate new [H3O+] from [HX] or [MOH]
pH
04/23/23 Chapter 17 33
Calculating pH in Acid-Base Reactions
• How do we calculate pH after adding a strong base to a strong acid? Considerations:
• Stoichiometry– Limiting reagent
• At what point in rxn does calculation take place?– Initial– Pre-equivalence– Equivalence– Post-equivalence
• Does dilution occur?– If mixing 2 solutions: Vtotal = V1 + V2
04/23/23 Chapter 17 34
Titration of 25 mL of 0.1 M HCl with 0.1 M NaOH
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50 55volume 0.1 M NaOH
pH
-All HX & MOH consumed
-NaX left (neutral)
-pH = 7.0
Strong Acid-Strong Base TitrationHX + NaOH NaX + H2O
-HX in excess
-HX/NaX left
-[H3O+] = CHX
-NaOH in excess
-NaOH/NaX left
-[OH-] = CMOH
Only HX
[H3O+] = CHX
pH
mL NaOH
04/23/23 Chapter 17 35
Example 5(5 of Example Problem Handout)
• Calculate the pH for a solution prepared by mixing 25.00 mL of 0.100 M HCl(aq) with a) 10.00 mL b) 25.00 mL and c) 35.00 mL of 0.100 M NaOH(aq). (ans.: a) 1.37, b) 7.0, c) 12.22)
04/23/23 Chapter 17 38
Titration of 25 mL of 0.1 M HOAc with 0.1 M NaOH
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0 5 10 15 20 25 30 35 40 45 50volume 0.1 M NaOH
pH
Weak Acid-Strong Base TitrationHA + NaOH NaA + H2O
Rxn is complete = equivalence point.
pH > 7.0 since only basic NaA present.
pH
mL NaOH
04/23/23 Chapter 17 39
Titration of 25 mL of 0.1 M HOAc with 0.1 M NaOH
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0 5 10 15 20 25 30 35 40 45 50volume 0.1 M NaOH
pHWeak Acid-Strong Base Titration
HA + NaOH NaA + H2O
-HA in excess
-HA/NaA left = buffer
-[H3O+] = Ka(CHA/CNaA)-All HA & MOH consumed
-NaA left (weak base)
-[OH-] (KbCNaA)1/2
-pH > 7.0
-NaOH in excess
-NaOH/NaA left
-[OH-] CMOH
-only HA
[H3O+] (KaCHA)1/2
pH
04/23/23 Chapter 17 40
Titration of 25 mL of 0.1 M ammonia with 0.1 M HCl
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0 5 10 15 20 25 30 35 40 45 50volume 0.1 M HCl
pH
Rxn is complete = equivalence point. pH < 7.0 since only acidic HB+ present.
Strong Acid-Weak Base TitrationHX + B HB+ + X-
pH