1. CH160 General Chemistry II Lecture Presentation Applications of Acid-Base Equilibria
Chapter 17
Sections 1-4
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2. HA + H2O <=> H3O+ + A-
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-
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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-].
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4. HA + H2O <=> H3O+ + A-
Common Ion Effect
Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + A-
A- is a “common ion”
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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 M NaC2H3O2.
(ans.: pH = 4.34, %)
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6. Calculations Using Ka Basic Steps for Weak Acid Calculations Using Ka
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.
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7. (problem 1a) (problem 8a) + CH3COONa 0.5 M CH3COOH + 0.200 M CH3COONa
pH = 4.3
(problem 1a)
0.5 M CH3COOH
(problem 8a)
pH = 2.5
+ CH3COONa
0.6%
0.009%
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8. HA + H2O <=> H3O+ + A-
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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9. 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”
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10. Buffer Solutions What is a buffer solution? Requirements
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.
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11. Buffer Action
How do buffers work? Consider buffer with weak acid HA and salt NaA:
Addition of acid
A- + H+  HA
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12. 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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13. 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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14. 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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15. 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
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16. 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]
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17. 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?)
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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]
Neither CHA nor CNaA change much since only a very small amount of HA ionizes.
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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]
Since % ionization is small: CA-  [A-] and CHA  [HA]
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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]
Solving for [H3O+] gives:
[H3O+] = KaCHA/CA-
pH = -log[H3O+]
(This is just a common ion effect problem.)
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21. 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-
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22. 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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23. Calculating pH Changes in Buffers
How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?
HA/NaA
+ H3O+ or OH-
H3O+ + A-  HA + H2O or
OH- + HA  A- + H2O
Buffer
Neutralization Rxn
Calculate new [H3O+] from [H3O+] = KaCHA/CA-
Calculate new [HA] & [A-] pH
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24. Calculating pH Changes in Buffers
How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?
HA/NaA
+ H3O+ or OH-
H3O+ + A-  HA + H2O or
OH- + HA  A- + H2O
Buffer
Neutralization Rxn
This much is a stoichiometry problem. (Oh, oh! General Chemistry I stuff here.).
Calculate new [HA] & [A-]
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25. Calculating pH Changes in Buffers
How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?
This part is an equilibrium calculation.
Calculate new [H3O+] from [H3O+] = KaCHA/CA- pH
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26. Example 3 (3a on Example Problem Handout)
Calculate the pH of the solution formed and the change in pH observed when (a) moles of HCl and (b) moles of NaOH are added to 500 mL of the buffer in example (2a). (c) Calculate the change in pH that occurs when 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)
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27. 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.
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28. Example 4 (4 on Example Problem Handout)
Starting with 1.0L of M CH3COOH (Ka = 1.8 x 10-5), how many grams of sodium acetate, CH3COONa (FW = g/mol), to give a buffer with a pH of 4.40? (Assume no volume change.)
(ans.: 3.7 g)
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29. 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
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30. Strong Acid-Strong Base Titration
HX + NaOH  NaX + H2O pH
Rxn is complete = equivalence point. pH = 7.0 since only neutral NaX present.
mL NaOH
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31. Calculating pH in Acid-Base Reactions
How do we calculate pH after adding a strong base to a strong acid? HX
+ MOH
HX + MOH  MX + H2O
Strong acid
Neutralization Rxn
Calculate new [H3O+] from [HX] or [MOH]
Calculate [HX] or [MOH] left (ignore neutral MX) pH
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32. 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
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33. Strong Acid-Strong Base Titration
HX + NaOH  NaX + H2O
-NaOH in excess
-NaOH/NaX left
-[OH-] = CMOH
-All HX & MOH consumed
-NaX left (neutral)
-pH = 7.0 pH
-HX in excess
-HX/NaX left
-[H3O+] = CHX
Only HX
[H3O+] = CHX
mL NaOH
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34. Example 5 (5 of Example Problem Handout)
Calculate the pH for a solution prepared by mixing mL of M HCl(aq) with a) mL b) mL and c) mL of M NaOH(aq). (ans.: a) 1.37, b) 7.0, c) 12.22)
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35. Weak Acid-Strong Base Titration
HA + NaOH  NaA + H2O
Rxn is complete = equivalence point.
pH > 7.0 since only basic NaA present. pH
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mL NaOH

36. Weak Acid-Strong Base Titration
HA + NaOH  NaA + H2O
-NaOH in excess
-NaOH/NaA left
-[OH-]  CMOH
-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 pH
-only HA
[H3O+]  (KaCHA)1/2
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37. Strong Acid-Weak Base Titration
HX + B  HB+ + X- pH
Rxn is complete = equivalence point. pH < 7.0 since only acidic HB+ present.
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