1. principles and modern applications
Chemistry 140 Fall 2002
ELEVENTH EDITION
GENERAL CHEMISTRY
principles and modern applications
PETRUCCI HERRING MADURA BISSONNETTE 17
Additional Aspects of Acid-Base Equilibria
PHILIP DUTTON
UNIVERSITY OF WINDSOR
DEPARTMENT OF CHEMISTRY AND BIOCHEMISTRY
General Chemistry: Chapter 17
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2. Additional Aspects of Acid-Base Equilibria
Chemistry 140 Fall 2002
Additional Aspects of Acid-Base Equilibria
CONTENTS
17-1
Common-Ion Effect in Acid-Base Equilibria
17-2
Buffer Solutions
17-3
Acid-Base Indicators
17-4
Neutralization Reactions and Titration Curves
17-5
Solutions of Salts of Polyprotic Acids
17-6
Acid-Base Equilibrium Calculations: A Summary
General Chemistry: Chapter 17
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3. 17-1 The Common-Ion Effect in Acid-Base Equilibria
The Common-Ion Effect describes the effect on an equilibrium by a second substance that furnishes ions that can participate in that equilibrium. The added ions are said to be common to the equilibrium.
General Chemistry: Chapter 17
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4. Solutions of Weak Acids and Strong Acids
Chemistry 140 Fall 2002
Solutions of Weak Acids and Strong Acids
Consider a solution that contains both M CH3CO2H and M HCl.
CH3CO2H + H2O CH3CO2− + H3O+
(17.1)
equilibrium
concentrations
(1−α)co αco αco
HCl H2O Cl− H3O+
(17.2)
0.100 M M
General Chemistry: Chapter 17
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5. A weak acid-strong acid mixture
for HCl, Ka ≈ 106
FIGURE 17-1
A weak acid-strong acid mixture
General Chemistry: Chapter 17
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6. General Chemistry: Chapter 17
CH3CO2H + H2O CH3CO2− + H3O+
Initial concs:
0.100 M 0 M M
Changes:
−x M +x M +x M
Equilibrium:
(0.100 −x) M +x M ( x) M
[CH3COO−] = 1.8×10−5 M and [CH3COOH] = M
x is only 0.018% of M so the assumption that x is small is valid.
General Chemistry: Chapter 17
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7. General Chemistry: Chapter 17
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8. General Chemistry: Chapter 17
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9. A mixture of a weak acid and its salt
Chemistry 140 Fall 2002
Solutions of Weak Acids and Their Salts
0.100 M CH3CO2
0.100 M CH3CO2H +
0.100 M CH3CO2Na
FIGURE 17-2
A mixture of a weak acid and its salt
General Chemistry: Chapter 17
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10. General Chemistry: Chapter 17
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11. A mixture of a weak base and its salt
Chemistry 140 Fall 2002
Solutions of Weak Bases and Their Salts
0.100 M NH3 +
0.100 M NH4Cl
0.100 M NH3
FIGURE 17-3
A mixture of a weak base and its salt
General Chemistry: Chapter 17
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12. General Chemistry: Chapter 17
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13. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
17-2 Buffer Solutions
Two component systems that change pH only slightly on addition of acid or base.
The two components must not neutralize each other but must neutralize strong acids and bases.
Recognizing a Buffer Solution
appreciable amounts of both a weak acid (HA) and it’s conjugate base (A−), or
appreciable amounts of both a weak base (B) and it’s conjugate acid (BH+)
General Chemistry: Chapter 17
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14. General Chemistry: Chapter 17
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15. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
Let [CH3CO2H] = [CH3CO2−] in a solution.
[H3O+] [CH3CO2−]
Ka= =
1.8×10−5
[C3CO2H]
[C3CO2H]
[H3O+] =
× Ka
= 1.8×10−5
(17.6)
[CH3CO2−]
pH = −log[H3O+] = −logKa = −log(1.8×10−5) = 4.74
General Chemistry: Chapter 17
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16. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
FIGURE 17-4
How A Buffer Works
General Chemistry: Chapter 17
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17. Calculating the pH of a Buffer Solution
CH3CO2H + H2O H3O+ + CH3CO2H−
Initial concs:
0.550 M − M
Changes:
−x M +x M +x M
Equilibrium:
(0.550 −x) M +x M ( x) M
x is only 0.003% of M, so the assumption that x is small is valid.
General Chemistry: Chapter 17
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18. An Equation for Buffer Solutions: The Henderson-Hasselbalch Equation
Chemistry 140 Fall 2002
An Equation for Buffer Solutions: The Henderson-Hasselbalch Equation
A variation of the ionization constant expression. Consider a hypothetical weak acid, HA, and its salt NaA:
[H3O+] [A−]
HA + H2O H3O+ + A−
Ka=
[HA]
[A−]
[A−]
Ka=
[H3O+] ×
−logKa =
−log[H3O+] − log
[HA]
[HA]
General Chemistry: Chapter 17
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19. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
[A−]
−logKa=
−log[H3O+] − log
[HA]
pH − log
[HA]
pKa =
[A−]
pKa + log
[HA]
pH =
[A−]
pKa + log
[acid]
pH =
[conjugate base]
(17.7)
General Chemistry: Chapter 17
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20. Useful when you initial concentrations of acid and salt.
Chemistry 140 Fall 2002
Useful when you initial concentrations of acid and salt.
Avoids the ICE table.
[conjugate base]initial
pH =
pKa + log
[acid] initial
A reasonable approach to ensure validity is that
[A−] 1.
0.1 <
< 10
(17.8)
[HA] 2.
[A−] > 100×Ka and [HA] > 100×Ka
General Chemistry: Chapter 17
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21. Preparing Buffer Solutions
If you want pH 5.09, you could choose to have 1:1 buffer ratio
pH = log 1 = 5.09
But, if you want a buffer at pH 5.09 finding an acid with the exact pKa is not practical.
One procedure to follow in making a buffer solution with a desired pH
General Chemistry: Chapter 17
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22. Six methods for preparing buffer solutions
FIGURE17-5
Six methods for preparing buffer solutions
General Chemistry: Chapter 17
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23. General Chemistry: Chapter 17
Calculating Changes in Buffer Solutions
FIGURE 17-6
Calculation of the new pH of a buffer after strong acid or base is added
General Chemistry: Chapter 17
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24. Buffer Capacity and Buffer Range
Buffer capacity refers to the amount of acid or base that a buffer can neutralize before its pH changes appreciably.
Maximum buffer capacity exists when [HA] and [A−] are large and approximately equal to each other.
Buffer range is the pH range over which a buffer effectively neutralizes added acids and bases.
Practically, the range of 2 pH units is the maximum range to which a buffer solution should be exposed.
General Chemistry: Chapter 17
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25. Applications of Buffer Solutions
Buffering in blood. Protein studies require buffers to maintain protein structure. Industrial processes, such as brewing, require specific pH.
General Chemistry: Chapter 17
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26. 17-3 Acid-Base Indicators
The color of acid-base indicators depends on the pH.
HIn + H2O In− + H3O+
Acid color
Base color
pKa + log
[InH]
pH =
[In−]
(17.9)
Bromthymol blue pKHIn = 7.1
pH < 6.1 (yellow)
pH ≈ 7.1 (green)
pH > 8.1 (blue)
General Chemistry: Chapter 17
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27. General Chemistry: Chapter 17
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28. pH and color changes for some common acid-base indicators
Chemistry 140 Fall 2002
FIGURE 17-8
pH and color changes for some common acid-base indicators
General Chemistry: Chapter 17
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29. General Chemistry: Chapter 17
Testing swimming pool water for its chlorine content and pH.
General Chemistry: Chapter 17
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30. 17-4 Neutralization Reactions and Titration Curves
Chemistry 140 Fall 2002
17-4 Neutralization Reactions and Titration Curves
The equivalence point of a neutralization reaction is the point at which both acid and base have been consumed and neither is present in excess. The end point of a titration is the point at which the indicator changes color. The titrant is a solution of known concentration added to a solution of unknown concentration (titrand) during a titration. A titration curve is a graph of pH vs. volume of titrant added.
General Chemistry: Chapter 17
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31. General Chemistry: Chapter 17
The Millimole
In a typical titration
Volume of titrant added is less than 50 mL.
Concentration of titrant is less than 1 mol/L.
Titration uses less than 1/1000 mole of acid and base.
L/1000
mol/1000 =
M = L
mol mL
mmol
General Chemistry: Chapter 17
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32. Titration of a Strong Acid with a Strong Base
General Chemistry: Chapter 17
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33. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
FIGURE 17-8
Titration curve for the titration of a strong acid with a strong base – 25 mL of M HCl with M NaOH
General Chemistry: Chapter 17
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34. Titration curve for the titration of a strong base with a strong acid
FIGURE 17-9
Titration curve for the titration of a strong base with a strong acid
General Chemistry: Chapter 17
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35. Titration of a Weak Acid with a Strong Base
for neutralization of a weak acid, there is direct proton transfer from the weak acid
CH3CO2H + OH− CH3CO2H + H2O
Kneutralization = Ka/Kw = 1.8 ×10+9 >>1
Because the equilibrium constant for the neutralization of CH3CO2H by OH− is extremely large, we can justifiably say that the neutralization reaction essentially goes to completion.
CH3CO2H + OH− CH3CO2H− + H2O
General Chemistry: Chapter 17
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36. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
for neutralization of a strong acid, the protons are transferred from H3O +
H3O + + OH− H2O
Kneutralization = 1/Kw = 1.0 ×10+14 >>1
For equal volumes of acid solutions of the same molarity, the volume of base required to titrate to the equivalence point is independent of the strength of the acid.
General Chemistry: Chapter 17
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37. General Chemistry: Chapter 17
Consider adding mol of CH3COOH and 0.03 mol of NaOH to water to make 1.0 L of solution
1. A stoichiometric calculation based on the neutralization reaction.
CH3COOH + OH− CH3COO− + H2O
Initial conc: mol mol 0 mol
Changes: −0.030 mol −0.030 mol mol
Equilibrium: mol 0 mol mol
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38. General Chemistry: Chapter 17
2. An equilibrium calculation involving the species produced by and left behind by the neutralization reaction.
CH3COOH + H2O CH3COO− + H3O+
Initial conc: M M 0 M
Changes: −x M + x M + x M
Equilibrium: (0.070 −x) M ( x) M + x M
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39. General Chemistry: Chapter 17
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40. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
FIGURE 17-10
Titration curve for the titration of a weak acid with a strong base – 25 mL of M CH3COOH with M NaOH
General Chemistry: Chapter 17
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41. Constructing the titration curve for a weak acid with a strong base
Chemistry 140 Fall 2002
FIGURE17-11
Constructing the titration curve for a weak acid with a strong base
General Chemistry: Chapter 17
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42. Titration of a Weak Polyprotic Acid
Chemistry 140 Fall 2002
Titration of a Weak Polyprotic Acid
H3PO NaH2PO Na2HPO Na3PO4
NaOH
NaOH
NaOH
Major species at
first
equivalence point
Major species at second
equivalence point
At the first equivalence point
H2PO4− + H2O H3O+ + HPO42−
Ka2 = 6.3 ×10−8 Kw
H2PO42− + H2O H3PO4 + HO−
Kb3 = = 1.4 ×10−12
Ka1
the solution is acidic
General Chemistry: Chapter 17
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43. the solution is basic H3PO4 NaH2PO4 Na2HPO4 Na3PO4
Chemistry 140 Fall 2002
H3PO NaH2PO Na2HPO Na3PO4
NaOH
NaOH
NaOH
Major species at
first
equivalence point
Major species at second
equivalence point
At the second equivalence point
HPO42− + H2O H3O+ + PO43−
Ka3 = 4.2 ×10−13 Kw
HPO42− + H2O H2PO4− + OH−
Kb2 = = 1.6 ×10−7
Ka2
the solution is basic
General Chemistry: Chapter 17
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44. the solution is strongly basic
Chemistry 140 Fall 2002
H3PO NaH2PO Na2HPO Na3PO4
NaOH
NaOH
NaOH
Major species at
first
equivalence point
Major species at second
equivalence point
The third equivalence point is not visible in the titration curve since the pH of the strongly hydrolyzed Na3PO4− approaching pH 13 − is higher than can be reached by adding 0.100M NaOH Kw
PO43− + H2O HPO42− + OH−
Kb1 = = 2.4 ×10−2
Ka3
the solution is strongly basic
General Chemistry: Chapter 17
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45. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
FIGURE 17-12
Titration of a weak polyprotic acid – 10.0 mL of M H3PO4 with M NaOH
General Chemistry: Chapter 17
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46. 17-5 Solutions of Salts of Polyprotic Acids
Chemistry 140 Fall 2002
17-5 Solutions of Salts of Polyprotic Acids
The third equivalence point of phosphoric acid can only be reached in a strongly basic solution. The pH of this third equivalence point is not difficult to calculate. It corresponds to that of Na3PO4 (aq) and PO43− can ionize only as a base as we have already seen. Kw
PO43− + H2O HPO42− + OH−
Kb1 = = 2.4 ×10−2
Ka3
General Chemistry: Chapter 17
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47. General Chemistry: Chapter 17
What is the pH of M Na3PO4(aq)
PO43− + H2O HPO42− + OH−
Initial conc: M 0 M 0 M
Changes: −x M + x M + x M
Equilibrium: (0.025 −x) M + x M + x M
x = [OH−] = M
pOH = −log[OH−] = log(0.015) = 1.82
pH = pKw − pOH = 14.00−1.82 = 12.18
General Chemistry: Chapter 17
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48. General Chemistry: Chapter 17
Chemistry 140 Fall 2002
NaH2PO4 and Na2HPO4 calculations require that two equilibria be considered simultaneously. The pH values prove to be independent of solution at reasonable concentrations (> 0.1 M).
for H2PO4−
pH = 0.5 (pKa1 + pKa2) = 0.5 ( ) = 4.68
(17.10)
for HPO42−
pH = 0.5 (pKa2 + pKa3) = 0.5 ( ) = 9.79
(17.11)
Exercise
17-1 ARE YOU WONDERING? works through this derivation.
General Chemistry: Chapter 17
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49. 17-6 Acid-Base Equilibrium Calculations: A Summary
Which species are potentially present in solution, and how large their concentrations are likely to be?
Are reactions possible among any of the solution components; if so, what is their stoichiometry.
Which equilibrium equations apply to the particular situation? Which are the most significant?
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50. General Chemistry: Chapter 17
End of Chapter
General Chemistry: Chapter 17
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