1. Chapter 17 Additional Aspects of Aqueous Equilibria
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 17 Additional Aspects of Aqueous Equilibria
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. Aim: What is the common ion effect, and how is it shown?
Do Now: In a reaction, A + B  C + D, if more of product D is added to the reaction, what happens to equilibrium?
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3. The Common-Ion Effect Consider a solution of acetic acid:
If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO−(aq)
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4. The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”
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5. The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Ka for HF is 6.8  10−4.
[H3O+] [F−]
[HF]
Ka =
= 6.8  10-4
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6. The Common-Ion Effect HF(aq) + H2O(l) H3O+(aq) + F−(aq)
Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M
[H3O+], M
[F−], M
Initially
0.20
0.10
Change −x +x
At Equilibrium
0.20 − x  0.20
x  0.10 x
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7. The Common-Ion Effect (0.10) (x) 6.8  10−4 = (0.20)
(0.20) (6.8  10−4)
(0.10)
= x
1.4  10−3 = x
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8. The Common-Ion Effect Therefore, [F−] = x = 1.4  10−3
[H3O+] = x =  10−3 = 0.10 M
So, pH = −log (0.10)
pH = 1.00
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9. Sample Exercise 17.1 (Pg. 726)
What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0L solution? The Ka of acetic acid is 1.8 x What if only .13 mol of sodium acetate was added?
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10. Aim: What is a buffer and how are they made?
Do Now: What is the common ion if a solution of Nitrous Acid and Potassium Nitrite were mixed? How would this common ion effect pH?
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11. Buffers Buffers are solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.
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12. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
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13. Buffers
Similarly, if acid is added, the F− reacts with it to form HF and water.
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14. Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
HA + H2O
H3O+ + A−
[H3O+] [A−]
[HA]
Ka =
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15. Buffer Calculations Rearranging slightly, this becomes
[HA]
Ka =
[H3O+]
Taking the negative log of both side, we get
base
[A−]
[HA]
−log Ka =
−log [H3O+] + −log
pKa pH
acid
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16. Buffer Calculations So pKa = pH − log [base] [acid]
Rearranging, this becomes
pH = pKa + log
[base]
[acid]
This is the Henderson–Hasselbalch equation.
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17. Henderson–Hasselbalch Equation Sample Exercise 17.3 (Pg. 731)
What is the pH of a buffer that is M in lactic acid, CH3CH(OH)COOH, and M in sodium lactate? Ka for lactic acid is 1.40  10−4.
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18. Henderson–Hasselbalch Equation
pH = pKa + log
[base]
[acid]
pH = −log (1.4  10−4) + log
(0.10)
(0.12)
pH = (−0.08) pH pH
pH = 3.77
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19. Preparing a Buffer Sample Exercise 17.5 (Pg. 733)
How many moles of NH4Cl must be added to 2.0L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution) Kb of NH3 is 1.8 x10-5.
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20. pH Range
The pH range is the range of pH values over which a buffer system works effectively.
It is best to choose an acid with a pKa close to the desired pH.
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21. Give it Some Thought (Pg. 735)
The Ka values for Nitrous Acid(HNO2) and Hypochlorous Acid(HClO) are 4.5 x 10-4 and 3.0 x 10-8 respectively. Which one would be more suitable for use in a solution buffered at pH = 7.0. What other substances would be needed to make the buffer?
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22. Aim: What is a titration, and how can we calculate pH at different points?
Do Now: What is the general reaction if a strong acid is mixed with a strong base? What could be the reason for performing this reaction?
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23. Titration
In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).
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24. Titration
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.
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25. Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly.
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26. Titration of a Strong Acid with a Strong Base
Just before (and after) the equivalence point, the pH increases rapidly.
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27. Titration of a Strong Acid with a Strong Base
At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
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28. Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off.
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29. Simulation
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30. Sample Exercise 17.7 (Pg. 739) Calculations for Strong Acid Base Titrations
Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of M NaOH solution have been added to 50.0 mL of M HCl solution?
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31. Practice
Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of M HNO3 have been added to 25.0mL of M KOH solution.
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32. Aim: How is the titration of a weak acid with a strong base, different from that of a strong acid, and what factors affect solubility?
Do Now: If equal moles of weak acid is added to a strong base, does a full neutralization reaction take place? (is pH 7)
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33. Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
At the equivalence point the pH is >7.
Phenolphthalein is commonly used as an indicator in these titrations.
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34. Titration of a Weak Acid with a Strong Base
At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.
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35. Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
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36. Sample Exercise 17.8 ( Pg. 742)
Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH3COOH. The Ka of CH3COOH = 1.8 x Look at exercise 17.9 at home.
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37. Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is < 7.
Methyl red is the indicator of choice.
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38. Titrations of Polyprotic Acids
When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.
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39. Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
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40. Solubility Products
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the solubility product.
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41. Writing Ksp expressions.
What are the correct expressions for:
CaF2 (s)
Ag3PO4 (s)
NH4ClO (s)
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42. Solubility Products Ksp is not the same as solubility.
Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
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43. Calculating Ksp from Solubility Sample Exercise 17.11 (Pg. 749)
Solid silver chromate is added to pure water at 25 degrees Celsius, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4 (s) and the solution. Analysis of the equilibrium solution shows that its silver ion concentration is 1.3 x 10-4 M. Assuming that the Ag2CrO4 solution is saturated and that there are no other important equilibria involving Ag+ or CrO4 2- ions in the solution, calculate the Ksp for this compound.
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44. Factors Affecting Solubility
The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
BaSO4(s)
Ba2+(aq) + SO42−(aq)
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45. Factors Affecting SolubilitypH
If a substance has a basic anion, it will be more soluble in an acidic solution.
Substances with acidic cations are more soluble in basic solutions.
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46. Factors Affecting Solubility
Amphoterism
Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
Examples of such cations are Al3+, Zn2+, and Sn2+.
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47. Will a Precipitate Form?
In a solution,
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.
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