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. The Common-Ion Effect
Calculate the pH of a 0.30 M solution of acetic acid, given Ka = 1.8 x 10
pH = 2.63
Calculate the pH of a solution that is 0.30 M acetic acid and 0.30 M sodium acetate
pH = 4.74
Sample 17.1
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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. 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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10. 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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11. Buffers
Similarly, if acid is added, the F− reacts with it to form HF and water.
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12. 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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13. 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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14. 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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15. Derivation of Henderson-Hasselbach Equation
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16. Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4.
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17. 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
Do Practice 17.3 (ds)
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18. Buffer Capacity
The amount of acid or base the buffer can neutralize before pH changes appreciably
The greater the amount of conjugate acid-base pair, the greater the buffering capacity
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19. 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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20. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.
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21. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use the Henderson–Hasselbalch equation to determine the new pH of the solution.
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22. Calculating pH Changes in Buffers
A buffer is made by adding mol HC2H3O2 and mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.
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23. Calculating pH Changes in Buffers
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8  10−5) = 4.74
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24. Calculating pH Changes in Buffers
The mol NaOH will react with mol of the acetic acid:
HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l)
HC2H3O2
C2H3O2−
OH−
Before reaction
0.300 mol
0.020 mol
After reaction
0.280 mol
0.320 mol
0.000 mol
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25. Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = log
(0.320)
(0.280) pH
pH = 4.80
pH =
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26. 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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27. Acid-Base Titration(ds)
Can be used to determine….
Unknown concentration of an acid or base
Ka or Kb
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28. Sample Problem
10.0 mL of a monoprotic acid of unknown concentration was placed in a beaker and diluted to 100 mL.
The acid was titrated with M NaOH
The equivalence point was determined to occur at mL NaOH
What was the concentration of the acid?
2.94 M
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29. 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.
Phenolpthalein and methyl red are common indicators
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30. 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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31. Titration of a Strong Acid with a Strong Base
Just before (and after) the equivalence point, the pH increases rapidly.
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32. 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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33. 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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34. 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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35. 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.
Sample 17.7
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36. Calculating Equivalence Point of Weak Acid (ds)
The equivalence point is > 7 because the conj. base increases [OH-] by hydrolysis
Stoichiometry calculation
determine concentration of conj. base at eq. point.
Consider moles CB formed and total volume of acid and base
Equilibrium calculation:
use Kb to determine equilibrium
determine pOH
determine pH
Sample 17.8
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37. Finding the Equivalence Point of a Weak Acid Titration
Determine the pH at the equivalence point when 50.0 mL of M CH3COOH is titrated with M NaOH.
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40. More Practice
Find pH at eq. pt. when 40 mL M benzoic acid (Ka = 6.3 x 10-5) is titrated with M NaOH
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41. Calculating pH at the Equivalence Point
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42. Calculating pH at the Equivalence Point
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43. 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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44. 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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45. 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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46. Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
This is a heterogeneous system. Not all species are in the same phase.
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47. 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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48. 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).
Review exercise 17.10, 17.11
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49. Classifying Solubility
Soluble ≥ 1 g/100mL
Slightly soluble 0.1 – 1.0 g/100 mL
Insoluble < 0.1 g/100 mL
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50. Factors Affecting Solubility
Temperature
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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51. Factors Affecting SolubilitypH
If a substance has a basic anion it will be more soluble in an acidic solution.
CB of a WA
Substances with acidic cations are more soluble in basic solutions.
CA of a WB
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52. Solubility & pH (ds) Mg(OH)2 (s) ↔ Mg2+ (aq) + 2 OH- (aq)
Ksp = [Mg2+][OH-]2 = 1.8 x 10-11
Adding acid will remove OH- ions, causing more dissolution of Mg(OH)2
See p. 743
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53. Factors Affecting Solubility
Complex Ions
Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.
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54. Factors Affecting Solubility
Complex Ions
The formation of these complex ions increases the solubility of these salts.
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55. AgCl (s) ↔ Ag+ (aq) + Cl- (aq) (Ksp= 1.8 x 10-10)
Ag+ (aq) + 2 NH3 (aq) ↔ Ag(NH3)2+
Silver chloride dissolves in ammonia solution because of the ability of Ag+ to act as a Lewis acid, forming a complex ion with NH3 molecules (Lewis base).
Nonbonding electrons of NH3 occupy vacant orbitals of the Ag+ ion.
The Lewis base is known as a ligand.
Formation constant quantifies this equilibrium
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56. Nomenclature of Coordination Compounds
The basic protocol in coordination nomenclature is to name the ligands attached to the metal as prefixes before the metal name.
Some common ligands and their names are listed above.
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57. Nomenclature of Coordination Compounds
As is the case with ionic compounds, the name of the cation appears first; the anion is named last.
Ligands are listed alphabetically before the metal. Prefixes denoting the number of a particular ligand are ignored when alphabetizing.
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58. 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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59. 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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60. Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.
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