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Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay,

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Presentation on theme: "Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay,"— Presentation transcript:

1 Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

2 Aqueous Equilibria March 11 COMMON ION EFFECT 17.11, 13, 15 BUFFER PROBLEMS 17, 19, 23, 25 TITRATIONS 31 to 39 odd, 43

3 Aqueous Equilibria 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. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq)

4 Aqueous Equilibria 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.”

5 Aqueous Equilibria 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. K a for HF is 6.8  10 −4. [H 3 O + ] [F − ] [HF] K a = = 6.8  10 -4

6 Aqueous Equilibria The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F − ], M Initially0.200.100 Change−x−x+x+x+x+x At Equilibrium 0.20 − x  0.200.10 + x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq)

7 Aqueous Equilibria The Common-Ion Effect = x 1.4  10 −3 = x (0.10) (x) (0.20) 6.8  10 −4 = (0.20) (6.8  10 −4 ) (0.10)

8 Aqueous Equilibria The Common-Ion Effect Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = 0.10 + x = 1.01 + 1.4  10 −3 = 0.10 M So,pH = −log (0.10) pH = 1.00

9 Aqueous Equilibria Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

10 Aqueous Equilibria Example Write the reaction that occurs in a buffer solution containing HF/F - when: 1.H + is addedH + + F -  HF 2.OH - is addedOH - + HF  F - + H 2 O

11 Aqueous Equilibria 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.

12 Aqueous Equilibria Buffers If acid is added, the F − reacts to form HF and water.

13 Aqueous Equilibria Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A −

14 Aqueous Equilibria Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log pKapKa pH acid base

15 Aqueous Equilibria Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

16 Aqueous Equilibria MARCH 14 BUFFER PROBLEMS Calculation of the ph of a buffer Buffer capacity Addition of strong acid or bases to buffers BUFFER PROBLEMS 17, 19, 23, 25 TITRATIONS a)Strong acid and Strong base b ) Weak acid – Strong Base Titration of polyprotic Acids TITRATIONS problems 31 to 39 odd, 43

17 Aqueous Equilibria Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4.

18 Aqueous Equilibria Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH = 3.85 + (−0.08) pH = 3.77

19 Aqueous Equilibria Buffer Capacity The amount of acid of base the buffer can neutralize before the pH begins to change to an appreciable degree. It depends on the amount of acid and base from which the buffer is made. The pH of the buffer depends on the Ka of the acid, and on the relative concentrations of acid and base.

20 Aqueous Equilibria Example A solution 1M in HA, 1M in NaA will have its pH=Ka. A solution.1M in HA and.1M in NaA will have SAME pH (Ka) but the first one will have greater buffer capacity.

21 Aqueous Equilibria 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 pK a close to the desired pH. In practice if the concentrations of one component of the buffer is 10 times the concentration of the other component the buffering action is poor.

22 Aqueous Equilibria Addition of Strong Acids or Bases to Buffers The amount of strong acid or base added results in a neutralization reaction: X - + H 3 O +  HX + H 2 O HX + OH -  X - + H 2 O. By knowing how much H 3 O + or OH - was added (stoichiometry) we know how much HX or X - is formed. With the concentrations of HX and X - (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation or the original equilibrium expression (ICE chart).

23 Aqueous Equilibria 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.

24 Aqueous Equilibria Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

25 Aqueous Equilibria Calculating pH Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74

26 Aqueous Equilibria Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O (l) HC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol

27 Aqueous Equilibria Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 200) pH = 4.74 + 0.06 pH = 4.80

28 Aqueous Equilibria Addition of Strong Acid or Base to a Buffer 1.Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2.Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

29 Aqueous Equilibria Examples 1.Consider a buffer prepared by placing 0.60 moles of HF (K a = 7.2 x 10 -4 ) and 0.48 moles of NaF in a 1.00 L solution. a. Calculate the pH of the buffer b. Calculte the pH after the addition of 0.08 moles of HCl c. Calculate the pH after the addition of a total of 0.16 moles of HCl d. Calculate the pH after the addition of 0.10 moles of NaOH.

30 Aqueous Equilibria 2.Consider a buffer containing 0.78 moles of HC 2 H 3 O 2 (Ka = 1.8 x 10-5) and 0.67 moles of NaC 2 H 3 O 2 in 1.00 L of solution. a. Calculate the pH of the buffer solution. b. Calculate the pH after the addition of 0.10 moles of HNO 3. c. Calculate the pH after the addtiion of a total of 0.20 moles of HNO 3. d. Calculate the pH after the addition of 0.10 moles of NaOH.

31 Aqueous Equilibria Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

32 Aqueous Equilibria 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.

33 Aqueous Equilibria 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.

34 Aqueous Equilibria Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

35 Aqueous Equilibria Titration of a Strong Acid with a Strong Base At the equivalence point, mol acid = mol base, the solution contains only water and the salt from the cation of the base and the anion of the acid.

36 Aqueous Equilibria Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

37 Aqueous Equilibria 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. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations.

38 Aqueous Equilibria 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.

39 Aqueous Equilibria 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.

40 Aqueous Equilibria 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.

41 Aqueous Equilibria Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.

42 Aqueous Equilibria Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq)

43 Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2− ] where the equilibrium constant, K sp, is called the solubility product.

44 Aqueous Equilibria Solubility Products K sp 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).

45 Aqueous Equilibria 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. BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq)

46 Aqueous Equilibria Factors Affecting Solubility pH  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.

47 Aqueous Equilibria Factors Affecting Solubility Complex Ions  Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

48 Aqueous Equilibria Factors Affecting Solubility Complex Ions  The formation of these complex ions increases the solubility of these salts.

49 Aqueous Equilibria 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 Al 3+, Zn 2+, and Sn 2+.

50 Aqueous Equilibria Will a Precipitate Form? In a solution,  If Q = K sp, the system is at equilibrium and the solution is saturated.  If Q < K sp, more solid will dissolve until Q = K sp.  If Q > K sp, the salt will precipitate until Q = K sp.

51 Aqueous Equilibria Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.


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