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Additional Aspects of Aqueous Equilibria. Aspects of Aqueous Equilibria: The Common Ion Effect Salts like sodium acetate are strong electrolytes NaC 2.

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Presentation on theme: "Additional Aspects of Aqueous Equilibria. Aspects of Aqueous Equilibria: The Common Ion Effect Salts like sodium acetate are strong electrolytes NaC 2."— Presentation transcript:

1 Additional Aspects of Aqueous Equilibria

2 Aspects of Aqueous Equilibria: The Common Ion Effect Salts like sodium acetate are strong electrolytes NaC 2 H 3 O 2(aq) Na + (aq) + C 2 H 3 O 2 - (aq) The C 2 H 3 O 2 - ion is a conjugate base of a weak acid HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) K a = [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ]

3 The Common Ion Effect K a = [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] Now, lets think about the problem from the perspective of LeChateliers Principle What would happen if the concentration of the acetate ion were increased? Q > K and the reaction favors reactant Addition of C 2 H 3 O 2 - shifts equilibrium, reducing H +

4 HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) Since the equilibrium has shifted to favor the reactant, it would appear as if the dissociation of the weak acid(weak electrolyte) had decreased. The Common Ion Effect K a = [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ]

5 HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) So where might the additional C 2 H 3 O 2 - (aq) come from? Remember we are not adding H +. So its not like we can add more acetic acid. Aspects of Aqueous Equilibria: The Common Ion Effect How about from the sodium acetate?

6 NaC 2 H 3 O 2(aq) Na + (aq) + C 2 H 3 O 2 - (aq) HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) In general, the dissociation of a weak electrolyte (acetic acid) is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte The shift in equilibrium which occurs is called the COMMON ION EFFECT

7 COMMON ION EFFECT Lets explore the COMMON ION EFFECT in a little more detail Suppose that we add 8.20 g or 0.100 mol sodium acetate, NaC 2 H 3 O 2, to 1 L of a 0.100 M solution of acetic acetic acid, HC 2 H 3 O 2. What is the pH of the resultant solution? HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) NaC 2 H 3 O 2(aq) Na + (aq) + C 2 H 3 O 2 - (aq)

8 Now you try it! Calculate the pH of a solution containing 0.06 M formic acid (HCH 2 O, K a = 1.8 x 10 -4 ) and 0.03 M potassium formate, KCH 2 O.

9 HCl Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of HCl and 0.20 mol HF in 1.0 L HF (aq) + H 2 O H 3 O + (aq) + F - (aq) HCl (aq) + H 2 O H 3 O + (aq) + Cl - (aq)

10 K b = [NH 4 + ] [OH - ] [NH 3 ] LeChateliers Principle Now, lets think about the problem from the perspective of LeChateliers Principle But this time lets deal with a weak base and a salt containing its conjugate acid. Q > K and the reaction favors reactant Addition of NH 4 + shifts equilibrium, reducing OH - NH 3(aq) + H 2 O NH 4 + (aq) + OH - (aq)

11 Calculate the pH of a solution produced by mixing 0.10 mol NH 4 Cl with 0.40 L of 0.10 M NH 3 (aq), pK b = 4.74? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - NH 4 Cl (aq) NH 4 + (aq) + Cl - (aq)

12 Common Ions Generated by Acid-Base Reactions The common ion that affects a weak-acid or weak-base equilibrium may be present because it is added as a salt, or the common ion can be generated by reacting an acid and base directly (no salt would be necessary….which is kind of convenient if you think about it) HC 2 H 3 O 2 (aq) + OH - (aq) H 2 O + C 2 H 3 O 2 - (aq) Suppose we react 0.20 mol of acetic acid (weak) with 0.10 mol of sodium hydroxide strong) 0.20 mol 0.10 mol -0.10 mol +0.10 mol-0.10 mol 0 0 0.10 mol

13 HC 2 H 3 O 2(aq) + OH - (aq) H 2 O + C 2 H 3 O 2 - aq) Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide 0.20 mol 0.10 mol -0.10 mol 0.10 mol -0.10 mol 0 0 0.10 mol HC 2 H 3 O 2 (aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) 0.10 M Lets suppose that all this is occurring in 1.0 L of solution 0

14 Sample problem: Calculate the pH of a solution produced by mixing 0.60 L of 0.10 M NH 4 Cl with 0.40 L of 0.10 M NaOH NH 4 Cl(aq) NH 4 + (aq) + Cl - (aq) NH 4 + + OH - NH 3 + H 2 O 0.06 mol 0.04 mol -0.04 mol 0.04 mol 0 0 -0.04 mol 0.02 mol Dont forget to convert to MOLARI TIES 0.02 M NH 4 + + H 2 O H 3 O + + NH 3 0.04 M 0

15 Now you try it! Calculate the pH of a solution formed by mixing 0.50 L of 0.015 M NaOH with 0.50 L of 0.30 M benzoic acid (HC 7 H 5 O 2, K a = 6.5 x 10 -5 )

16 adding acid or base calculate the pH of a solution that has.2 mol of NaOH added to a solution that is.25 M HC 2 H 3 O 2 and.32M NaC 2 H 3 O 2 HC 2 H 3 O 2(aq) + OH - (aq) H 2 O + C 2 H 3 O 2 - aq..25.20.32 -.20 -.20 +.20.05 0.52

17 HC 2 H 3 O 2(aq) H + + C 2 H 3 O 2 - aq.05 0.52 -X XX X (.52+X) =1.8 x 10 -5.05-X

18 [H + ] = [HX] [X - ] KaKa BUFFERED SOLUTIONS A buffered solution is a solution that resists change in pH upon addition of small amounts of acid or base. Suppose we have a salt: MX M + (aq) + X - (aq) And weve added the salt to a weak acid containing the same conjugate base as the salt, HX: HX +H 2 O H 3 O + + X - And the equilibrium expression for this reaction is K a = [H + ] [ X - ] [HX] Note that the concentration of the H + is dependent upon the K a and the ratio between the HX and X - (the conjugate acid-base pair)

19 Two important characteristics of a buffer are buffering capacity and pH. Buffering capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. This capacity depends on the amount of acid and base from which the buffer is made The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and hence the pH, to change The pH of the buffer depends upon the K a [H + ] = [HX] [X - ] KaKa -log[H + ] = [HX] [X - ] -log K a Henderson-Hasselbalch Equation pH = pK a + log [X - ] [HX]

20 HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - (aq) NaC 2 H 3 O 2 (aq) Na + (aq) + C 2 H 3 O 2 - (aq) 0.1 M 0 0.1 0.1 M -x x x x 0.1 - x 0.1 + x 1.8 x 10 -5 = x(0.1 + x ) 0.1 - x x = 1.8 x 10 -5 pH = 4.74 Using the Henderson-Hasselbalch Equation pH = 4.74 + log [.1] Note that these are initial concentrations

21 after 0.020 mol NaOH is addedadding 0.020 mol HCl A liter of solution containing 0.100 mol of HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is added. HC 2 H 3 O 2(aq) + OH - H 2 O + C 2 H 3 O 2 - (aq) 0.1 M 0.02 M -0.02 M 0.1 M NaC 2 H 3 O 2(aq) Na + (aq) + C 2 H 3 O 2 - (aq) 0.1 M 0.02 M 0.12 M 0.00 M 0.08 M Henderson-Hasselbalch Equation pH = 4.74 + log [.12] [.08] Note that these are initial concentrations pH = 4.92 Step 1 Step 2

22 0.020 mol NaOH after adding 0.020 mol HCl is added A liter of solution containing 0.100 mol of HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is added. C 2 H 3 O 2 - (aq) + H + HC 2 H 3 O 2 Step 1 0.10 M 0.02 M -0.02 M 0.00 M -0.02 M 0.02 M 0.12 M 0.08M Henderson-Hasselbalch Equation pH =4.74 + log [.08] [.12] Note that these are initial concentrations pH = 4.56 Step 2

23 Now consider, for a moment, what would have happened if I had added 0.020 mol of NaOH or 0.02 mol HCl to.1 M HC 2 H 3 O 2. HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - aq 0.1 M 0 0 -x x x x 0.1 - x x 1.8 x 10 -5 = x2x2 0.1 - x x = 0.0013 pH = 2.9

24 HC 2 H 3 O 2(aq) + OH - H 2 O + C 2 H 3 O 2 - (aq) 0.1 M 0.02 M -0.02 M 0.00 0.02 M 0.00 M 0.08 M Henderson-Hasselbalch Equation pH = 4.74 + log [.02] [.08] pH = 4.13

25 HC 2 H 3 O 2(aq) + H 2 O H 3 O + (aq) + C 2 H 3 O 2 - aq 0.10 0 0.02 M pH = 1.7 H + from HCl Completely dissociates therefore the pH is calculated without regard for the weak acid pH = -log [0.02]

26 add 2 ml 10 M HCl to 1.8 x 10 -5 M HCl pH = 4.74 add 2 ml 10 M HCl to.1M HC 2 H 3 O 2 +.1M NaC 2 H 3 O 2 pH = 4.74.02 M HCl pH= 1.7 a drop of 3.04 to.12M HC 2 H 3 O 2 +.06M NaC 2 H 3 O 2 pH = 4.56 a drop of.18

27 Sample exercise: Consider a litter of buffered solution that is 0.110 M in formic acid (HC 2 H 3 O) and 0.100 M in sodium formate (NaC 2 H 3 O ). Calculate the pH of the buffer (a)before any acid or base are added, (b) after the addition of 0.015 mol HNO 3, (c) after the addition of 0.015 mol KOH

28 Titration Curves HCl(aq) + NaOH(aq) H 2 O + NaCl Stoichiometrically equivalent quantities of acid and base have reacted End points

29 Titration of a weak acid and a strong base results in pH curves that look similar to those of a strong acid-strong base curve except that the curve (a) begins at a higher pH, (b) the pH rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point pH is not 7.0

30

31 Calculating pHs from Titrations Calculate the pH in the titration of acetic acid by NaOH after 30.0 ml of 0.100 M NaOH has been added to 50 mL of 0.100 acetic acid HC 2 H 3 O 2(aq) + OH - H 2 O (l) + C 2 H 3 O 2 - 0.005 mol 0.003 mol 0 -0.003 mol -0.002 mol 0 pH = 4.74 + log [.0370] [.0250] pH = 4.91

32 Determining the K a From the Titration Curve pK a = pH = 4.74

33 Solubility Equilibria K sp The equilibrium expression for the following reaction is CaF 2(s) Ca 2+ (aq) + 2F - (aq) [Ca 2+ ] [F - ] 2 [CaF 2 ] Look at the table on page 759 or the appendices A26, where you will find the value of the k sp to be 4.1 x 10 -11 = K sp

34 Calculating K sp from solubility The solubility of Bi 2 S 3 is 1.0 x 10 -15 M what would be the K sp ? Bi 2 S 3(s) 2Bi 3+ (aq) + 3S 2- (aq) 1.0 x 10 -15 2(1.0 x 10 -15 ) 3(1.0x 10 -15 ) K sp = [Bi 3+ ] 2 [S 2- ] 3 K sp = (2.0 x 10 -15 ) 2 (3.0 x 10 -15 ) 3 = 1 x 10 -73

35 Calculating solubility from K sp The K sp of Cu(IO 3 ) 2 is 1.4 x 10 -7 what would be the solubility? Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2IO 3 1- (aq) X X 2X K sp = 1.4 x 10 -7 = (X)(2X) 2 = 4X 3 X = 3.3 x 10 -3 mol/L

36 Common ion effect What would be the solubility of Ag 2 CrO 4 in a solution that is.1M AgNO 3, the K sp = 9.0 x 10 -12 Ag 2 CrO 4 2Ag 1+ + CrO 4 2- X 2X +.1 X 9.0 x 10 -12 = (2X+.1) 2 (X) Drop the 2X as insignificant X = 9.0 x 10 -10 mol/L

37 pH and Solubility Mg(OH) 2 Mg 2+ + 2OH - If the pH is raised (the OH - is raised) then we have the common ion effect and the solubility is decreased. If the pH is lowered (the H + is raised) then OH - is reacted with H + to make water and the solubility is increased.

38 pH and solubility of salts Ag 3 PO 4 3Ag 1+ + PO 4 3- If the pH is lowered (the H + is raised) the PO 4 3- reacts with H + to make HPO 4 2-, which essentially removes PO 4 3- from the equation, shifting the reaction to the right and the solubility is increased.

39 Will a precipitate form? If 750 mL of 4.00E-3M Ce(NO 3 ) 3 is mixed with 300 mL of 2.00E-2M KIO 3, will a precipitate form? Ce(NO 3 ) 3(aq) + KIO 3 (aq) Ce(IO 3 ) 3(s) + KNO 3(aq) (750)(4 x 10 -3 ) = (1050) X; X= 2.86 x 10 -3 M for Ce 3+ (300)(2 x 10 -2 ) = (1050) Y; Y= 5.71 x 10 -3 M for IO 3 1- Ce(IO 3 ) 3(s) Ce 3+ (aq) + 3IO 3 1- (aq) K sp = 1.9 x 10 -10 Q = (2.86 x 10 -3 )(5.71 x 10 -3 ) 3 = 5.32 x 10 -10 Q is greater than K so a precipitate will form


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