1. Aspects of Aqueous Equilibria

2. Aspects of Aqueous Equilibria: The Common Ion Effect Recall that salts like sodium acetate are strong electrolytes NaC 2 H 3 O 2 (aq)  Na + (aq) + C 2 H 3 O 2 - (aq) …and that 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. Aspects of Aqueous Equilibria: 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 LeChatelier’s 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 + 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. Aspects of Aqueous Equilibria: 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 it’s 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) Aspects of Aqueous Equilibria: The Common Ion Effect 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. Aspects of Aqueous Equilibria: The Common Ion Effect COMMON ION EFFECT Let’s 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! Aspects of Aqueous Equilibria: The Common Ion Effect 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 Note who the strong electrolyte is this time! Aspects of Aqueous Equilibria: The Common Ion Effect HF(aq) + H 2 O  H 3 O + (aq) + F - (aq) HCl+ H 2 O(aq)  H 3 O + (aq) + Cl - (aq)

10. Aspects of Aqueous Equilibria: The Common Ion Effect K b = [NH 4 + ] [OH - ] [NH 3 ] Now, lets think about the problem from the perspective of LeChatelier’s 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 - reducing OH - NH 3 (aq) + H 2 O  NH 4 + (aq) + OH -

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 - Aspects of Aqueous Equilibria: The Common Ion Effect NH 4 Cl(aq)  NH 4 + (aq) + Cl - (aq)

12. Aspects of Aqueous Equilibria: 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) Let’s take a look at a weak acid-strong base combination first: HC 2 H 3 O 2 (aq) + OH -  H 2 O + C 2 H 3 O 2 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 0 0.10 mol

13. HC 2 H 3 O 2 (aq) + OH -  H 2 O + C 2 H 3 O 2 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 Let’s suppose that all this is occurring in 1.0 L of solution Aspects of Aqueous Equilibria: Common Ions Generated by Acid-Base Reactions 0

14. Aspects of Aqueous Equilibria: Common Ions Generated by Acid-Base Reactions 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 Don’t forget to convert toMOLARITIES 0.02 M NH 4 + + H 2 O  H 3 O + + NH 3 0.04 M 0

15. Now you try it! Aspects of Aqueous Equilibria: Common Ions Generated by Acid-Base Reactions 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. [H + ] = [HX] [X - ] KaKa Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS A buffered solution is a solution that resists change in pH upon addition of small amounts of acid or base. HERE’S HOW IT WORKS ! Suppose we have a salt: MX  M + (aq) + X - (aq) And we’ve 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) CONTINUED ON NEXT SLIDE

17. Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS 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 pH =pK a - log [HX] [X - ] Henderson-Hasselbalch Equation pH =pK a + log [X - ] [HX]

18. 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 Henderson-Hasselbalch Equation pH =4.74 + log [.1] Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS Note that these are initial concentrations

19. Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS 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 after forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH is addedadding 0.020 mol HCl 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 Note that the OH - reacts with the HX

20. Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS 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 NaOHafter adding 0.020 mol HCl is added 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

21. 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 Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS

22. 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 Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS

23. C 2 H 3 O 2 - (aq) + H +  HC 2 H 3 O 2 0.00 0.10 M 0.02 M pH = 1.7 Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS Complete dissociation pH = -log [0.02]

24. Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS

25. Aspects of Aqueous Equilibria: BUFFERED SOLUTIONS 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, (3) after the addition of 0.015 mol KOH

26. Aspects of Aqueous Equilibria: Titration Curves HCl(aq) + NaOH(aq)  H 2 O + NaCl Stoichiometrically equivalent quantities of acid and base have reacted

27. 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

28. Aspects of Aqueous Equilibria: Titration Curves of Both Strong and Weak Acids

29. Aspects of Aqueous Equilibria: Calculating pH’s 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

30. Aspects of Aqueous Equilibria: Determining the K a From the Titration Curve pK a = pH = 4.74

31. Na 2 CO 3  2Na + (aq) + CO 3 2- H + (aq) + CO 3 2-  HCO 3 - (aq) H + (aq) + HCO -  H 2 CO 3 (aq) Aspects of Aqueous Equilibria: Titrations of Polyprotic Acids