1. Previously in Chem104: Polyprotic acids Titrations Buffers Is there any more we need to say about these????

2. Titrations: a summary strong acid + strong base titrations weak acid or base titrations (by strong base or acid) Have pH 7 at equivalence pt have flat slopes at beginning and end have pH at equivalence pt determine by conjugate weak acid titrations have basic pH at eq. pt. weak base titrations have acidic pH at eq. pt. have more pronounced slope at beginning have pH = pKa at ½ volume to equivalence point have buffer region where [AH] ~ [A], i.e., where conjugate species have about the same concentrations

3. Buffers: a summary 1. Resist change in pH 2. Made from conjugates in ~equal concentrations Acid form [AH] reacts with added base Base form [A] reacts with added acid pH = pKa + log [A] / [AH] But you don’t need to memorize this: you can derive it! Fast! 4. Buffer pH determined from the Henderson-Hasselbalch equation 3. An acid or base may have multiple buffer regions Give me 2 examples

4. Buffers: one new point Buffer capacity: how much acid or base can it “absorb”, or compensate for before pH changes Consider these two buffer solutions and answer, “Which has higher buffer capacity?” 0.100 M Acetic acid + 0.100 M sodium acetate 0.001 M Acetic acid + 0.001 M sodium acetate

5. Buffers: how would you make one? My research methodology students have that very problem this week. Let’s do it and I can report to them that my Gen Chem students can help them out! How would you make 1 L of a 0.100 M phosphate buffer at pH 7?” 1st: find the K a ’s for the acid/base system 2nd: determine the conjugate pair appropriate for the pH 3rd: use the HH equation (or derive it) or the Ka expression to find the relative proportions of conjugates

6. Phosphoric acid, H 3 PO 4 …which conjugate pair to use at pH 7? Step 1. H 3 PO 4 + H 2 OH 2 PO 4 - + H 3 O+ K a1 = 7.6 x 10 -3 Step 2. H 2 PO 4 - + H 2 OHPO 4 2- + H 3 O+ K a2 = 6.2 x 10 -8 Step 3. HPO 4 2- + H 2 OPO 4 3- + H 3 O+ K a3 = 2.12 x 10 -13

7. Let’s do it! Step 2. H 2 PO 4 - + H 2 OHPO 4 2- + H 3 O+ K a2 = 6.2 x 10 -8 pH = pKa + log [A] / [AH] 7.00 = 7.21 + log [A] / [AH] -0.21 = log [A] / [AH] 0.62 = [A] / [AH]Or 0.62 = mol A / mol AH For 1L of 0.100 M: mol A + mol AH = 0.100mol So: (0.62 mol AH) + mol AH = 0.100mol So 0.62 mol AH = 1.00 mol A 1.62 mol AH = 0.100mol mol AH = 0.100 / 1.62 mol = 0.0617 mol AH mol A = 0.62 x 0.0617 mol AH = 0.0383 mol A

8. Let’s do it! Step 2. H 2 PO 4 - + H 2 OHPO 4 2- + H 3 O+ K a2 = 6.2 x 10 -8 To make the buffer solution: 0.0617 mol AH = 0.0617 mol NaH 2 PO 4 0.0617 mol NaH 2 PO 4 x 119.98 g/mol = 7.40 g NaH 2 PO 4 0.0383 mol A = 0.0383 mol Na 2 HPO 4 0.0383 mol Na 2 HPO 4 x 141.96 g/mol = 5.44 g Na 2 HPO 4 Dissolved in 1 L water

9. All Definitions of Acid and Base use Donor /Acceptor Bronsted Acid/Base: proton H + donor/acceptor Remember this reaction? Lewis Acid/Base: electron pair donor/acceptor CuCl 2 (H 2 O) 2 (s) + 3H 2 O[CuCl(H 2 O) 5 ] + + Cl- Cu 2+  :OH 2 e- acceptor  :e- donor Lewis Acid  :Lewis Base

10. All ionic solids dissolve using Lewis A/B interactions NaCl(s) + 6H 2 O[Na(H 2 O) 6 ] + + Cl- Na +  :OH 2 e- acceptor  :e- donor Lewis Acid  :Lewis Base

11. All ionic solids dissolve using Lewis A/B interactions AgCl(s) + 2H 2 O[Ag(H 2 O) 2 ] + + Cl- K sp = 1.8 x10 -10 K sp = [Ag+][Cl-] 1.8 x10 -10 = [Ag+][Cl-] 1.3 x10 -5 M = [Ag+] = [Cl-] AgCl(s)Ag+ + Cl- Written simply: This is typical expression for solubility equilibrium Given by the Solubility Product K sp Very low solubility due to weak Lewis A/B interactions which does not compensate for large lattice energy 1.3 x10 -5 M = [Ag+] = [Cl-] This is the molar solubility of AgCl