Previously in Chem104: more acid/base reactions: weak / weak strong / strong strong / weak calculations Polyprotic acids Today in Chem104: Titrations Buffers calculations

Titrations: a summary 1) strong acid + strong base titrations 2) 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

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

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?” M Acetic acid M sodium acetate M Acetic acid M sodium acetate

Buffers: how would you make one? My research students have that very problem in research lab. 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 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

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 H 3 O+ K a1 = 7.6 x Step 2. H 2 PO H 2 OHPO H 3 O+ K a2 = 6.2 x Step 3. HPO H 2 OPO H 3 O+ K a3 = 2.12 x

Let’s do it! Step 2. H 2 PO H 2 OHPO H 3 O+ K a2 = 6.2 x pH = pKa + log [A] / [AH] 7.00 = log [A] / [AH] = log [A] / [AH] 0.62 = [A] / [AH]Or 0.62 = mol A / mol AH For 1L of 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 = / 1.62 mol = mol AH mol A = 0.62 x mol AH = mol A

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

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

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

Written simply: This is typical expression for solubility equilibrium Given by the Solubility Product K sp All ionic solids dissolve using Lewis A/B interactions AgCl(s) + 2H 2 O[Ag(H 2 O) 2 ] + + Cl- K sp = 1.8 x K sp = [Ag+][Cl-] 1.8 x = [Ag+][Cl-] 1.3 x10 -5 M = [Ag+] = [Cl-] AgCl(s)Ag+ + Cl- 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

Ionic solids which completely dissolve are highly soluble and cannot be described with a K sp NaCl(s) + 6H 2 O[Na(H 2 O) 6 ] + + Cl-

Solubility obeys AgCl(s)Ag + + Cl- + excess Cl- K sp = 1.8 x Solubility =1.3 x10 -5 M = [Ag+] = [Cl-] AgCl(s)Ag+ + Cl- If more chloride is added the equilbirum shifts left, and Solubility Product K sp requires that less AgCl dissolves K sp = 1.8 x Solubility, [Ag+] <1.3 x10 -5 M

otherwise called The Common Ion Effect obeys AgCl(s)Ag + + Cl- + excess Cl- K sp = 1.8 x Solubility =1.3 x10 -5 M = [Ag+] = [Cl-] AgCl(s)Ag+ + Cl- If more chloride is added the equilbirum shifts left, and Solubility Product K sp requires less AgCl dissolves K sp = 1.8 x but Solubility, [Ag+] <1.3 x10 -5 M because [Cl- ] >>1.3 x10 -5 M

Cleanliness is next to Godliness So controlling solubility can make you more holy? Let’s see how…

The pH Effect obeys K sp = 3.7 x10 -9 Ca(CO 3 )(s)Ca 2+ + CO 3 2- If pH is lowered by adding acid, more CaCO 3 dissolves …. and cleans the dishwasher: + AH HCO 3 - H 2 CO 3 + AH H 2 O + CO 2 + AH

The Chelate Effect obeys K sp = 3.7 x10 -9 Ca(CO 3 )(s)Ca 2+ + CO 3 2- If Ca2+ is removed by adding a ligand, more CaCO 3 dissolves …. and also cleans the dishwasher: + citric acid Ca(citrate)

Making better (stronger) Lewis A/B interactions can improve solubility and clean, too We have seen: AgCl(s) + 2H 2 O [Ag(H 2 O) 2 ] + + Cl- K sp = 1.8 x AgCl can be completely dissolved! Very low solubility due to weak Lewis A/B interactions which do not compensate for large lattice energy But if ammonia is Lewis base: AgCl(s) + 2 NH 3 [Ag(NH 3 ) 2 ] + + Cl-