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Previously in Chem104: more acid/base reactions: weak / weak strong / strong strong / weak

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. Have pH 7 at equivalence pt

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Previously in Chem104: more acid/base reactions: weak / weak strong / strong strong / weak

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

  2. Titrations: a summary 1) strong acid + strong base titrations • Have pH 7 at equivalence pt • have flat slopes at beginning and end 2) weak acid or base titrations (by strong base or acid) • 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 3. An acid or base may have multiple buffer regions • Give me 2 examples 4. Buffer pH determined from the Henderson-Hasselbalch equation • pH = pKa + log [A] / [AH] • But you don’t need to memorize this: • you can derive it! Fast!

  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 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 0.100 M phosphate buffer at pH 7?” 1st: find the Ka’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, H3PO4 …which conjugate pair to use at pH 7? Step 1. H3PO4 + H2O H2PO4 -+ H3O+ Ka1 = 7.6 x 10-3 Step 2. H2PO4-+ H2O HPO4 2-+ H3O+ Ka2 = 6.2 x 10-8 Step 3. HPO42-+ H2O PO4 3-+ H3O+ Ka3 = 2.12 x 10-13

  7. Let’s do it! pH = pKa + log [A] / [AH] 7.00 = 7.21 + log [A] / [AH] -0.21 = log [A] / [AH] Step 2. H2PO4-+ H2O HPO4 2-+ H3O+ 0.62 = [A] / [AH] Or 0.62 = mol A / mol AH Ka2 = 6.2 x 10-8 So 0.62 mol AH = 1.00 mol A For 1L of 0.100 M: mol A + mol AH = 0.100mol So: (0.62 mol AH) + mol AH = 0.100mol 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! To make the buffer solution: 0.0617 mol AH = 0.0617 mol NaH2PO4 0.0617 mol NaH2PO4 x 119.98 g/mol = 7.40 g NaH2PO4 Step 2. H2PO4-+ H2O HPO4 2-+ H3O+ Ka2 = 6.2 x 10-8 0.0383 mol A = 0.0383 mol Na2HPO4 0.0383 mol Na2HPO4 x 141.96 g/mol = 5.44 g Na2HPO4 Dissolved in 1 L water

  9. All Definitions of Acid and Base use Donor /Acceptor Bronsted Acid/Base: proton H+ donor/acceptor Lewis Acid/Base: electron pair donor/acceptor Remember this reaction? CuCl2(H2O)2 (s) + 3H2O [CuCl(H2O)5]+ + Cl- Cu2+:OH2 e- acceptor :e- donor Lewis Acid :Lewis Base

  10. All ionic solids dissolve using Lewis A/B interactions NaCl(s) + 6H2O [Na(H2O)6]+ + Cl- Na+:OH2 e- acceptor :e- donor Lewis Acid :Lewis Base

  11. All ionic solids dissolve using Lewis A/B interactions AgCl(s) + 2H2O [Ag(H2O)2]+ + Cl- Very low solubility due to weak Lewis A/B interactions which does not compensate for large lattice energy Written simply: This is typical expression for solubility equilibrium Given by the Solubility Product Ksp AgCl(s) Ag++ Cl- Ksp = 1.8 x10-10 Ksp = [Ag+][Cl-] 1.8 x10-10 = [Ag+][Cl-] 1.3 x10-5 M = [Ag+] = [Cl-] 1.3 x10-5 M = [Ag+] = [Cl-] This is the molar solubility of AgCl

  12. Ionic solids which completely dissolve are highly soluble and cannot be described with a Ksp NaCl(s) + 6H2O [Na(H2O)6]+ + Cl-

  13. Solubility obeys Le Chatelier’s Principle AgCl(s) Ag++ Cl- Ksp = 1.8 x10-10 Solubility =1.3 x10-5 M = [Ag+] = [Cl-] If more chloride is added the equilbirum shifts left, and Solubility Product Ksp requires that less AgCl dissolves AgCl(s) Ag+ + Cl- + excess Cl- Ksp = 1.8 x10-10 Solubility, [Ag+] <1.3 x10-5 M

  14. otherwise called The Common Ion Effect obeys Le Chatelier’s Principle AgCl(s) Ag++ Cl- Ksp = 1.8 x10-10 Solubility =1.3 x10-5 M = [Ag+] = [Cl-] If more chloride is added the equilbirum shifts left, and Solubility Product Ksp requires less AgCl dissolves AgCl(s) Ag+ + Cl- + excess Cl- Ksp = 1.8 x10-10but Solubility, [Ag+] <1.3 x10-5 M because [Cl- ] >>1.3 x10-5 M

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

  16. The pH Effect obeys Le Chatelier’s Principle …. and cleans the dishwasher: Ca(CO3)(s) Ca2+ + CO32- Ksp = 3.7 x10-9 + AH HCO3- H2O + CO2 If pH is lowered by adding acid, more CaCO3 dissolves + AH + AH H2CO3

  17. The Chelate Effect obeys Le Chatelier’s Principle …. and also cleans the dishwasher: Ca(CO3)(s) Ca2+ + CO32- Ksp = 3.7 x10-9 + citric acid Ca(citrate) If Ca2+ is removed by adding a ligand, more CaCO3 dissolves

  18. Making better (stronger) Lewis A/B interactions can improve solubility and clean, too We have seen: AgCl(s) + 2H2O [Ag(H2O)2]+ + Cl- Very low solubility due to weak Lewis A/B interactions which do not compensate for large lattice energy Ksp = 1.8 x10-10 But if ammonia is Lewis base: AgCl(s) + 2 NH3 [Ag(NH3)2]+ + Cl- AgCl can be completely dissolved!

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