CH204  Expt 9 acid base buffers fall09
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CH204 Expt 9 acid base buffers fall09

Course Number: CH 204, Spring 2010

College/University: University of Texas

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This Week EXPERIMENT 9: ACID-BASE EQUILIBRIA Weak acid titration. Determine Ka of acetic acid by a couple of different methods. Witness the power of a buffer solution to resist changes in pH. 1 2 Non-Equilibrium Reaction Ionization is complete in dilute aqueous solution. H 2O HCl H 3O + + Cl 100% in dilute solution Equilibrium Reaction Ionization is incomplete in dilute aqueous solution. H 2O CH 3COOH...

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Week EXPERIMENT This 9: ACID-BASE EQUILIBRIA Weak acid titration. Determine Ka of acetic acid by a couple of different methods. Witness the power of a buffer solution to resist changes in pH. 1 2 Non-Equilibrium Reaction Ionization is complete in dilute aqueous solution. H 2O HCl H 3O + + Cl 100% in dilute solution Equilibrium Reaction Ionization is incomplete in dilute aqueous solution. H 2O CH 3COOH CH 3COO + H 3O + less than 1% ionization H 2O NaOH Na + + OH 100% in dilute solution H 2O + NH 3 NH 4 + OH less than 1% ionization 3 4 Equilibrium Expression For any equilibrium reaction, aA + bB cC + dD [C]c[D]d products Keq = [A]a[B]b reactants For a weak acid dissociation, HA H3O+ + A Ka = [H3O+][A] [HA] 5 Seven Strong Acids HCl HI HBr (binary) HClO3 HClO4 H2SO4 HNO3 (oxoacids or ternary acids) Weak Acids All the rest! CH3COOH, R-COOH, HNO2 H3PO4, HF, NH4+, HCN, HBrO3 HA abbreviation 6 1 Four-Part Lab (part 1) - Calibration Calculating [H3O+] and pH If you are given the [H3O+] pH = log[H3O+] If you are given the pH [H3O+] = 10pH [H+] = [H3O+] Calibrate pH meter Make sure youre in CALIBRATION mode. Calibrate the pH meter in the order in the lab manual: pH 7.00 first, then pH 4.01, then pH 10.01 Press ENTER or CON to confirm calibration. 7 8 Part 2 Titration of Weak Acid with Strong Base Work with a partner. Titrate 25 ml of 0.1 M acetic acid using 0.1 M NaOH DO NOT add water! No indicator this time. Titrate in a beaker, not a flask, because you need room for the pH electrode. Record pH after the addition of every 1.0 ml of NaOH at first, and as the pH begins to change more quickly, record smaller volume increments, down to 0.2 or 0.1 ml. Switch roles with your lab partner and repeat the titration a second time. Graph pH (y-axis) versus mL added (x-axis) in Excel. 9 Equivalence Point All of the HA has been consumed by the OH to form water.. Moles of HA = moles of OH HA start rxn equil hydrolysis : NaA Na + + A A + H 2 O HA + OH (basic solution ) 11 Half-Equivalence Point At the half-equivalence point, [HA] = [A]. Ka = [H3O+][A] [HA] + NaOH NaA 1.0 mol 1.0 mol + H 2O 1.0 mol 1.0 mol 0 + 1.0 mol 1.0 mol [A] Ka = [H3O+] [HA] log Ka = log [H3O+] pH = pKa 12 2 Henderson-Hasselbalch pH = pK a + log = pK a + log when [ A ] = [ HA] pH = pK a + log (1) pH = pK a [ salt ] [acid ] [ A ] [ HA] To determine the pKa: Draw a straight vertical line through the equivalence point to the xaxis. This will give you the volume of base required to reach the equivalence point. On the graph, this volume is 17.5 mL. Divide the equivalence volume by 2. This will give you the volume of base required to reach the point in the titration where the HA=NaA. On the graph, this volume is 8.75 mL. The point where the volume and pH intersect is the pH that is equal to pKa. On the graph, the pH = 3.9 at 8.75 mL. Part 3- Ionization of Acetic Acid Ionization of Acetic Acid Measure the pH of 1.2 M acetic acid and two buffer solutions. Use measured pH and known acetic acid and acetate concentrations to calculate Ka [10-pH][A] [HA] 15 CH 3COOH start rxn equil 1.2 M xM CH 3COO + H 3O + 0 + xM xM 0 + xM xM ( 1.2 x ) M [CH 3COO ][ H 3O + ] [CH 3COOH ] Ka = Ka = x = 10 pH assume x << 1.2 M Ionization of Acetic Acid If x << 1.2 M, then Buffer Solutions When you mix a weak acid (HA) and the salt of a weak acid (NaA), very little shift in the equilibrium takes place (Le Chatliers Principle). CH3COOH + NaCH3COO (completely dissociates) M 1.0 1.0 M CH 3COOH CH 3COO + H 3O + 1.0 M ( HA) 1.0 M ( A ) Ka = [ x ][ x] x2 = [1.2 x] 1.2 [ H 3O + ] = [CH 3CHOO ] = x [ H 3O + ] = 10 pH Ka = (10 pH ) 1.2 2 17 The number of moles of HA and of A remain essentially constant. 18 3 Part 4 - Buffering Action Mixing solutions of HA and A will dilute both of them. Calculate the new [HA] and [A] after the dilution, then use that value in the equilibrium expression. E.g. 50.0 mL of 0.750 M CH3COOH is added to 40.0 mL of 0.500 M NaCH3COO. What is the molarity of each species in the mixture? The volume of the mixture is 90.0 mL. Part 4 - Buffering Action Add strong acid & base to buffers and to water and compare the changes in pH. Remember that buffers have high concentrations of both HA and A. ( 50.0 mL ) 0.750 mmol HA 37.5 mmol HA = 0.417 M = mL 90.0 mL ( 40.0 mL ) 0.500 mmol A 20.0 mmol A = 0.222 M = mL 90.0 mL Adding a small amount of a strong acid increases the hydronium ion and shifts the equilibrium to the left. Adding a small amount of a strong base decreases the hydronium ion and shifts the equilibrium to the right. 19 Example: 3.0 mL of 1.0 M HCl is added to 50.0 mL of the buffer solution that contains 0.417 M HA and 0.222 M A. What is the change of pH of the buffer solution after the 5. strong acid is added? The Ka = 1.76 1. Determine the pH of the buffer solution before the HCl is added. 5 = 4.75 pKa = log 1.76 2. Determine the number of moles of HA, A- and H3O+ in the buffer solution. ( 50 mL ) [ A ] pH = pK a + log [ HA] 0.222 You can use the mmol of = 4.75 + log salt and acid. 0.417 = 4.48 0.417 mmol HA = 20.85 mmol HA mL 0.222 mmol A (50 mL) = 11.1 mmol A mL 1.0 mmol + (3.0 mL) = 3.0 mmol HCl = 3.0 mmol H 3O mL 3. Set up the reaction summary after HCl is added. HA start rxn equil 20.35 mmol + 3.0 mmol 23.35 mmol A + H 3O + 3.0 mmol ( shifts to the left ) 0 mmol 11.1 mmol 8.1 mmol 3.0 mmol 3.0 mmol 4. Determine pH of solution after HCl added and the change of pH. [ A ] pH = pK a + log [ HA] 8.1 = 4.75 + log 23.35 = 4.29 pH = 4.29 4.48 = 0.19 01 01 Note: The [H3O+] added by the strong acid is >>> than the [H3O+] in the buffer solution before the strong acid is added. Example: 3.0 mL of 1.0 M NaOH is added to 50.0 mL of the buffer solution that contains 0.417 M HA and 0.222 M A. What is the change of pH of the buffer solution after the strong base is added? The Ka = 1.76 105. 1. Set up the reaction summary after 3 mmol of NaOH added to buffer solution. HA start rxn equil + OH A + H 2O (shifts to the right ) 20.35 mmol 3.0 mmol 3.0 mmol 3.0 mmol 17.35 mmol 0 mmol 11.1 mmol + 3.0 mmol 14.3 mmol 4 2. Determine pH of solution after NaOH added and the change of pH. [ A ] pH = pK a + log [ HA] 14.3 = 4.75 + log 17.35 = 4.66 pH = 4.66 4.48 = + 0.19 pH Meters Glass bulb is very thin Remove carefully from storage bottle turn the bottle, not the cap Rinse well between samples, dab dry with KimWipes Keep the bulb wet between readings Swirl samples to get better readings 26 Next Weeks Quiz Last Quiz Given [H3O+] calculate pH Given pH, calculate [H3O+] Know how to recognize a buffer solution. Know how to make up a buffer solution. Know how to do equilibrium calculations for both weak acid and buffers. Dilution problems. Next Week Final Lecture (Except for Wednesday make-up section.) Course/Instructor Surveys TA Evaluations Kinetics lab Lab check-out 27 28 This Weeks Quiz Quiz 8 29 5

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