AP Chemistry Buffers and Titration (not done).pptx

buffers and titration lab

1. AP Chemistry Buffers and Titration

2. Chapter Objectives Preview Buffered Solutions Describe the composition of a buffer and explain how it works to maintain a constant pH Explain how to prepare a buffer with a specific pH Calculate the pH of a buffer, before or after adding H+ or OH- . Acid Base Titration Choose the proper indicator for an acid-base titration Calculate the pH during the titration of: strong acid and strong base, weak acid and strong base, and strong acid and weak base. Interpret the titration curves for the above solutions

3. Introduction Equilibrium established within mixtures of acids and bases Buffers Weak acid (HB) and conjugate base (B - ) Weak base and conjugate acid Titration Reaction of acid and base Acid–base indicators Polyprotic acids

4. Buffers Highly resistant to changes in pH caused by addition of strong acid (H + ) or strong base (OH - ) pH is close to pK a of weak acid used Consists of weak acid and its conjugate base Reacts with (neutralizes) strong acid or strong base HB (aq) + OH - (aq) => B - (aq) + H 2 O B - (aq) + H + (aq) => HB (aq) We will determine How to produce a buffer pH of a buffer system Appropriate buffer system to maintain desired pH Small change in pH that occurs on addition of acid or base Buffer capacity

5. Determination of [H + ] in a Buffer System Concentration of H + based on concentrations of weak acid HB and conjugate base B - . Assume [HB] = [HB] and [B - ] = [B - ] at equilibrium i.e. ignore the little bit that dissociates/hydrolyzes Concentration ratio = mole ratio (same solution)

6. Henderson-Hasselbach Equation Previous equation Take log both sides Multiply by -1 Simplify to final form Henderson-Hasselbach Equation

7. Example 1: [H + ] in Buffer Solution A buffer is prepared by dissolving 1.00 mol of lactic acid HLac (K a = 1.4 x 10 -4 ) and 1.00 mole of sodium lactate, NaLac, in enough water to form 550 ml of solution. Calculate [H + ] and the pH of the buffer.

8. Choosing a Buffer System pH of buffer produced depends on Ionization constant of weak acid (K a ) Usually HB and B - present in nearly equal amounts [H + ] ≈ K a pH ≈ pK a Concentration ratio of HB to B - Add more HB to lower pH slightly Add more B - to raise pH slightly Choose acid with pK a within ±1 pH unit of the desired value.

9. Buffer Systems and pH Example 2: pH of Buffer System To prepare a buffer with a pH of 9.00, which of the systems from the table above would you use? What should be the ratio of concentrations of [HB]/[B - ]?

10. Formation of Buffers Mixing weak acid and conjugate base As in previous examples Partial neutralization of weak acid or base H + (aq) + NH 3 (aq) ⬄NH 4 + (aq) Add 0.10 mol HCl to 0.20 mol NH 3 in solution Final mixture contains 0.10mol NH 3 and 0.10 mol NH 4 + (which is a buffer!)

11. Distribution Curves Pt. 1 Plot of fraction of HB and B - as a function of pH Definition of fractions of HB and B - At pH 5.0 f (HB) = f (B - ) = 0.50 pH = pK a of HB Between pH 4.0 and 6.0 Appreciable amounts of HB and B - (acts as buffer) pH range = pK a ± 1 Below pH 4.0 or above pH 6.0 system consists of essentially single component (HB at low pH, B - at high pH) no longer acts as buffer

12. Effect of Added H + or OH - on Buffer Systems pH changes slightly on addition of (moderate) amount of strong acid or base Addition of strong acid H + (aq) + B - (aq) => HB (aq) Converts weak base to conjugate acid Addition of strong base HB (aq) + OH - (aq) => B - (aq) + H 2 O Converts weak acid to conjugate base

13. Example 3: pH Changes in Buffer Systems upon Adding Strong Acid or Base Consider the lactic acid/sodium lactate buffer prepared by dissolving 1.00 mol of lactic acid HLac (K a = 1.4 x 10 -4 ) and 1.00 mole of sodium lactate, NaLac, in enough water to form 550 ml of solution and whose pH was 3.85. Calculate the pH after addition of a) 0.10 mol of HCl b) 0.10 mol of NaOH

14. Acid-Base Indicators Derived from weak acid HIn HIn (acid form) and In - (base form) have different colors Observed color depends on ratio: Three cases mostly HIn: “acid” color is visible mostly In - : “base” color is visible color observed is mixture of the two

15. Acid-Base Indicators (cont) Color of indicator depends on two factors [H + ] or pH of solution Low pH = acid color High pH = base color K a of indicator Color change occurs when [H + ] ≈ K a pH ≈ pK a Different indicators change colors at different pHs

16. Acid Bases Indicators (cont.)

17. Distribution Curves Pt. 2 Bromothymol Blue Example pH = 7.0 f (HIn) = f (In - ) Green is 50:50 blend of HIn (yellow) and In - (blue) pH 6.0 - 8.0 Appreciable amounts of HIn and In - Smooth change from pH 6.0 (yellow) to pH 8.0 (blue) pH range = pK a of HIn ± 1 pH below 6.0 or above 8.0 All HIn below pH 6.0, all In - above pH 8.0 No color change (no longer useful as indicator)

18. Distribution Curves Pt. 2 Bromothymol Blue Example Continued pH below 6 => acid form (yellow) dominates pH 6 – 8 => mixture (color change to green) pH above 8 => base form (blue) dominates

19. Example 4: Indicators Consider bromothymol blue (K a = 1 x 10 -7 ). At pH 7.5 a) Calculate the ratio of [In - ]/[HIn] b) Calculate the fractions of In - and HIn c) What is the color of the indicator at this point?

20. Acid-Base Titrations Determine unknown concentration of acid or base using measured volume of known standard acid or base Equivalence point Reaction is complete Equivalent amounts of acid and base have reacted Detected using indicator End point Point at which indicator changes color Should coincide with equivalence point

21. Strong Acid-Strong Base Titration of HCl with NaOH Neutralization reaction: H 3 O + (aq) + OH - (Aq) => 2H 2 O K = 1/K w = 1x 10 14 (reaction goes to “completion”) Equivalence point All acid neutralized by base to produce neutral salt solution (NaCl) pH = 7 Near equivalence point pH rises rapidly (as much as six units) with a single drop Beyond equivalence point pH increases slightly with added base

22. Strong Acid-Strong Base (cont) Titration Apparatus Titration Curve for Strong Acid-Strong Base

23. Example 5: Strong Acid-Base Titration When 50.00 ml of 1.000 M HCl is titrated with 1.000 M NaOH, the pH increases. Find the pH of the solution after the following volumes of 1.000M NaOH have been added. A)49.99 ml B)50.01 ml

24. Example 6: Strong Acid-Base Titration 2 What mass of HClO 4 is present in 20.00 ml of solution if 50.00 ml of 0.25 M Ca(OH) 2 is required to reach the equivalence point?

25. Weak Acid–Strong Base Titration Strategy Initial point => weak acid solution Buffer region => mixture of HA and A - Apply Henderson-Hasselbach equation Half equivalence point => pH = pK a Equivalence point => solution of conjugate base of the weak acid (i.e. basic salt solution) Beyond equivalence point => more OH - added to solution

26. Weak Acid-Strong Base Titration of 0.10 M acetic acid with 0.10 M sodium hydroxide HC 2 H 3 O 2 (aq) + OH - => C 2 H 3 O 2 - (aq) + H 2 O Initial pH = 2.9 Buffer region pH changes slowly with added base Weak acid and its conjugate are present together Equivalence point Solution of sodium acetate pH > 7 since acetate is a weak base

27. Weak Acid-Strong Base (cont.) Comparison of Titration of Weak and Strong Acid with Strong Base and Choice of Indicator

28. Various Weak Acid-Strong Base pH Titration Curves Titration of 40.0 mL of 0.100 M solutions of various weak acids with 0.100 M NaOH. In each case, the equivalence point comes after addition of 40.0 mL of 0.100 M NaOH, but the increase in pH at the equivalence point gets smaller and more difficult to detect as the Ka value of the weak acid decreases.

29. Example 7: Weak Acid-Strong Base Titration 50.00 ml of 1.000 M acetic acid, HC 2 H 3 O 2 , is titrated with 1.000 M NaOH. Find the pH of the solution after the following volumes of NaOH have been added. A) 0.00 ml B) 25.00 ml C) 50.00ml

30. Strong Acid-Weak Base Titration of 50.0ml 1.00 M NH 3 with 1.00 M HCl H 3 O + (aq) + NH 3 (aq) => NH 4 + (aq) + H 2 O Initial pH = 11.62 After 25.0 ml HCl added pH = pK a (NH 4 + ) = 9.25 Equivalence point 0.50M NH 4 + (solution of salt of weak acid) pH = 4.77 Suitable indicator => methyl red

31. A Weak Base-Strong Acid Titration Curve Titration of 40.0 mL of 0.100 M NH3 with 0.100 M HCl. The pH is 11.12 at the start of the titration, 9.25 (the pK a value for NH 4 + ) in the buffer region halfway to the equivalence point, and 5.28 at the equivalence point. Note that methyl red is a good indicator for this titration, but phenolphthalein is unacceptable.

32. Acid-Base Titration Summary

33. Example 8: Titration Consider the titration of nitrous acid, HNO 2 , with barium hydroxide A) Write a balanced net ionic equation for the reaction B) Calculate K for the reaction C) Is the solution at the equivalence point acidic, basic, or neutral D) What would be an appropriate indicator for the titration?

34. Polyprotic Weak Acids Contain more than one ionizable hydrogen atom Ionize in steps Separate ionization constants for each step Examples Oxalic acid H 2 C 2 O 4 (aq) ⬄ H + (aq) + HC 2 O 4 - (aq) K a1 = 5.9 x 10 -2 HC 2 O 4 - (aq) ⬄ H + (aq) + C 2 O 4 2- (aq) K a2 = 5.2 x 10 -5 Phosphoric acid H 3 PO 4 (aq) ⬄ H + (aq) + H 2 PO 4 - (aq) K a1 = 7.1 x 10 -3 H 2 PO 4 - (aq) ⬄ H + (aq) + HPO 4 2- (aq) K a2 = 6.2 x 10 -8 HPO 4 2- (aq) ⬄ H + (aq) + PO 4 3- (aq) K a3 = 4.5 x 10 -13 K a1 > K a2 > K a3

35. Polyprotic Weak Acids (cont.) To a first approximation, essentially all hydrogen ions in solution come from the first ionization of the acid.

36. Example 9: Polyprotic Acids The distilled water you use ion the lab is slightly acidic because of dissolved CO 2 , which reacts to form carbonic acid. Calculate the pH of a 0.0010 M solution of H 2 CO 3 and the concentration of CO 3 -2 ions in solution.

37. Titration Curve for Polyprotic Acid Change in the pH of 1.00 L of a 1.00 M solution of H2A+ on addition of solid NaOH. The protonated form of alanine, H2A+, is a diprotic acid, so the titration curve exhibits two equivalence points, at pH 6.02 and pH 11.85, and two buffer regions, near pH 2.34 and pH 9.69.

38. Distribution Curves Pt. 3 Two buffer systems H 2 CO 3 /HCO 3 - centered at pK a1 (H 2 CO 3 ) = 6.4 HCO 3 - /CO 3 2- centered at pK a2 (H 2 CO 3 ) = 10.3 Predominant species at various pHs H 2 CO 3 at low pH (below 6) HCO 3 - at intermediate pH (7-10) CO 3 2- at high pH (>11)

39. Titration Simulation

40. Buffer Simulation