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3. Acid Content of Beverages A titration exercise SUSB - 010 2

4. characterize theacidity How can we characterize the acidity of a natural mixture of unknown acids? strongweak What do chemists mean when they talk about acids being strong or weak? ? QUESTIONS ? quantitativemeasure What is a quantitative measure of the strength of an acid? titrations How do we conduct titrations and what can we learn from them? 3

5. Concepts: Strong/Weak Acids Acid Dissociation / K a Concentration Titration Titration curve Equivalence point End point Indicator Mole Relationships Polyprotic acids Total available acid pH & pK a Logarithms Techniques: Apparatus: BuretpH Meter Graduated Cylinder Titration 4 pH Measurement

6. MOLES, LITERS & CONCENTRATION UNITS mL mgsmall numbers of grams. 1 kg1 mole1 Liter In lab, we generally measure volumes in mL & weights in mg or small numbers of grams. We don’t normally use even close to 1 kg, 1 mole or 1 Liter of anything. 0 To avoid writing quantities with zer0s after decimal point, procedures and data sheets often specify: volumes in m mL (= 1 / 1000 liter) and molar quantitites in m mmol (= 1 / 1000 mol) & weights in m mg (= 1 / 1000 g) 0.03421 L 00 0.002845 mol m 34.21 mL m 2.845 mmol 0 0.0757 g m 75.7 mg 5

7. E.g., 6 M HCl has6 mmol of HCl in 1 mL, also 1.0 m mol of NaOH weighs 40 m g Similarly, for atomic and molecular weights: atomic weight of carbon is 12 m g / m mol molar mass of vanillin is 152 mg/mmol same numerical value For molar concentrations, M (molarity) has same numerical value in mol / L and in mmol / mL mm HCl 6.0 M 6 mol ——— mol 1000 mmol Y ——— = Y ————— = Y ———— L L mL ——— 1000

8. Background Strong and Weak Acids (and bases) Acids and bases can be characterized by the extent to which they dissociate in solution STRONG  Fully Dissociated E.g., in water, HCl dissolves to give a STRONG acid HCl (aq)  H + (aq) + Cl - (aq) WEAK  Partially Dissociated 7 HCl (g) + H 2 O (l) proceeds virtually to completion (no undissociated HCl )

9. Acetic Acid ( CH 3 COOH), dissolves in water to give a WEAK acid CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) Almost all CH 3 COOH exists in ASSOCIATED form Associated proceeds only slightly in forward direction. Dissociated 8 group defines an organic acid The

10. Analogously, NaOH dissolves in water to give a STRONG base: NaOH (aq)  Na + (aq) + OH - (aq) 9 NaOH (s) + H 2 O (l) proceeds virtually to completion (no undissociated NaOH ), dissolves in water to give a WEAK base, NH 3 (aq) NH 4 + (aq) + OH - (aq) NH 3 (g) + H 2 O (l) whereas, ammonia (NH 3 ) proceeds only slightly in forward direction.

11. A QUANTITATIVE measure of strength or weakness of an acid ( or base ) is its DISSOCIATION CONSTANT, K a For reaction HA  H + + A - acid dissociation constant, K a, is defined as [ H + ] [ A - ] K a = ------------- [ HA ] LARGE K a ( >> 1) STRONG ACID SMALL K a ( << 1) WEAK ACID DISSOCIATION CONSTANTS 10 concentration of X [ X ] means concentration of X, normally in mol/L

12. weak _________ A weak acid is one which exists mostly in its _________ form in aqueous solution 11 A. Dissociated B. Associated

13. weak __________ A weak acid is one which exists mostly in its __________ form in aqueous solution associated B Associated 12 strong A strong acid is one that dissociates completely in aqueous solution

14. TITRATION A reaction conducted by slow addition of a precisely measured volume of a reagent solution of known concentration SIGNALCOLOR CHANGE SIGNAL is often a COLOR CHANGE, but may be an observable change in another property. 13 until to an amount of another substance* SIGNAL a SIGNAL indicates that reaction between reagent and substance is complete * with which it is known to react

15. desired point If reaction itself does not produce a SIGNAL when the desired point is reached, INDICATOR may be added to produce an indicator signal. An indicator is a substance that reacts with a reactant or product to produce an observable signal. INDICATORS e.g., from colorless to a color, or from one color to another, or observable change producing a gas 14 producing a precipitate or other indication of reaction

16. Point at which observable signal occurs is called END POINT. Point at which stoichiometric amount of one reagent is consumed by second is called EQUIVALENCE POINT If signal is accurate indication of completion of reaction, it signals the EQUIVALENCE POINT Indicators are chosen so that, as closely as possible, End Points & Equivalence Points signal END POINT = EQUIVALENCE POINT Stoichiometric: Stoichiometric: having consumed the appropriate number of moles 15 HCl e.g.,in the titration of a fixed amount of HCl HCl + NaOH = NaCl + H 2 O Equiv point is when: mmol NaOH added = mmol HCl

17. Consider titration of STRONG ACID with STRONG BASE, e.g., HCl and NaOH Reaction we conduct is basically: HCl + NaOH  NaCl + H 2 O Reactants are strong and therefore fully dissociated. The products are water and salt, NaCl, strong electrolyte (i.e., fully dissociated) If we include only reacting species, the equation becomes: H + (aq) + OH - (aq)  H 2 O (l) H + + Cl - Na + + OH - Na + + Cl - Net Ionic Equation 16

18. When we have added the stoichiometric amount (50 mL containing 5.0 mmol) of NaOH, we have made 100 mL of a solution containing 5 mmol NaCl, which is neither acidic nor basic. Start with 50 mL of 0.10 M HCl (5.0 mmol) What about the hydrogen ion concentration, [ H + ], in between?? and 0.10 M NaOH in buret. Before we have added any NaOH, beaker has [ H + ] = 0.10 M 50 mL 0.10M HCl 100 mL 0.05M NaCl 50 X 0.10 = 17

19. Calculating concentrations of [ H + ] etc. as we add BASE to ACID is a problem in STOICHIOMETRY Stoichiometry Stoichiometry: The accounting of the relative amounts of chemical substances that can react or combine with one another. The basic unit of stoichiometry is the mmole 18

20. Suppose we add 1 mL of 0.1 M NaOH ( 0.1 mmol ) to the 50 mL of 0.10 M HCl ( 5.0 mmol of HCl ) 0.1 mmol NaOH reacts with 0.1 mmol HCl to produce 0.1 mmol NaCl Result is 50 + 1 = 51 mL containing 5.0 – 0.1 = 4.9 mmol HCl & 0.1 mmol NaCl Continuing this calculation as we add more 0.1 M NaOH, we produce the following table [ H + ] = 4.9 / 51 = 0.096 M 1 mL After 2.0 mL [ H + ] = 4.8 / 52 = 0.092 M 19

21. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED Vol NaOH Added mmol NaOH Added mmol HCl Remain mmol NaCl Product mmol NaOH Unreact Vol mL Tot [ H + ]pH 00.05.00.0 500.1001.00 10.14.90.10.0510.0961.02 101.04.01.00.0600.0671.18 202.03.02.00.0700.0431.37 303.02.03.00.0800.0251.60 404.01.04.00.0900.0111.95 494.90.14.90.0990.0013.00 49.94.990.014.990.099.90.0001 4.00 50.05.00.05.00.01000.0(?)7.00 20

22. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED Vol NaOH Added mmol NaOH Added mmol HCl Remain mmol NaCl Product mmol NaOH Unreact Vol mL Tot [ H + ]pH 00.05.00.0 500.1001.00 10.14.90.10.0510.0961.02 101.04.01.00.0600.0671.18 202.03.02.00.0700.0431.37 303.02.03.00.0800.0251.60 404.01.04.00.0900.0111.95 494.90.14.90.0990.0013.00 49.94.990.014.990.099.90.0001 4.00 50.05.00.05.00.01000.0(?)7.00 as a function ofLet’s Plot [ H + ]the volume of added NaOH 21

23. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED Vol NaOH Added mmol NaOH Added mmol HCl Remain mmol NaCl Product mmol NaOH Unreact Vol mL Tot [ H + ]pH 00.05.00.0 500.1001.00 10.14.90.10.0510.0961.02 101.04.01.00.0600.0671.18 202.03.02.00.0700.0431.37 303.02.03.00.0800.0251.60 404.01.04.00.0900.0111.95 494.90.14.90.0990.0013.00 49.94.990.014.990.099.90.0001 4.00 50.05.00.05.00.01000.0(?)7.00 Let’s Plot [ H + ]as a function ofvolume of added NaOH 22

24. Equivalence Point mmol NaOH = mmol HCl 23 0.1 ( 50 – V b ) [ H + ] = -------------------- 50 + V b

25. Graph shows an apparent discontinuity (break) at 50 mL. What else is in the solution at the equivalence point? H 2 O  H + +OH - K a = [ H + ] [ OH - ] / K w = [ H + ] [ OH - ]= 1.0 x 10 -14 (@ 298 o K) When no other acids or bases are present, [ H + ] = [ OH - ] =  (1.0 x 10 -14 ) = 1.0 X 10 -7 [ H + ] varies widely from one aqueous solution to another, e.g., in 0.1 M HCl,[ H + ] = 1.0 X 10 -1 mol/L in H 2 O,[ H + ] = 1.0 X 10 -7 mol/L in 0.1 M NaOH,[ H + ] = 1.0 X 10 -13 mol/L [ H 2 O ] In pure water or dilute solutions, [ H 2 O ]= (1000 g / 18 g/mol) / L = 55.5 M 24

26. In an aqueous solution, it is possible to have [ H + ] = 0.1 & [ OH - ] = 0.1 at the same time. 25 A.True B.False

27. In an aqueous solution, it is possible to have [ H + ] = 0.1 & [ OH - ] = 0.1 at the same time. B False 26 [H + ] and [OH - ] are dependent on one another in aqueous solutions. [ H + ] [ OH - ] = 1.0 X 10 -14

28. LOGARITHMS large ranges By introducing LOGARITHMS, can define quantities that enable making plots with very large ranges. For [ H + ], chemists (Sörenson, 1907) chose to define: pH= - log [ H + ] Or, [ H + ]= 10 -pH (Why negative sign was chosen is of historical interest. Consequently, large [ H + ] corresponds to low pH.) If [ H + ] = 3.0 X 10 -8 pH = - log ( 3.0 X 10 -8 ) = 7.5 If pH = 6.1 [ H + ] = 10 -6.1 = 7.9 X 10 -7 Let’s include pH in our table of data 27

29. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED Vol NaOH Added mmol NaOH Added mmol HCl Remain mmol NaCl Product mmol NaOH Unreact Vol mL Tot [ H + ]pH 00.05.00.0 500.1001.00 10.14.90.10.0510.0961.02 101.04.01.00.0600.0671.18 202.03.02.00.0700.0431.37 303.02.03.00.0800.0251.60 404.01.04.00.0900.0111.95 494.90.14.90.0990.0013.00 49.94.990.014.990.099.90.00014.00 50.05.00.05.00.01001x10 -7 7.00 If continue to add NaOH to this solution, we add base to a solution which begins as neutral 28

30. Vol NaOH Added mmol NaOH Added mmol HCl Remain mmol NaCl Product mmol NaOH Unreact Vol mL Tot [ OH - ]pH 515.10.05.00.11010.00111.00 606.00.05.01.01100.00911.96 707.00.05.02.01200.01712.22 808.00.05.03.01300.02312.36 909.00.05.04.01400.02912.46 10010.00.05.0 1500.03312.52 Let us plot pH vs the Volume of NaOH Continuing our table to include pH with added base past the equivalence point 29

31. 30 Equivalence Point (pH = 7)

32. orders of magnitude Just as [ H + ] varies over many orders of magnitude (powers of 10), the strength of acids also varies over a wide range, e.g., ACID K a Trichloroacetic acid3.1 x 10 -1 Acetic acid1.7 x 10 -5 Folic acid5.0 x 10 -9 Phenolphthalein4.0 x 10 -10 Sorbitol2.5 x 10 -14 Following convention for [ H + ], define, for an acid with an ionization constant K a pK a = - log K a Strong Acids have Low (or even negative) pK a pK a 0.51 4.8 8.3 9.4 13.6 31

33. How does pH vary with added NaOH for a weak acid? How sharp is the pH rise at the equivalence (end) point? Volume of added NaOH pH What is the pH at the equivalence (end) point? 32

34. 33 pH pK a = 2 pK a = 4 pK a = 6 pK a = 8 Note that pH at equivalence point changes with pK a strong acid That depends on the pK a of the weak acid

35. In titrating strong acid with strong base, the pH ______ as we add base through the equivalence point. 34 A.Increases slowly B.Decreases slowly C.Increases rapidly D.Decreases rapidly E.Changes very little

36. In titration of strong acid with strong base, the pH ______ as we add base through the equivalence point. 35 C Increases rapidly

37. Some acids have more than 1 available hydrogens. H 2 SO 4 has 2,H 3 PO 4 has 3 Such acids are called POLY-PROTIC (or specifically, DI-PROTIC, TRI-PROTIC, etc.) Many acids in this exercise are polyprotic. Have more than one ionization constant & more than one corresponding pK a Citrus Beverages owe acidity primarily to CITRIC ACID, a TRI-PROTIC acid. pK a ' s are: 3.09 4.7 5.41 making it a relatively strong acid 36

38. Acidity of Cola Flavored Beverages is normally due to PHOSPHORIC ACID, H 3 PO 4, also a TRI-PROTIC ACID. pK a ' s are: 2.127.2112.32 In Carbonated Beverages, CARBON DIOXIDE acts as a DI-PROTIC ACID in water: CO 2 (aq) + H 2 O  HCO 3 - + H + HCO 3 -  CO 3 = + H + pK a ' s are: 6.3710.25 Acids in this exercise are COLORLESS. Need SIGNAL when the reaction between acid and base is “complete”, i.e. AN INDICATOR 37

39. PHENOLPTHALEIN Use PHENOLPTHALEIN, which is itself, a weak DI-PROTIC ACID. ColorlessPink PHENOLPHTHALEIN Color change for PHENOLPHTHALEIN occurs when pH increases from 9. pH > 9   pH < 9 38

40. Before end point At end point Past end point 39 If you bring in your own beverage, remember that you will need to be able to tell when the color changes from its original color to pink.

41. 1.) Determine TOTAL AMOUNT OF ACID in measured sample of beverage.HOW? by titrating the sample with NaOH of known concentration. NaOH reacts with ALL OF THE ACID*, dissociated, or not. Will not detect any acids with pK a ’s > 9 (pH at which phenolphthalein signals end point) We make two measurements: This tells us TOTAL DISSOCIATED + ASSOCIATED ACID in the beverage. 40

42. 2. Measure pH of beverage to determine H + concentration of beverage using a pH METER – This tells us CONCENTRATION OF DISSOCIATED H + in beverage, [ H + ] 41 Electronic device designed to measure hydrogen ion concentration in aqueous solutions

43. PROCEDURE Beverages you bring in vary widely in acid content. Cannot predict how much of it to use. In such instances, we always do a: To get reasonable precision from buret readings, always try to use net volumes between 20 mL – 30 mL in a titration. PRELIMINARY TITRATION Objective: To determine how much of the beverage we need to use to consume the desired amount (20 – 30 mL) of NaOH. 42

44. If 30.0 mL requires 15.45 mL of base X39 X = 20 X 30.0 / 15.45 = 39 3958 So, if we use between 39 and 58 mL of our beverage, we will assure that the amount of NaOH required will be in the correct range (20-30 mL) X 30.0 X -------- = ----- 15.45 20 Y 30.0 Y -------- = ----- 15.45 30 X X mL requires 20 mL of base Y Y mL will require 30 mL Y58 Y = 30 X 30.0 / 15.45 = 58 43

45. Example: 20.0 mL of a beverage requires 10.0 mL of NaOH to reach the phenolphthalein end point. How much beverage will use 30.0 mL of the NaOH solution? 44 A.20 mL B.30 mL C.60 mL

46. Example: How much beverage 20.0 mL of a beverage requires 10.0 mL of NaOH to reach the phenolphthalein end point. How much beverage will use 30.0 mL of the NaOH solution? C C60 mL 45 X mL beverage X mL beverage 20.0 mL beverage = 30.0 mL NaOH 10.0 mL NaOH 20.0 mL beverage X mL beverage X mL beverage = 30.0 mL NaOH 10.0 mL NaOH

47. one Having determined how much juice you need to conduct one titration, you can compute how much beverage you will need to complete the exercise. You are asked to do no more than 4 titrations (in addition to the preliminary titration) and report the best 3 out of those 4. So, if x mL is required to do one titration, you need, at most, (4x + 20) mL of beverage. If you do not have enough beverage to complete exercise, use juice provided in laboratory. (But you must do another preliminary titration.) Based on the preliminary titration, Calculate this quantity Based on the preliminary titration, Calculate this quantity. If you brought sufficient beverage, continue using your beverage. 46

48. For simplicity, assume beverage contains single weak monoprotic acid HA  H + + A - with acid dissociation constant K a For this, we need pH of untitrated (and undiluted) beverage. I.e., [ H + ] = 10 -2.7 = 2.0 X 10 -3 All that remains it to calculate (effective) K a for acid(s) in beverage. [ H + ] [ A - ] K a = --------- [ HA ] ( 2.0 X 10 -3 ) 2 = ----------------- 0.124 = 3.2 X 10 -5 Suppose result of titration is Total Acid = 0.124 M & measured pH = 2.7 47 From pH Total Acid [ H + ] = [ A - ] Should subtract 2.0 X 10 -3 But to 2 sig figs, can ignore

49. REMEMBER: Second QUIZ will be given at the beginning of the acid in beverages lab period. (15 minutes) It will cover: SUSB-030,SUSB-004,SUSB-039 48

51. Stamp out Chemophobia U CHEMISTRY TELL ‘EM This message is approved by the USB Department of Chemistry

52. NEXT EXERCISE Strength of Vinegar by Acid Base Titration Read SUSB-011 and Do Prelab A TEST EXERCISE – 105 Points 51

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