Acid Content of Beverages

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

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

4. Concepts: Strong/Weak Acids Acid Dissociation / Ka Concentration Titration Titration curve Equivalence point End pointIndicator Mole Relationships Polyprotic acids Total available acid pH & pKa Logarithms Techniques: pH Measurement Titration Apparatus: Buret pH Meter Graduated Cylinder

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

6. For molar concentrations, M (molarity) has same numerical value in mol / L and inmmol / mL HCl 6.0 M E.g., 6 M HCl has 6mmol of HCl in 1mL, 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 mol ——— mol 1000 mmol Y ——— = Y ————— = Y ———— L L mL ——— 1000 m m

7. 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 WEAK Partially Dissociated E.g., in water, HCl dissolves to give a STRONG acid HCl (g) + H2O (l) HCl (aq) H+ (aq) + Cl-(aq) proceeds virtually to completion (no undissociated HCl )

8. The group defines an organic acid Acetic Acid ( CH3COOH), dissolves in water to give a WEAK acid CH3COOH (aq) CH3COO-(aq) + H+(aq) Associated Dissociated proceeds only slightly in forward direction. Almost all CH3COOHexists in ASSOCIATEDform

9. Analogously, NaOH dissolves in water to give a STRONGbase: NaOH (s) + H2O (l) NaOH (aq) Na+(aq) + OH-(aq) proceeds virtually to completion (no undissociated NaOH ), whereas, ammonia (NH3) dissolves in water to give a WEAK base, NH3 (g) + H2O (l) NH3 (aq) NH4+(aq) + OH-(aq) proceeds only slightly in forward direction.

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

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

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

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

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

15. End Points & Equivalence Points signal • Point at which observable signal occurs is called • END POINT. • Point at which stoichiometricamount of one reagent is consumed by second is called • EQUIVALENCE POINT • If signalis accurate indication of completion of reaction, it signals the EQUIVALENCE POINT • Indicators are chosen so that, as closely as possible, Stoichiometric: having consumed the appropriate number of moles e.g.,in the titration of a fixed amount of HCl HCl + NaOH = NaCl + H2O Equiv point is when: mmol NaOH added = mmol HCl HCl END POINT = EQUIVALENCE POINT

16. Consider titration of STRONG ACID with STRONG BASE, e.g., HCl and NaOH Reaction we conduct is basically: HCl + NaOH  NaCl + H2O Reactants are strong and therefore fully dissociated. The products are water and salt, NaCl, strong electrolyte (i.e., fully dissociated) H+ + Cl- Na+ + OH- Na+ + Cl- If we include only reacting species, the equation becomes: H+ (aq) + OH- (aq)  H2O (l) Net Ionic Equation

17. 50 X 0.10 = Start with 50 mL of 0.10 M HCl (5.0 mmol) and 0.10 M NaOHin buret. Before we have added any NaOH, beakerhas[ H+ ] = 0.10 M When we have added the stoichiometric amount (50 mL containing5.0 mmol) of NaOH, we have made 100 mL of a solution containing 5 mmol NaCl, which is neither acidic nor basic. 100 mL 0.05M NaCl 50 mL 0.10M HCl What about the hydrogen ion concentration, [ H+], in between??

18. Calculating concentrations of [ H+ ] etc. as we add BASE to ACID is a problem in 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

19. Suppose we add 1 mL of 0.1 M NaOH( 0.1mmol ) to the 50 mL of 0.10 M HCl ( 5.0mmol of HCl ) 0.1 mmolNaOH 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 1 mL After 2.0 mL [ H+ ] = 4.8 / 52 = 0.092 M [ H+ ] = 4.9 / 51 = 0.096 M

20. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED

21. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED Let’s Plot [ H+ ] as a function of the volume of added NaOH

22. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOH ARE ADDED Let’s Plot [ H+ ] as a function of volume of added NaOH

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

24. Graph shows an apparent discontinuity (break) at 50 mL. What else is in the solution at the equivalence point? H2O  H+ + OH- Ka= [ H+ ] [ OH- ] / Kw = [ H+ ] [ OH- ]= 1.0 x 10-14 (@ 298oK) 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 H2O, [ H+ ] = 1.0 X 10-7 mol/L in 0.1 M NaOH, [ H+ ] = 1.0 X 10-13 mol/L [ H2O ] In pure water or dilute solutions, [ H2O ] = (1000 g / 18 g/mol) / L = 55.5 M

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

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

27. 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 negativesign 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

28. SPECIES IN A BEAKER CONTAINING 50.0 mL 0.1 M HCl TO WHICH SPECIFIED VOLUMES OF 0.1 M NaOHARE ADDED If continue to add NaOH to this solution, we add base to a solution which begins as neutral

29. Continuing our table to include pH with added base past the equivalence point Let us plot pHvs the Volume of NaOH

30. Equivalence Point (pH = 7)

31. Just as [ H+ ]varies over many orders of magnitude (powers of 10), the strength of acids also varies over a wide range, e.g., • ACIDKa • Trichloroacetic acid 3.1 x 10-1 • Acetic acid 1.7 x 10-5 • Folic acid 5.0 x 10-9 • Phenolphthalein 4.0 x 10-10 • Sorbitol2.5 x 10-14 • Following convention for [ H+], define, • for an acid with an ionization constant Ka • pKa = - log Ka • Strong Acids have Low (or even negative) pKa pKa 0.51 4.8 8.3 9.4 13.6

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

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

34. In titrating strong acid with strong base, the pH ______ as we add base through the equivalence point. • Increases slowly • Decreases slowly • Increases rapidly • Decreases rapidly • Changes very little

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

36. Some acids have more than 1 available hydrogens. H2SO4 has 2, H3PO4 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 pKa Citrus Beveragesowe acidity primarily toCITRIC ACID, a TRI-PROTIC acid. pKa' s are: 3.09 4.7 5.41 making it a relatively strong acid

37. Acidity ofCola Flavored Beveragesis normally due toPHOSPHORIC ACID, H3PO4, also a TRI-PROTIC ACID. pKa' s are: 2.12 7.21 12.32 In Carbonated Beverages, CARBON DIOXIDE acts as a DI-PROTIC ACID in water: CO2(aq) + H2O HCO3-+ H+ HCO3- CO3=+H+ pKa' s are: 6.37 10.25 Acids in this exercise are COLORLESS. Need SIGNALwhen the reaction between acid and base is “complete”, i.e. AN INDICATOR

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

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. Before end point At end point Past end point

40. We make two measurements: • 1.)Determine TOTAL AMOUNTOFACID 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. This tells us TOTAL DISSOCIATED + ASSOCIATED ACID in the beverage. Will not detect any acids with pKa’s > 9 (pH at which phenolphthalein signals end point)

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

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

43. If 30.0 mL requires 15.45 mL of base X mL requires 20 mL of base 30.0 X -------- = ----- 15.4520 X = 20 X 30.0 / 15.45 = 39 Y mL will require 30mL 30.0 Y -------- = ----- 15.4530 Y = 30X 30.0 / 15.45 = 58 So, if we use between 39 and 58mL of our beverage, we will assure that the amount of NaOH required will be in the correct range (20-30 mL)

44. 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? • 20 mL • 30 mL • 60 mL

45. Example: 20.0 mL of a beverage requires 10.0 mL of NaOH to reach the phenolphthalein end point. How much beveragewill use 30.0 mL of the NaOH solution? X mL beverage 20.0 mLbeverage = 30.0 mL NaOH 10.0 mL NaOH 20.0 mL beverage XmL beverage = 30.0 mL NaOH 10.0 mL NaOH C 60 mL

46. Having determined how much juice you need to conduct onetitration, 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. Based on the preliminary titration, Calculate this quantity. If you brought sufficient beverage, continue using your beverage. If you do not have enough beverage to complete exercise, use juice provided in laboratory. (But you must do another preliminary titration.)

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

48. 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

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