1. Acids, Bases, Salts and Buffers Experiment 8

2. #8 Acids, bases, salts and buffers Goals:  Understand weak acid (base) equilibria and conjugate acid-base pairs  Investigate the acidity/basicity of solutions containing aqueous ions  Perform calculations involving ionic equilibria  Understand buffer solutions and how they work Method:  Approximate the pH of aqueous salt solutions using acid-base- indicators  Make buffer solutions  Compare the effect of added base in weak acid, weak base, buffer, and dilute strong acid solutions

3. Acid-Base Definitions Acids  generate H + in water  H + donors  excess H + Bases  generate OH - in water  H + acceptors  Excess OH -

4. Equilibrium in Water Small K  equilibrium favors reactants

5. As [H + ] rises, [OH - ] falls H+H+ H+H+ H+H+ OH - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Acidic Neutral Basic

6. H + and pH [H + ] 1  10 0 to 1  10 -14 in water pH 1to 14in water

7. Relationships H+H+ OH - [H + ] >[OH - ] Acidic solution Basic solution [H + ] =[OH - ] [H + ] <[OH - ] Neutral solution H2OH2O pH < 7 pH > 7 pH = 7

10. Strong Acids (exp. 4) 100% dissociation / good H + donor equilibrium lies far to right HA  H + + A - HCl H + Cl - Relative # moles Before dissociation After dissociation

11. Weak Acids (exp. 5) <100% dissociation / not-as-good H + donor equilibrium lies far to left HA H + + A - HA H + A - Relative # moles Before dissociation After dissociation HA H + A -

12. Acid Dissociation Constant For a weak acid: 10 -2 < K a < 10 -7 2 < pK a < 7 ←amount dissociated ←amount undissociated

13. Henderson-Hasselbach Equation log(xy) = log x + log y –log(x/y) = +log (y/x) For buffer system only considerable [HA],[A - ] Smaller K a → weaker acid Larger pK a → weaker acid

14. [H + ], pH and K a, [A - ], [HA] 1)[A - ] < [HA] → 2)[A - ] > [HA] → 3)[A - ] = [HA] → → pH < pK a → pH > pK a → pH = pK a

15. Chemical Equations 1) Weak acid HA dissociation 2) Conjugate base reaction with water 3) Water autoionization

16. Buffer Characteristics - Contain relatively large amounts of weak acid and corresponding base. - Added H + reacts to completion with weak base, A -. - Added OH  reacts to completion with weak acid, HA. - pH is determined by ratio of concentrations of weak acid and weak base.

17. Hydrolysis – Anions of Weak Acids Anion of weak acids can react with proton sources. In water, the anions react with water to some extent to form OH - (and the conjugate acid). The OH - causes the solution pH to be greater than 7. Ex.: NO 2 - (aq) + H 2 O (l)  HNO 2(aq) + OH - (aq) The K a of HNO 2 is 4.5  10 -4, so K b for NO 2 - is:

18. Hydrolysis – Anions of Weak Acids Ex.: OCl - (aq) + H 2 O (l)  HOCl (aq) + OH - (aq) Find the pH of 0.10 M NaOCl. K a of HOCl is 3.0  10 -8 K b for OCl - is: K b is small so x can be neglected relative to 0.10; 0.10– x  0.10 OCl - +H2OH2O  HOCl+OH - 0.10---0~0 -x---+x 0.10 - x---xx

19. Another example – finding K b from pH A solution of 0.10 M NaOBr has a pH of 10.85.  Na + spectator ion  OBr - conjugate base of a weak acid (HOBr) Hydrolysis equation:OBr - (aq) + H 2 O (l)  HOBr (aq) + OH - (aq)  pOH of the solution is 14.00 – 10.85 = 3.15  [OH - ] is 10 -3.15 = 7.1  10 -4 OBr - +H2OH2O  HOBr+OH - 0.10---0~0  7.1  10 -4 --- +7.1  10 -4 ~0.10--- 7.1  10 -4

20. Hydrolysis – Cations of Weak Bases Cations derived from weak bases react with water to increase the H 3 O + concentration (acidic). Consider NH 4 + in water: NH 4 + (aq) + H 2 O (l)  NH 3(aq) + H 3 O + (aq) or:NH 4 + (aq)  NH 3(aq) + H + (aq) K a for NH 4 +, the conjugate acid of NH 3, can be determined using the K b of NH 3 and K w : Cations of the group 1A metals (Li +, Na +, K +, Rb +, Cs + ) and the group 2A metals (Ca 2+, Sr 2+, Ba 2+ ) do not react with water and are nonacids. They do not affect the pH of the solution. Hydrated cations of many other metals do hydrolyze to produce acidic solutions. For example:Fe(H 2 O) 6 3+ (aq) + H 2 O (l)  Fe(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) The waters of hydration are sometimes omitted: Fe(H 2 O) 6 3+ (aq) ≡ Fe 3+ (aq) ; Fe(H 2 O) 5 (OH) 2+ (aq) ≡ Fe (OH) 2+ (aq)

21. Summary of Hydrolysis of Salts The acidity, basicity, or neutrality of an aqueous salt solution can be predicted based on the strengths of the acid and base from which the salt was derived. 1.Cation from strong base; anion from strong acid Ex. NaCl, KNO 3 Solution has pH = 7 (neutral) 2.Cation from weak base; anion from strong acid Ex. NH 4 Cl, Zn(NO 3 ) 2 Solution has pH < 7 (acidic) due to the hydrolysis of the cation. 3.Cation from strong base; anion from weak acid Ex. NaF, KNO 2 Solution has pH > 7 (basic) due to the hydrolysis of the anion. 4.Cation from weak base; anion from weak acid Ex. NH 4 F, NH 4 C 2 H 3 O 2 Solution pH is determined by the relative K a and K b of the cation and anion.

22. CO 2 and experimental pH The solutions will generally be more acidic than predicted primarily due to the presence of dissolved CO 2. CO 2 reacts with water to generate H 3 O + (aqueous protons, H + (aq) ): CO 2(g) + H 2 O (l)  H 2 CO 3(aq)  HCO 3 - (aq) + H + (aq) The solubility of CO 2 is greatest in basic solutions; intermediate in neutral; and least in acidic. Boiling can help remove the CO 2

23. Part 1 In part 1 of this experiment, the pH of water and several salt solutions will be tested. Use pH and initial concentration of each solution to obtain approximate value of K a or K b  Approximation: extent of dissociation is small relative to initial concentration  A set of acid-base indicators will be used to determine pH

24. Testing solutions In each well: 1 drop indicator and a few drops solution Set 1:  boiled H 2 O (H 2 O)  unboiled H 2 O (H 2 O, CO 2 )  NaCl (H 2 O, CO 2 )  NH 4 Cl (NH 4 + ) Set 2:  NaC 2 H 3 O 2 (C 2 H 3 O 2 - )  ZnCl 2 (Zn(H 2 O) 6 2+ )  KAl(SO 4 ) 2 (Al(H 2 O) 6 3+ )  Na 2 CO 3 (CO 3 2- )

25. Example colors set 1 Approximate pH values: 5.8 7.0 5.8 8.1

26. Example colors set 2 Approximate pH values: 4.4 3.8 10.4

27. Results – Part 1 Liquid or SolutionHydrolysis Equation unboiled H 2 O H 2 O (l)  H + (aq) + OH - (aq) CO 2(g) + H 2 O (l)  H 2 CO 3(aq)  HCO 3 - (aq) + H + (aq) boiled H 2 O H 2 O (l)  H + (aq) + OH - (aq) 0.1 M NaCl H 2 O (l)  H + (aq) + OH - (aq) 0.1 M C 2 H 3 O 2 - 0.1 M NH 4 + 0.1 M Zn(H 2 O) 6 2+ 0.1 M Al(H 2 O) 6 3+ 0.1 M CO 3 2-

28. Part 1 results Solution Species (i.e., what ion?) K a or K b ? Exp. Value K a or K b Lit. Value of K a or K b %error 0.1 M NaClH2OH2OKwKw 0.1 M C 2 H 3 O 2 - C2H3O2-C2H3O2- KbKb See next slide 5.6  10 -10 0.1 M NH 4 + 0.1M Zn(H 2 O) 6 2+ 0.1M Al(H 2 O) 6 3+ 0.1 M CO 3 2-

29. Example – Part 1 As an example, a solution of 0.10 M NaC 2 H 3 O 2 has a pH of 8.1. It appears  It appears:yellow in methyl orange; yellow in methyl red; blue in bromothymol blue; red in phenol red; and, colorless in phenolphthalein (no need to go farther).  Since the top of the phenol red pH range is 8.0 and the bottom of the phenophthalein range is 8.2, an estimate of 8.1 is reasonable  Na + spectator ion  C 2 H 3 O 2 - conjugate base of a weak acid (HC 2 H 3 O 2 ) Hydrolysis equation:C 2 H 3 O 2 - (aq) + H 2 O (l)  HC 2 H 3 O 2 (aq) + OH - (aq)  pOH of the solution is 14.0 – 8.1 = 5.9  [OH - ] is 10 -5.9 = 1.2  10 -6 Literature K a for HC 2 H 3 O 2 is 1.8  10 -5 so expected K b is 5.6  10 -10. Notice how small differences in concentrations can lead to large ‘error’ in experimental values (vs. literature values). Remember the pH scale is logarithmic. C2H3O2-C2H3O2- +H2OH2O  HC 2 H 3 O 2 +OH - 0.10---0~0  1.2  10 -6 --- + 1.2  10 -6 ~0.10--- 1.2  10 -6

30. Part 2 – buffers Expected pH of 0.05 M HAc, Ac -, and HAc/Ac - solutions Initial concentration calculations: use M 1 V 1 = M 2 V 2  HAc:5 mL of 0.05 M of HAc to final volume of 100 mL  Ac - :5 mL of 0.05 M of Ac - to final volume of 100 mL (you do)  HAc/Ac - :5 mL each of 0.05 M of HAc and of Ac - to final volume of 100 mL (you do)

31. Part 2 – buffers Expected pH of 0.05 M HAc, Ac -, and HAc/Ac - solutions Use ICE tables to find expected equilibrium [H + ] and then pH  HAc: HAc - (aq)  H + (aq) +Ac - (aq)  Ac - :Ac - (aq) +H 2 O  OH - (aq) +HAc (aq)  HAc/Ac - :What equilibrium expression should be used? How do your experimental results compare?

32. Part 2 - buffers Theoretical pH after OH - addition  a stoichiometry problem for the neutralization look at moles added and resulting ‘initial’ concentrations  an equilibrium problem for the new concentrations use ICE tables to find expected values Which solution should show the smallest change in pH – HAc, Ac -, or HAc/Ac - ?

33. Part 2 - buffers Example, HAc after NaOH addition – if initial HAc had pH = 3.02:  Stoichimetry:HAc (aq) + OH - (aq)  Ac - (aq) + H 2 O  Equilibrium:HAc (aq)  H + (aq) + Ac - (aq) HAc - +OH -  Ac - +H2OH2O Before rxn0.005mol0.0010.000--- Change-0.001 +0.001--- After rxn~0.004~00.001--- HAc (aq)  H + (aq) +Ac - (aq 0.04~00.01 -x+x+x+x+x ~0.04x~0.01

34. Report Abstract Results including  Indicator colors of salt solutions (part 1)  Expected and actual pH of salt solutions (part 1)  Expected and actual pH of solutions in parts 2a and 2b Sample calculations  Expected and actual pH of salt solutions (part 1)  Expected and actual pH of solutions in parts 2a and 2b  Percent error* (generally will be large – what are possible reasons?). Discussion/review questions