Unit 18 Acid-Base Equilibria: Buffers & Hydrolysis

1. CHM 1046: General Chemistry and Qualitative Analysis Unit 18Acid-Base Equilibria: Buffers & Hydrolysis Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL • Textbook Reference: • Chapter 19 (sec. 1-4) • Modules # 7-8

2. HC2H3O2(aq)+ H2O(l) H3O+ (aq) + C2H3O2−(aq) The Common-Ion Effect A solution composed of two substances, each containing a same ion in common. One is a weak electrolyte (equilibrium) the other strong(not equil). Examples: (1) HF(aq) + NaF(s) (2) HC2H3O2(aq) + NaC2H3O2 (s) (3) HF(aq) + HCl(aq) “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” Add strong electrolyte NaC2H3O2 Consider a solution of acetic acid, a weak electrolyte: What will Le Châtelier predict will happen to the equilibrium? It will shift to the left.

3. Ka = [H3O+] [F−] [HF] HF(aq) + H2O(l) H3O+(aq) + F−(aq) = 6.8  10-4 The Common-Ion Effect Problem: Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.Ka for HF is 6.8  10−4. Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

4. Ka = [H3O+] [F−] [HF] (0.10+x) (x) (0.20-x) = 6.810−4 = (0.20) (6.8  10−4) (0.10) The Common-Ion Effect X= x = 1.4  10−3 What is [F−] ? And pH? Therefore, [F−] = x = 1.4  10−3 [H3O+] = 0.10 + x= 0.10 + 1.4  10−3= 0.10 M • So, pH = −log (0.10) • pH = 1.00

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7. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Buffers • Solutions that are particularly resistant to pH changes, even when strong acid or base is added. • Buffers are solutions of a weak acid mixed with a saltof its conjugate base(a weak conjugate acid-base pair solution). i.e., acetic acid + sodium acetate [ ] [ ] Which is less acidic, buffer (WA + Salt of c.base) or same conc. of weak acid alone?

8. Add base (OH-) Add acid (H3O+) Buffers [HF] + H2O H3O+ + [F- ] OH- + HF  H2O + F- HF  H3O+ + F- If an acid is added, the F− reacts with H+ to form HF and water. If a hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

9. Ka = [H3O+] [A−] [HA] [base] [acid] pKa = pH − log [A−] [HA] [H3O+] Ka = [base] [acid] HA + H2O H3O+ + A− pH = pKa + log [A−] [HA] −log [H3O+] + −log −log Ka = pH of Buffer Calculations: The Henderson–HasselbalchEquation For the dissociation of a generic acid, HA: Rearranging slightly, this becomes Taking the negative log of both side, we get Rearranging, this becomes  Henderson–HasselbalchEquation.

10. Ka = [H3O+] [A−] [HA] [base] [acid] (0.10) (0.12) pH = pKa + log pH = −log (1.4  10−4) + log Henderson–Hasselbalch Equation Problem: What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka = 1.4  10−4. pH = 3.85 + (−0.08) pH = 3.77

11. [base] [acid] [1M] [1M] pH = pKa + log pH = pKa + log pH Range of Buffers Systems pKa 3. 3. 4. 4. 7. 9. 9. • The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH. pH = pKa

12. Ka = [H3O+] [A−] [HA] [base] [acid] pH = pKa + log [ ] [ ]

13. pH= pKa + log [base] [acid]

14. [HX] + H2O H3O+ + [X−] When Strong Acids or Bases Are Added to a Buffer… You end up with two problems: (1) an acid base neutralization, and (2) an equilibrium problem Strong base (OH-) Strong acid (H3O+)

15. [base] [acid] [HA] + H2O H3O+ + [A−] pH = pKa + log Addition of Strong Acid or Base to a Buffer Add base (OH-) Add acid (H3O+) Add some OH- (HA + OH-A-+ H2O) (H3O+ + A-HA+ H2O) • Neutralization is NOT an equilibrium reaction. • However, it affects the amounts of the weak acid [HA] and its conjugate base [A-] left over, and a new equilibrium will be established. What are the new equilibrium conc. after acid or base is added? (1) Use: MoleICEnd Table: *used for neutralization Rx* Add some H3O+ Moles (η) Molarity (η/L) For a buffer: (2)

16. [base] [acid] pH = pKa + log Calculating pH Changes in Buffers Problem: HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l) A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8  10−5) = 4.74 The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: Use: MoleICEnd Table: *used for neutralization Rx*

17. (0.320) (0.280) [base] [acid] pH = pKa + log pH = 4.74 + log Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.058 pH = 4.80

19. Hydrolysishydro = water lysis = breaking down Module 8 Electrochemistry(also called electrolysis or electrohydrolysis) elect. 2 H2O 2 H2 + O2 Organic/Biochemistry(enzymatic splitting of organic molecules by using water) hydrolysis dehydration synthesis Acid Base Chemistry(splitting water by cations or anions to form acidic or basic solutions. Cation+ + HOHCationOH + H+ Anion- + HOHHAnion + OH-

20. The Acid-Base Properties of Salt Solutions and Hydrolysis Salt (Ionic Compound) = Metal + Nonmetal = Cation+ + Anion- Hydrolysis: splitting water by cations or anions to form acidic or basic solutions Cation+ + HOHCationOH + H+ Cations of Weak Bases Al 3+ Al3+ Anion- + HOHHAnion + OH- Anions of Weak Acids

21. conjugate The Effect of Anions • An anion that is the conjugate base of a strong acid will not affect the pH. • An anion that is the conjugate base of a weak acid will increase the pH (behave as bases).

22. Effect of Cations • Cations of the strong bases will not affect the pH SOLUBILITY RULES: for Ionic Compounds (Salts) All OH- are insoluble, except for IA metals, NH4+, Ca2+, Ba2+ , and Sr2+ (heavy IIA). • Cations of a weak bases will lower the pH (behave as acids) • Greater charge and smaller size make a cation more acidic.

23. Ka of Weak Acid andKb of its Conjugate Base The Ka of an acidandKb of its conjugate base are related in this way: Ka Kb = Kw Therefore, if you know one of them, you can calculate the other. Problem: Is NaC2H3O2 acidic or basic? Calculate its Ka or Kb. = 5.6 x 10-10 Ka Kb = Kw (1.0 x 10-14) Kb = (1.8 x 10-5)  Kb = (1.0 x 10-14) (1.8 x 10-5)

24. Effect of Cations and Anions ANION CATION • Cations of the strong bases will not affect the pH When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Kaand Kbvalues. SOLUBILITY RULES: for Ionic Compounds (Salts) All OH- are insoluble, except for IA metals, NH4+, Ca2+, Ba2+ , and Sr2+ (heavy IIA). • Cations of a weak bases will lower the pH (behave as acids)

27. Acidic and Basic Salts Acid Salts: salts of weak polyprotic acids. Are not necessarily acidic, but do neutralize bases. Examples: AcidSaltAcid Salt(s) HNO3 NaNO3 ----------- H2CO3 Na2CO3 NaHCO3 H3PO4 Na3PO4 Na2HPO4 NaH2PO4 Basic Salts: salts of weak polyhydroxy bases. Are not necessarily basic, but do neutralize acids. Mg(OH) 2(s) Fe(OH)3(s) Cr(OH) 3(s) {Milk of Magnesia} • Solubility Rule: • All OH- are insolubleexcept for IA metals, NH4+ & slightly soluble Ca 2+Ba2+ & Sr2+

28. Buffers can act as either…acids or bases… ...they are amphiprotic . HCO3−(aq) H2PO4−(aq) H2O(l) Add acid Add base H3O+ OH- ↔CO32- + H2O H2O + H2CO3↔ H2O +H3PO4↔ ↔HPO42-+H2O ↔OH- +H2O H2O +H3O+↔ Not a buffer accepts protons / donates protons

29. 2007 (B)

30. Making Solution Molarity

31. 2007B Q5 Carboxylic acid Organic Acids: carboxylic acid functional group -COOH  -COO- + H+

32. 2005A Q1

34. Review of Strong Acid & Bases Strong acidshave very weak conjugate bases HX + H2O H3O+ + X- acid base conj. a conj. b Has no affinity for H+ Example: HCl Strong baseshave very weak conjugate acids MOH  M+ + OH- baseconj. a conj. b Has no affinity for OH- Example: NaOH Salts of Strong acids & baseshave very weak conjugate acids & bases MX + H2O M+(aq) + X- (aq) salt conj. a conj. b Example: NaCl Have no affinity for H+ orOH-

35. Review of Weak Acid & Bases Weak acids(HA) have very strong conjugate bases (A-) HA + H2O↔ H3O+ + A- acidbase conj. a conj. b Has affinity for H+ Example: HF Salts: (1) of strong base & weak acid: MA  M+(aq) + A-(aq) A- + HOH  HA + OH- (hydrolysis) (2) of a weak base & strong acid: BHX  BH+(aq) + X-(aq) BH+ + HOH  B + H3O+ (hydrolysis) (3) of a weak base & weak acid: BHA ↔ BH+(aq) + A-(aq) Weak bases(B) have very strong conjugate acids (BH+) B + HOH↔ BH+ + OH- baseacidsconj. a conj. b Example: NH3 Has affinity for OH- Example: NaF Example: NH4Cl Example: NH4F

36. Reactions of Cations with Water • Cations with acidic protons (like NH4+) will lower the pH of a solution: NH4+ + H2O  H3O+ + NH3 H O • Most metal cations that are hydrated in solution also lower the pH of the solution, acting as Lewis acids: • The attraction between nonbonding electrons on waters’ oxygen and the metal cation causes a shift of the electron density in water. • This makes the O-H bond more polar and the water more acidic. H H O H Greater charge and smaller size make a cation more acidic.

37. conjugate The Effect of Cations