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

2. Acids and Bases The Common-Ion Effect “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.” HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq) Add strong electrolyte 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. 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) HC 2 H 3 O 2(aq) + NaC 2 H 3 O 2 (s) (3) HF (aq) + HCl (aq) NaC 2 H 3 O 2

3. Acids and Bases The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is 6.8  10 −4. [H 3 O + ] [F − ] [HF] K a = = 6.8  10 -4 Problem: Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F − ], M Initially0.200.100 Change−x−x+x+x+x+x At Equilibrium 0.20 − x  0.200.10 + x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq)

4. Acids and Bases The Common-Ion Effect X =X = x = 1.4  10 −3 (0.10+x) (x) (0.20-x) = 6.8  10 −4 = (0.20) (6.8  10 −4 ) (0.10) So,pH = −log (0.10) pH = 1.00 [H 3 O + ] [F − ] [HF] K a = Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = 0.10 + x = 0.10 + 1.4  10 −3 = 0.10 M What is [F − ] ? And pH?

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7. Acids and Bases 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 salt of its conjugate base (a weak conjugate acid- base pair solution). i.e., acetic acid + sodium acetate HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq) [ ] Which is less acidic, buffer (WA + Salt of c.base) or same conc. of weak acid alone?

8. Acids and Bases Buffers 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. [HF] + H 2 O H 3 O + + [F - ] Add acid (H 3 O + )Add base (OH - ) OH - + HF  H 2 O + F-HF  H 3 O + + F-

9. Acids and Bases pH of Buffer Calculations: The Henderson–Hasselbalch Equation Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log For the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A − pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] Henderson–Hasselbalch Equation.

10. Acids and Bases Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a = 1.4  10 −4. pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH = 3.85 + (−0.08) pH = 3.77 Problem: [H 3 O + ] [A − ] [HA] K a =

11. Acids and Bases pH Range of Buffers Systems 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 pK a close to the desired pH. pH = pK a + log [base] [acid] pK a 3. 4. 7. 9. pH = pK a + log [1M] pH = pK a

12. Acids and Bases [H 3 O + ] [A − ] [HA] K a = pH = pK a + log [base] [acid] [ ]

13. Acids and Bases pH= pK a + log [base] [acid]

14. Acids and Bases 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 [HX] + H 2 OH 3 O + + [X − ] Strong acid (H 3 O + ) Strong base (OH - )

15. Acids and Bases Addition of Strong Acid or Base to a Buffer (1)Neutralization is NOT an equilibrium reaction. (2)However, it affects the amounts of the weak acid [HA] and its conjugate base [A - ] left over, and a new equilibrium will be established. pH = pK a + log [base] [acid] [HA] + H 2 OH 3 O + + [A − ] Mole (η)HAA-A- H 3 O + or OH - Initially ηη+ η Change (H 3 O + or OH - ) + η+ η+ η+ η- η Add acid (H 3 O + ) Add base (OH - ) End (after)reaction ηη0 η (1) Use: Mole ICEnd Table: * used for neutralization Rx * Moles (η)  Molarity (η/L) Add some OH - Add some H 3 O + (HA + OH -  A - + H 2 O) (H 3 O + + A -  HA + H 2 O) What are the new equilibrium conc. after acid or base is added? (2) For a buffer:

16. Acids and Bases Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 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 HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74 The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: MoleHC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Initially 0.300 η 0.020 η Change- 0.020 η+ 0.020 η- 0.020 η HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O (l) Problem: End (after) reaction 0.280 η 0.320 η 0.000 η Use: Mole ICEnd Table: *used for neutralization Rx* pH = pK a + log [base] [acid]

17. Acids and Bases Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0.280) pH = 4.74 + 0.058 pH = 4.80 pH = pK a + log [base] [acid] MoleHC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Initially 0.300 η 0.020 η Change- 0.020 η+ 0.020 η- 0.020 η End (After) reaction 0.280 η 0.320 η 0.000 η

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19. Acids and Bases Hydrolysis hydro = water lysis = breaking down Organic/Biochemistry (enzymatic splitting of organic molecules by using water) hydrolysis dehydration synthesis Electrochemistry (also called electrolysis or electrohydrolysis) Acid Base Chemistry (splitting water by cations or anions to form acidic or basic solutions. 2 H 2 O 2 H 2 + O 2 elect. Cation + + HOH CationOH + H + Anion - + HOH HAnion + OH - Module 8

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

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

22. Acids and Bases 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, NH 4 +, Ca 2+, Ba 2+, and Sr 2+ (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. Acids and Bases K a of Weak Acid and K b of its Conjugate Base The K a of an acid and K b of its conjugate base are related in this way:K a  K b = K w Therefore, if you know one of them, you can calculate the other. Problem: Is NaC 2 H 3 O 2 acidic or basic? Calculate its K a or K b. K a  K b = K w (1.8 x 10 -5 )  K b = (1.0 x 10 -14 ) K b = (1.0 x 10 -14 ) (1.8 x 10 -5 ) = 5.6 x 10 -10

24. Acids and Bases Effect of Cations and Anions 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 K a and K b values. Cations of the strong bases will not affect the pH SOLUBILITY RULES: for Ionic Compounds (Salts) All OH - are insoluble, except for IA metals, NH 4 +, Ca 2+, Ba 2+, and Sr 2+ (heavy IIA). Cations of a weak bases will lower the pH (behave as acids) CATION ANION

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27. Acids and Bases Acidic and Basic Salts Acid Salts: salts of weak polyprotic acids. Are not necessarily acidic, but do neutralize bases. Basic Salts: salts of weak polyhydroxy bases. Are not necessarily basic, but do neutralize acids. Examples: AcidSaltAcid Salt(s) HNO 3 NaNO 3 ----------- H 2 CO 3 Na 2 CO 3 NaHCO 3 H 3 PO 4 Na 3 PO 4 Na 2 HPO 4 NaH 2 PO 4 Solubility Rule: All OH - are insoluble except for IA metals, NH 4 + & slightly soluble Ca 2+ Ba 2+ & Sr 2+ Mg(OH) 2(s) Fe(OH) 3(s) Cr(OH) 3(s) {Milk of Magnesia}

28. Acids and Bases Buffers can act as either…acids or bases…...they are amphiprotic. HCO 3 − (aq) H 2 PO 4 − (aq) H 2 O (l) ↔ CO 3 2- + H 2 O H 2 O + H 2 CO 3 ↔ H 2 O + H 3 PO 4 ↔ H 2 O + H 3 O + ↔ ↔ HPO 4 2- + H 2 O ↔ OH - + H 2 O accepts protons /donates protons OH - H3O+H3O+ Add acidAdd base Not a buffer

29. Acids and Bases 2007 (B)

30. Acids and Bases Making Solution Molarity

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

32. Acids and Bases 2005A Q1

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34. Acids and Bases Review of Strong Acid & Bases Strong acids have very weak conjugate bases HX + H 2 O  H 3 O + + X - acid base conj. a conj. b Strong bases have very weak conjugate acids MOH  M + + OH - base conj. a conj. b Salts of Strong acids & bases have very weak conjugate acids & bases MX + H 2 O  M + (aq) + X - (aq) salt conj. a conj. b Has no affinity for H + Has no affinity for OH - Have no affinity for H + or OH - Example: HCl Example: NaOH Example: NaCl

35. Acids and Bases 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 + H 3 O + (hydrolysis) (3) of a weak base & weak acid: BHA ↔ BH + (aq) + A - (aq) Weak acids (HA) have very strong conjugate bases (A - ) HA + H 2 O ↔ H 3 O + + A - acid base conj. a conj. b Review of Weak Acid & Bases Weak bases (B) have very strong conjugate acids (BH + ) B + HOH ↔ BH + + OH - base acids conj. a conj. b Has affinity for H + Has affinity for OH - Example: HF Example: NH 3 Example: NaF Example: NH 4 Cl Example: NH 4 F

36. Acids and Bases Reactions of Cations with Water 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. Cations with acidic protons (like NH 4 + ) will lower the pH of a solution: NH 4 + + H 2 O  H 3 O + + NH 3 O H H O H H Greater charge and smaller size make a cation more acidic.

37. Acids and Bases conjugate The Effect of Cations