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Weak acids & Bases

Weak acids & Bases. SCH 4U. Weak acids. Ionize only partially in water, exist primarily in molecule form Dynamic equilibrium established between unreacted molecules and ions formed from rxn with water Like all equilibria, can be shifted by removal/ addition of reactants or products.

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Weak acids & Bases

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  1. Weak acids & Bases SCH 4U

  2. Weak acids • Ionize only partially in water, exist primarily in molecule form • Dynamic equilibrium established between unreacted molecules and ions formed from rxn with water • Like all equilibria, can be shifted by removal/ addition of reactants or products

  3. Weak bases • Have a weak attraction for protons • Recall: the conjugate base of a strong acid is a weak base HA + H2O A– + H3O+ • Usually non-hydroxide bases (Recall Arrhenius vs. Bronsted-Lowry definitions)

  4. Percent ionization For weak acids: p = [H3O+] x 100% [HA] For weak bases: p = [OH-] x 100% [B] Ex. The pH of a 0.10 mol/L methanoic acid solution is 2.38. What is the percent ionization of methanoic acid?

  5. Percent ionization Example Ex. The pH of a 0.10 mol/L methanoic acid solution is 2.38. What is the percent ionization of methanoic acid? HCO2H(aq) H+(aq) + HCO2-(aq) [H+(aq)] = 10-pH = 10-2.38 = 4.2 x 10-3 mol/L

  6. Percent ionization Example p = conc. of acid ionized x 100% conc. of acid solute p = 4.2 x 10-3 mol/L x 100% 0.10 mol/L p = 4.2 % Therefore methanoic acid ionizes 4.2% in a 0.10 mol/L solution i.e. HCO2H(aq) H+(aq) + HCO2–(aq) 4.2 %

  7. Ionization constants (Ka) for weak acids • Equilibrium constant found as before, called “acid ionization constant,” Ka • E.g. for acetic acid: HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Ka = [H+][C2H3O2-] [HC2H3O2]

  8. Calculating KaFROM percent ionization Calculate the acid ionization constant, Ka, of acetic acid if a 0.1000 M solution at equilibrium at SATP has a percent ionization of 1.3% HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Ka = [H+][C2H3O2-] [HC2H3O2]

  9. A percent ionization of 1.3% means initial [HC2H3O2] is diminished by 1.3% by time system reaches equilibrium • Note in this example molar ratio is 1:1:1 • Use an ICE table: • Solve for x: x = 0.1 mol/L X 0.013 x = 0.0013 mol/L

  10. Now we can calculate Ka: Ka = [H+][C2H3O2-] [HC2H3O2] = (0.0013) (0.0013) (0.0987) Ka = 1.7 x 10-5

  11. Percent ionization & concentration • Ka values provide a means of comparing relative strengths of acids • Can also compare % ionization values, but only when acids are equal in initial conc. Data for acetic acid

  12. Percent ionization & concentration • More dilute the solution, greater the degree of ionization • Can explain using Le Chatelier’s Principle: HA(aq) A–(aq) + H+(aq)

  13. Ionization constants (Kb) for weak Bases • Equilibrium constant called “base ionization constant,” Kb • E.g. for ammonia: NH3(aq) + H2O(l) OH-(aq) + NH4+(aq) Kb = [OH-][NH4+] [NH3] • Note: many weak bases contain one or more N atom, others are conjugate bases of strong acids

  14. Relationship between Ka and kb • Consider an acetic acid solution at equilibrium: HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) • As a base, the acetate ion also reacts with water, establishing an equilibrium: C2H3O2-(aq) + H2O(l) OH-(aq) + HC2H3O2(aq) Ka = [H3O+][C2H3O2-] [HC2H3O2] Kb = [OH-][HC2H3O2] [C2H3O2-]

  15. Relationship between Ka and kb Ka x Kb = [H3O+] [C2H3O2-] x [OH-] [HC2H3O2] [HC2H3O2] [C2H3O2-] = [H3O+] [OH-] Kw = Ka x Kb Ka = Kw Kb Kb = Kw Ka

  16. Strength generalizations • The conjugate base of a strong acid is a ______________ base • The conjugate base of a weak acid is a ______________ base • The conjugate base of a very weak acid is a ______________ base * Go over page 562 together very weak weak strong

  17. Strength generalizations • The conjugate acid of a strong base is a ______________ acid • The conjugate acid of a weak base is a ______________ acid • The conjugate acid of a very weak base is a ______________ acid very weak weak strong

  18. Calculating conc & pH of a weak acid given Ka Calculate the hydrogen ion conc. and pH of a 0.10 mol/L acetic acid solution. Ka for acetic acid is 1.8 x 10-5. *First need to compare Ka’s of all equilibria that may contribute H+ to the system... Ka = 1.8 x 10-5Kw = 1.0 x 10-14 HC2H3O2(aq) H+(aq) + C2H3O2-(aq) I 0.10 0 0 C- x +x +x E 0.10 – x xx

  19. Ka = [H+][C2H3O2-] [HC2H3O2] 1.8 x 10-5 = x2 0.10 –x Could solve quadratic or could make it simpler... FINISH... on p.563!

  20. Finding Kb, given conc & pH For diethylamine (CH3CH2)2NH, the pH of a 2.6 x 10-2 M solution is 11.56. What is Kb for diethylamine? B(aq) + H2O(l) OH-(aq) + HB+(aq) Can use pH to det. [OH]... pH = 11.56 therefore pOH = 14.00 – 11.56 = 2.44 pOH = -log[OH-] [OH-] = 10-2.44 [OH-] = 3.6 x 10-3 M

  21. B(aq) + H2O(l) OH-(aq) + HB+(aq) I 2.6 x 10-2 0 0 C -x +x +x E 2.6 x 10-2 – x xx x = 3.6 x 10-3 Kb = [OH-][HB+] [B] Kb = (3.6 x 10-3)2 2.6 x 10-2 – 3.6 x 10-3 = 5.8 x 10-4 See summary of problem-solving steps p.574

  22. Polyprotic acids • E.g. Sulfuric acid H2SO4, boric acid H3BO3 • Different Ka values (Ka1, Ka2, etc.) • In general, Ka1 > Ka2 > Ka3 ... • See p. 574, 575 * Because Ka1 is usually >> Ka2, Ka3, etc., typically use just Ka1 to determine pH

  23. Homework • p. 579 # 3-6, 11-13

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