Ch17_Weak_3 - 1 2 3 Equilibria Involving Weak Acids and...

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Unformatted text preview: 1 2 3 Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, Ka = 3.2 x 10 -4 3.2 Equilibria Involving Involving Weak Acids and Bases Acid Acid acetic, CH3CO2H ammonium, NH 4+ bicarbonate, HCO 3Conjugate Base CH3CO2-, acetate NH3, ammonia CO32-, carbonate Acid Equilibria Involving Involving Weak Acids and Bases Consider acetic acid HOAc + H2 O ¸ H3O+ + OAcOAc Conj. base Conj. Ka [H3 O+ ][OAc- ] [HOAc] 1.8 x 10-5 A weak acid (or base) is one that ionizes to a VERY small extent (< 5%). (K is designated Ka for ACID) for ACID) Because [H 3O+] and [OAc-] are SMALL, Ka << 1. [OAc << 4 Equilibria Involving Involving Weak Acids and Bases Values of Ka for acid and Kb for bases are for for found in TABLE 17.4 — page 799 Notice the relation of TABLE 17.4 to the table of relative acid/base strengths (Table 17.3). 0.0001 M 5 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the HOAc. Calc. equilibrium concs. of HOAc, H3O+, OAc-, concs. HOAc, and the pH. Step 1. Define equilibrium concs. concs. [HOAc] HOAc] initial change equilib 1.00 1.00 -x -x 1.00-x 1.00-x [H3O+] 0 0 +x +x x x [OAc-] 0 0 +x +x x x 6 0.003 M Equilibria Involving A Weak Acid Determining the pH Determining the pH of an acetic acid of an acetic acid solution. solution. See Screen 17.8. See Screen 17.8. 0.06 M 2.0 M a pH meter Note that we neglect [H 3O+] from H 2O. Page 1 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. HOAc. Calc. concs. of HOAc, H3O+, OAc-, and the pH. HOAc, 7 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. HOAc. Calc. concs. of HOAc, H3O+, OAc-, and the pH. HOAc, 8 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. HOAc. Calc. concs. of HOAc, H3O+, OAc-, and the pH. HOAc, 9 Step 2. Write K a expression expression Ka [H3O+ ][OAc- ] 1.8 x 10 -5 = [HOAc] x2 1.00 - x Step 3. Solve K a expression approximately expression approximately Ka [H3O+ ][OAc- ] 1.8 x 10 -5 = [HOAc] x2 1.00 - x Step 3. Solve K a approximate expression approximate Ka 1.8 x 10 -5 = x2 1.00 This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A). First assume x is very small because Ka is so is small. Ka 1.8 x 10 -5 = x2 1.00 x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 [K 1.00] x = [H3O+] = [OAc-] = 4.2 x 10 -3 M pH = - log [H3O+] = -log (4.2 x 10 -3) = 2.37 And so x = [ H3O+ =[ OAc- = [Ka • [K 1.00]1/2 1.00] Is approximate solution valid? Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO2H. HCO2H + H2O ¸ HCO2 - + H3O+ HCO Ka = 1.8 x 1.8 10 -4 Approximate solution [H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.37 [H 4.2 M, Exact Solution [H3O+] = [HCO2-] = 3.4 x 10-4 M [H [HCO [HCO2H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M 0.0007 pH = 3.47 10 11 Exact or Approximate Solution? Ka 1.8 x 10 -5 = [H3O+ ][OAc- ] [HOAc] x2 1.00 - x Equilibria Involving A Weak Acid Consider the approximate expression Ka 1.8 x 10 -5 = x2 1.00 x [H3O+ ] = [K a • 1 .00]1/2 12 For many weak acids When do you use the exact solution and when the approximate solution? [H3O+] = [conj. base] = [Ka • Co]1/2 [H [conj. [K where C0 = initial conc. of acid initial conc. The Great SUCO Rule of Thumb: If 100 • Ka < Co, then [H3O+] = [Ka • Co]1/2 [K Ka 1.8 x 10 -5 = x 1.00 2 Page 2 ...
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