Ch17_Weak_3

Ch17_Weak_3 - 1 Equilibria IInvolving nvolving Weak Acids...

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