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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 105 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 1.00x 1.00x [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 104 M, pH = 3.37 [H 4.2 M, Exact Solution [H3O+] = [HCO2] = 3.4 x 104 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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 Fall '08
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