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# Ab Lec2

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### Ab Lec2

1. 1. Auto Ionization of water 2H2O (l)  H3O+ + OH- KW = [H3O+][OH-] = 1.0 x 10-14 (ion product constant for water at 250)  NEUTRAL SOL’N: [H3O+] = [OH-]  ACIDIC SOL’N: [H3O+] > [OH-]  BASIC SOL’N: [H3O+] < [OH-]  What is the concentrations of H3O+ & OH- in neutral solutions?
2. 2. Exercises  Indicate whether the solution with the following ion concentration is neutral, acidic, or basic.  [H+] = 4 x 10-9  [OH-] = 1.01 x 10-7  [OH-] = 7 x 10-13  Given the following concentrations of OH-; determine [H+].  2 x 10-6  0.010  1.8 x 10-9
3. 3. Ionization of Strong Acids and Bases  Strong acids and bases completely ionize in aqueous solution HA + H2O  H3O+ + A- B + H2O  OH- + BH+  If [HNO3] = 0.05 M, [H+] = ?  If [Mg(OH)2] = 0.25 M, [OH-] = ? Consider the mole ratio of acid to H+ and base to OH-
4. 4. pH Box Inter-converts pH to pOH to [OH-] to [H+] and to pH again.
5. 5. Examples [H+] [OH-] pH pOH 0.012 M HCl 0.012 8.3 e -13 1.92 12.08 0.082 M NaOH 1.2 e-13 0.082 12.91 1.09 0.37 M 1.4e-14 0.74 13.87 0.13 Ba(OH)2
6. 6. Exercise  If a lemon juice has a concentration of 3.8 x 10-4 M, what is its pH?  A feminine wash has a pH of 5.5, what is the concentration of H3O+ present in this soap?
7. 7. Ionization of Weak Acids & Bases  Weak acids and bases partially ionize  The extent of dissociation depends on KA and KB of each acid and base.  KA = acid ionization constant For the reaction: HA + H2O  H3O+ + A-
8. 8. Ionization of Weak Acids & Bases  KB = base ionization constant For the reaction: XOH (aq)  OH- + X+ % Ionization = ( conc. Of H+ / conc of HA) x 100 = ( conc. Of OH-/ conc of XOH x 100)
9. 9. pKA Box
10. 10. Calculating KA from pH  A student prepared a 0.10 M solution of formic acid & measured its pH using a pH meter. The pH at 250C was found to be 2.38.  Calculate KA for formic acid at this temperature  What is the % ionization of the solution?
11. 11. Step by Step Solution  Write the reaction, KA expression and the equation  Calculate [H+] from pH  Tabulate (ICE)  Substitute Answer: KA = 1.8 x 10-4 % I = 4.2%
12. 12. Using KA to calculate pH  Calculate the pH of 0.3M solution of acetic acid at 250C given KA = 1.8 x 10-5
13. 13. Step by Step Solution 1. Write the reaction 2. Write the dissociation constant expression and its value 3. Express the concentrations of the unknown using the table (ICE) 4. Substitute the equilibrium conc’n into the expression Answer: pH = 2.64