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Tang 05 ionization + kb 2

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Transcript of "Tang 05 ionization + kb 2"

1. 1. Percent Ionization + K b
2. 2. Percent Ionization Percentage ionization may be determined if the K a value of a weak acid is known. % ionization = amount of acid ionized x 100 amount of initial acid
3. 3. Percent Ionization Example #1 What is the percentage ionization of acetic acid (HC 2 H 3 O 2 ) with a concentration of 0.20 M. K a = 1.8 x 10 -5 K a = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] HC 2 H 3 O 2(aq) <==> H + (aq) + C 2 H 3 O 2 - (aq) I 0.20M 0 0 C -x +x +x E 0.20-x x x 1.8 x 10 -5 = [x][x] [0.20-x]  use assumption 1.8 x 10 -5 = [x][x] [0.20]
4. 4. Percent Ionization Example #1 What is the percentage ionization of acetic acid (HC 2 H 3 O 2 ) with a concentration of 0.20 M. K a = 1.8 x 10 -5 HC 2 H 3 O 2(aq) <==> H + (aq) + C 2 H 3 O 2 - (aq) 1.8 x 10 -5 = [x][x] [0.20] 1.897 x 10 -3 = x % ionization = amount of acid ionized x 100% amount of initial acid % ionization = 1.897X10 -3 x 100% 0.20 % ionization = 0.94868% .: the % ionization is 0.95%
5. 5. Percent Ionization Example #1 <ul><li>What is the percentage ionization of acetic acid in solutions with concentration of: </li></ul><ul><ul><li>i) 0.010 M </li></ul></ul><ul><ul><li>ii) 0.0010 M </li></ul></ul>4.2% 13%
6. 6. Percent Ionization Example #1 What do your calculated values indicate? Percentage ionization increases as the solution of the weak acid becomes more dilute.
7. 7. Percent Ionization Example #2 The pH of a 0.10 M methanoic acid (HCO 2 H) solution is 2.38. Calculate the percent ionization of methanoic acid. HCO 2 H (aq) <==> H + (aq) + CO 2 H - (aq) I 0.10M 0 0 C -x +x +x E 0.10-x x x [H + ] = 10 -pH [H + ] = 10 –(2.38) [H + ] = 4.16869x10 –3 4.16869x10 -3 % ionization = acid ionized x 100% initial acid % ionization = 4.16869x10 -3 x 100% 0.10 % ionization = 4.2% .: the % ionization is 4.2%
8. 8. Percent Ionization Example #3 Calculate K a of acetic acid if a 0.100 M solution has a percent ionization of 1.3%. HC 2 H 3 O 2(aq) <==> H + (aq) + C 2 H 3 O 2 - (aq) K a = [0.0013][0.0013] [0.0987] I 0.100M 0 0 C -x +x +x E 0.100-x x x % ionization = acid ionized x 100% initial acid 1.3 = acid ionized x 100% 0.100 0.0013 = acid ionized 0.0013 0.0013 0.0987 K a = 1.71x10 -5 .: K a is 1.7x10 -5
9. 9. K b K b K b – dissociation constant for weak bases NH 3 + H 2 O <===> NH 4 + + OH - The larger K b , the stronger the base. The smaller K b , the weaker the base.
10. 10. K b Similar to K a values, pK b values can be calculated. The larger the pK b value, the weaker the base and the smaller the pK b , the stronger the base.
11. 11. K b How are K a and K b related? Write out the equilibrium eq n for acetic acid. Write out the equilibrium eq n for the conjugate base of acetic acid. Perform Hess’ Law. HC 2 H 3 O 2(aq) + H 2 O (l) <==> H 3 O + (aq) + C 2 H 3 O 2 - (aq) C 2 H 3 O 2 - (aq) + H 2 O (l) <==> HC 2 H 3 O 2(aq) + OH - (aq) 2 H 2 O (l) <==> H 3 O + (aq) + OH - (aq) K a x K b = K w
12. 12. K b RECALL: Equilibrium Law 2 N 2(g) + O 2(g) <=> 2 N 2 O (g) 2 N 2 O (g) + 3 O 2(g) <=> 4 NO 2(g) k eq = [N 2 O (g) ] 2 [N 2(g) ] 2 [O 2(g) ] k eq = [NO 2(g) ] 4 [N 2 O (g) ] 2 [O 2(g) ] 3 k eq = [NO 2(g) ] 4 [N 2(g) ] 2 [O 2(g) ] 4 = [NO 2(g) ] 4 [N 2(g) ] 2 [O 2(g) ] 4 What is the equilibrium law of the sum of the following reactions? K eq final = [N 2 O (g) ] 2 [N 2(g) ] 2 [O 2(g) ] [NO 2(g) ] 4 [N 2 O (g) ] 2 [O 2(g) ] 3 X
13. 13. K b RECALL: Equilibrium Law When chemical equilibria are added together, the equilibrium constants are multiplied together. K eq final rxn = K eq rxn 1 x K eq rxn 2
14. 14. K b K b Calculations Two types of calculations may also be completed: 1) Calculate the values of K b and pK b from the pH of a solution of a weak base of known initial concentration. 2) Calculate the pH of a solution where pK b and initial concentration are known.
15. 15. K b Example #4 Methylamine, CH 3 NH 2 , is one of several substances that give herring brine its pungent odor. In 0.100 M CH 3 NH 2 , the pH is 11.80. What is the K b of methylamine? CH 3 NH 2(aq) + H 2 O (l) <==> OH - (aq) + CH 3 NH 3 + (aq) pH + pOH = 14 [OH - ] = 10 -pOH pOH = 14 – 11.80 pOH = 2.2 [OH - ] = 10 -(2.2) [OH - ] = 6.30957x10 -3 I 0.100M 0 0 C -x +x +x E 0.100-x x x 0.0063 0.0063 0.0937 K b = [0.0063][0.0063] [0.0937] K b = 4.235859x10 -4 .: K b is 4.24x10 -4
16. 16. K b Example #5 Morphine is an alkaloid (an alkaline compound obtained from plants), which is a weak base. The pH of 0.010 M morphine is 10.10. Calculate K b and pK b morphine. morphine (aq) + H 2 O (l) <==> OH - (aq) + morphine-H + (aq) C 17 H 19 NO 3 I 0.010M 0 0 C -x +x +x E 0.010-x x x pOH = 14 – 10.10 pOH = 3.9 [OH - ] = 10 -pOH [OH - ] = 10 -(3.9) [OH - ] = 1.2589x10 -4 1.2589x10 -4 1.2589x10 -4 9.8741x10 -3 K b = [1.2589x10 -4 ][1.2589x10 -4 ] [9.8741x10 -3 ] K b = 1.605x10 -6
17. 17. K b Example #5 Morphine is an alkaloid (an alkaline compound obtained from plants), which is a weak base. The pH of 0.010 M morphine is 10.10. Calculate K b and pK b morphine. C 17 H 19 NO 3 morphine (aq) + H 2 O (l) <==> OH - (aq) + morphine-H + (aq) I 0.010M 0 0 C -x +x +x E 0.010-x x x 1.2589x10 -4 1.2589x10 -4 9.8741x10 -3 K b = 1.605x10 -6 pK b = -log K b pK b = -log (1.605x10 -6 ) pK b = 5.79 .: pK b = 5.8, k b = 1.6x10 -6
18. 18. K b Example #6 Calculate the values of pH, pOH and [OH - ] of a 0.20 M solution of ammonia. K b of ammonia is 1.8x10 -5 NH 3(aq) + H 2 O (l) <==> OH - (aq) + NH 4 + (aq) I 0.20M 0 0 C -x +x +x E 0.20-x x x K b = [OH - (aq) ][NH 4 + (aq) ] [NH 3(aq) ] 1.8x10 -5 = [x][x] [0.20-x] 1.8 x 10 -5 = [x][x] [0.20] 1.897 x 10 -3 = x  use assumption pOH = -log [OH - ] pOH = -log [1.897x10 -3 ] pOH = 2.72 pH = 14-2.72 pH = 11.28 .: pH = 11.3, pOH = 2.7, [OH - ] = 1.9x10 -3 M