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Chemistry:

A 1.00 L solution is prepared by placing 0.10 moles of a weak acid (HA) and 0.050 moles of its conjugate base NaA. Determine the pH of the solution. The Ka of the HA is 1.4 x 10 -5. Chemistry:. HA + H 2 O ↔ H 3 O + + A -. I. C. E. 0.10 0 0.050.

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Chemistry:

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  1. A 1.00 L solution is prepared by placing 0.10 moles of a weak acid (HA) and 0.050 moles of its conjugate base NaA. Determine the pH of the solution. The Ka of the HA is 1.4 x 10-5 . Chemistry: HA + H2O ↔ H3O+ + A- I. C. E. 0.10 0 0.050 -x +x +x 0.10-x x 0.050+x Try dropping the +x and the –x terms (5 % rule). X = [H3O+] = 2.8 x 10-5 pH = -log[x] = 4.55

  2. A solution is 0.095 M in ascorbic acid and 0.055 M in sodium hydrogen ascorbate. Determine the pH of the solution. Ka1 of the H2C6H6O6 is 6.8 x 10-5, Ka2 = 2.8 x10-12 . Chemistry: H2C6H6O6 + H2O ↔ H3O+ + HC6H6O6- I. C. E. 0.095 0 0.055 -x +x +x 0.095-x x 0.055+x Try dropping the +x and the –x terms (5 % rule). X = [H3O+] = 1.17 x 10-4 pH = -log[x] = 3.95

  3. A 1.00 L solution is prepared by placing 0.10 moles of a weak base (B) and 0.050 moles of its conjugate acid BHCl. Determine the pH of the solution. The Kb of the B is 1.4 x 10-5 . Chemistry: B + H2O ↔ OH- + BH+ I. C. E. 0.10 0 0.050 -x +x +x 0.10-x x 0.050+x Try dropping the +x and the –x terms (5 % rule). X = [OH-] = 2.8 x 10-5 pOH = -log[x] = 4.55 and pH = 9.45

  4. A solution is prepared that has a trimethlyamine, (CH3)3N, concentration of 0.100 M and a concentration of (CH3)3NHCl of 0.050 M. Determine the pH of the solution. The Kb of trimethylamine =6.4 x 10-5. Chemistry: (CH3)3N + H2O ↔ (CH3)3NH+ + OH- I. C. E. 0.100 0.050 0 -x +x +x 0.100-x 0.050+x +x Try dropping the +x and the –x terms (5 % rule). X = [OH-] = 1.28 x 10-5 pOH = -log[x] = 3.89 and pH = 10.11

  5. Determine the Ka of a 0.10 M acid solution that has a pH of 3.45. Chemistry HA + H2O ↔ H3O+ + A- 0.10 NA 0 0 I. C. E. -x +x +x 0.10 – x x x Since x = [H3O+] and pH = 3.45; [H3O+] = 10-pH = 3.548 x 10-4

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