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Acids Lesson 12 Calculating Ka From pH

Acids Lesson 12 Calculating Ka From pH. 1. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. 1. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + + HC 2 O 4 - I 0.100 0 0 C E.

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Acids Lesson 12 Calculating Ka From pH

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  1. Acids Lesson 12 Calculating Ka From pH

  2. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid.

  3. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C E

  4. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C E pH = 1.28

  5. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C E pH = 1.28 [H+] = 10-1.28

  6. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C E pH = 1.28 [H+] = 10-1.28 [H+] = 0.05248 M

  7. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C E0.052480.05248 pH = 1.28 [H+] = 10-1.28 [H+] = 0.05248 M

  8. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C - 0.052480.05248 0.05248 E 0.04752 0.052480.05248 pH = 1.28 [H+] = 10-1.28 [H+] = 0.05248 M

  9. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C - 0.052480.05248 0.05248 E 0.04752 0.052480.05248 pH = 1.28 [H+] = 10-1.28 [H+] = 0.05248 M Ka = [H+][HC2O4-] [H2C2O4]

  10. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C - 0.052480.05248 0.05248 E 0.04752 0.052480.05248 pH = 1.28 [H+] = 10-1.28 [H+] = 0.05248 M Ka = [H+][HC2O4-] = (0.05248)2 [H2C2O4] 0.04752

  11. 1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. H2C2O4⇄ H+ + HC2O4- I 0.100 0 0 C - 0.052480.05248 0.05248 E 0.04752 0.052480.05248 pH = 1.28 [H+] = 10-1.28 [H+] = 0.05248 M Ka = [H+][HC2O4-] = (0.05248)2 = 5.8 x 10-2 [H2C2O4] 0.04752

  12. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb.

  13. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄NH4+ + OH- I 0.40 0 0 C E

  14. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄NH4+ + OH- I 0.40 0 0 C E pH = 11.427

  15. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C E pH = 11.427 pOH = 2.573

  16. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C E pH = 11.427 pOH = 2.573 [OH-] = 10-2.573

  17. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C E pH = 11.427 pOH = 2.573 [OH-] = 10-2.573 [OH-] = 0.002673 M

  18. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C E0.0026730.002673 pH = 11.427 pOH = 2.573 [OH-] = 10-2.573 [H+] = 0.002673 M

  19. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C - 0.0026730.002673 0.002673 E0.39730.0026730.002673 pH = 11.427 pOH = 2.573 [OH-] = 10-2.573 [OH-] = 0.002673 M

  20. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C - 0.0026730.002673 0.002673 E0.39730.0026730.002673 pH = 11.427 pOH = 2.573 [OH-] = 10-2.573 [OH-] = 0.002673 M Kb = [NH4+][OH-] [NH3]

  21. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C - 0.0026730.002673 0.002673 E0.39730.0026730.002673 pH = 11.427 pOH = 2.573 [OH-] = 10-2.573 [OH-] = 0.002673 M Kb = [NH4+][OH-] = (0.002673)2 [NH3] 0.3973

  22. 2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb. NH3 + H2O ⇄ NH4+ + OH- I 0.40 0 0 C - 0.0026730.002673 0.002673 E0.39730.0026730.002673 pH = 11.427 pOH = 2.573 [OH-] = 10-2.573 [OH-] = 0.002673 M Kb = [NH4+][OH-] = (0.002673)2 = 1.8 x 10-5 [NH3] 0.3973

  23. 3. The pH of a 1.0 M triprotic weak acid is 4.568. Calculate the Ka and identify the acid. Ka = 7.3 x 10-10 Boric acid

  24. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.

  25. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH-

  26. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I 0.20 0 0

  27. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I 0.20 0 0 C 0.002858 0.002858 0.002858

  28. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I 0.20 0 0 C 0.002858 0.002858 0.002858 E 0.1971 0.002858 0.002858

  29. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I 0.20 0 0 C 0.002858 0.002858 0.002858 E 0.1971 0.002858 0.002858 [OH-] = 0.002858 M

  30. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I 0.20 0 0 C 0.002858 0.002858 0.002858 E 0.1971 0.002858 0.002858 [OH-] = 0.002858 M Kb = [HCN][OH-] = [CN-]

  31. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I 0.20 0 0 C 0.002858 0.002858 0.002858 E 0.1971 0.002858 0.002858 [OH-] = 0.002858 M Kb = [HCN][OH-] = (0.002858)2 = 4.1 x 10-5 [CN-] 0.1971

  32. 5. Calculate the pH of 0.020 M H3BO3 H3BO3⇄H+ + H2BO3- I 0.020 M 0 0 C x x x E 0.020 - x x x 0 small ka x2 = 3.8 x 10-10 0.020 x = [H+] = 2.76 x 10-6 M pH = -Log[2.76 x 10-6] pH = 5.42 2 sig figs due to molarity and Ka

  33. 6. Calculate the pH of a solution made by mixing 100.0 mL of 0.050 M HCl with 100 mL of water. HCl → H+ + Cl- 1(0.050 M) 0.025 M 0.025 M 2 pH = 1.60

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