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Acetic acid, has a K a of 1.7 x 10 -5 . Determine the pH of a 0.10 M solution of acetic acid.

Acetic acid, has a K a of 1.7 x 10 -5 . Determine the pH of a 0.10 M solution of acetic acid. CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -. Hint: First write out the equilibrium expression of the acid in water.

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Acetic acid, has a K a of 1.7 x 10 -5 . Determine the pH of a 0.10 M solution of acetic acid.

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  1. Acetic acid, has a Ka of 1.7 x 10-5. Determine the pH of a 0.10 M solution of acetic acid. CH3COOH + H2O ↔ H3O+ + CH3COO- Hint: First write out the equilibrium expression of the acid in water.

  2. Acetic acid, has a Ka of 1.7 x 10-5. Determine the pH of a 0.10 M solution of acetic acid. Now make a I.C.E. Chart and fill in the values CH3COOH + H2O ↔ H3O+ + CH3COO- 0 0 0.10 I. C. E. +x +x -x 0.10-x +x +x

  3. Acetic acid, has a Ka of 1.7 x 10-5. Determine the pH of a 0.10 M solution of acetic acid. CH3COOH + H2O ↔ H3O+ + CH3COO- Create an equilibrium expression from the “E.” term. 0 0 0.10 I. C. E. +x +x -x 0.10-x +x +x Try dropping the -x term. If the value of x comes out to less than 5% of 0.10 dropping the term is justified. so: x = 1.30 x 10-3 so dropping the term was valid. since x = [H3O+] , pH = -log(1.30 x 10-3) = 2.88 and pH = 14 – 2.87 = 11.12

  4. A weak acid, HA, has a pKa of 5.82. Determine the pH of a 0.050 M solution of HA. HA + H2O ↔ H3O+ + A- Hint: First write out the equilibrium expression of the acid in water.

  5. A weak acid, HA, has a pKa of 5.82. Determine the pH of a 0.050 M solution of HA. HA + H2O ↔ H3O+ + A- Now make a I.C.E. Chart and fill in the values 0 0 0.050 I. C. E. +x +x -x 0.050-x +x +x

  6. A weak acid, HA, has a pKa of 5.82. Determine the pH of a 0.050 M solution of HA. HA + H2O ↔ H3O+ + A- Determine the Ka and create an equilibrium expression from the “E.” term. 0 0 0.050 I. C. E. +x +x -x 0.050-x +x +x Try dropping the -x term. If the value of x comes out to less than 5% of 0.050 dropping the term is justified. Since pKa = 5.82. The value of Ka = 10-5.82 = 1.51 x 10-6 so: x = 2.75 x 10-4 so dropping the term was valid. since x = [H3O+] , pH = -log(2.75 x 10-4) = 3.56

  7. Ammonia, has a Kb of 1.8 x 10-5. Determine the pH of a 0.10 M solution of ammonia. NH3 + H2O ↔ NH4+ + OH- Hint: First write out the equilibrium expression of the base in water.

  8. Ammonia, has a Kb of 1.8 x 10-5. Determine the pH of a 0.10 M solution of ammonia.. NH3 + H2O ↔ NH4+ + OH- Now make a I.C.E. Chart and fill in the values 0 0 0.10 I. C. E. +x +x -x 0.10-x +x +x

  9. Ammonia, has a Kb of 1.8 x 10-5. Determine the pH of a 0.10 M solution of ammonia. NH3 + H2O ↔ NH4+ + OH- Create an equilibrium expression from the “E.” term. 0 0 0.10 I. C. E. +x +x -x 0.10-x +x +x Try dropping the -x term. If the value of x comes out to less than 5% of 0.10 dropping the term is justified. so: x = 1.34 x 10-3 so dropping the term was valid. since x = [OH-] , pOH = -log(1.34 x 10-3) = 2.87 and pH = 14 – 2.87 = 11.13

  10. A weak base, B, has a pKb of 5.82. Determine the pH of a 0.050 M solution of B. B + H2O ↔ BH+ + OH- Hint: First write out the equilibrium expression of the base in water.

  11. A weak base, B, has a pKb of 5.82. Determine the pH of a 0.050 M solution of B. B + H2O ↔ BH+ + OH- Now make a I.C.E. Chart and fill in the values 0 0 0.050 I. C. E. +x +x -x 0.050-x +x +x

  12. A weak base, B, has a pKb of 5.82. Determine the pH of a 0.050 M solution of B. B + H2O ↔ BH+ + OH- Determine the Kb and create an equilibrium expression from the “E.” term. 0 0 0.050 I. C. E. +x +x -x 0.050-x +x +x Try dropping the -x term. If the value of x comes out to less than 5% of 0.050 dropping the term is justified. Since pKb = 5.82. The value of Kb = 10-5.82 = 1.51 x 10-6 so: x = 2.75 x 10-4 so dropping the term was valid. since x = [OH-] , pOH = -log(2.75 x 10-4) = 3.56 and pH = 14 – 3.56 = 10.44

  13. Trimethlyamine, has a Kb of 6.4 x 10-5. Determine the pH of a 0.10 M solution of trimethylamine. (CH3)3N (CH3)3N + H2O ↔ (CH3)3NH+ + OH- Hint: First write out the equilibrium expression of the base in water.

  14. Trimethlyamine, has a Kb of 6.4 x 10-5. Determine the pH of a 0.10 M solution of trimethylamine. (CH3)3N (CH3)3N + H2O ↔ (CH3)3NH+ + OH- Now make a I.C.E. Chart and fill in the values 0 0 0.10 I. C. E. +x +x -x 0.10-x +x +x

  15. Trimethlyamine, has a Kb of 6.4 x 10-5. Determine the pH of a 0.10 M solution of trimethylamine. (CH3)3N (CH3)3N + H2O ↔ (CH3)3NH+ + OH- Create an equilibrium expression from the “E.” term. 0 0 0.10 I. C. E. +x +x -x 0.10-x +x +x Try dropping the -x term. If the value of x comes out to less than 5% of 0.10 dropping the term is justified. so: x = 2.53 x 10-3 so dropping the term was valid. since x = [OH-] , pOH = -log(2.53 x 10-3) = 2.60 and pH = 14 – 2.60 = 11.40

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