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1 Acetic acid, has a K a of 1.7 x 10 -5. Determine the pH of a 0.10 M solution of acetic acid. Hint: First write out the equilibrium expression of the.

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Presentation on theme: "1 Acetic acid, has a K a of 1.7 x 10 -5. Determine the pH of a 0.10 M solution of acetic acid. Hint: First write out the equilibrium expression of the."— Presentation transcript:

1 1 Acetic acid, has a K a of 1.7 x Determine the pH of a 0.10 M solution of acetic acid. Hint: First write out the equilibrium expression of the acid in water. CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -

2 2 Now make a I.C.E. Chart and fill in the values CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO - I. C. E x +x 0.10-x+x Acetic acid, has a K a of 1.7 x Determine the pH of a 0.10 M solution of acetic acid.

3 3 Acetic acid, has a K a of 1.7 x Determine the pH of a 0.10 M solution of acetic acid. Create an equilibrium expression from the “E.” term. CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO - I. C. E x +x 0.10-x+x so: Try dropping the -x term. If the value of x comes out to less than 5% of 0.10 dropping the term is justified. x = 1.30 x so dropping the term was valid. since x = [H 3 O + ], pH = -log(1.30 x ) = 2.88 and pH = 14 – 2.87 = 11.12

4 4 A weak acid, HA, has a pK a of Determine the pH of a M solution of HA. Hint: First write out the equilibrium expression of the acid in water. HA + H 2 O ↔ H 3 O + + A -

5 5 A weak acid, HA, has a pK a of Determine the pH of a M solution of HA. Now make a I.C.E. Chart and fill in the values HA + H 2 O ↔ H 3 O + + A - I. C. E x +x x+x

6 6 A weak acid, HA, has a pK a of Determine the pH of a M solution of HA. Determine the K a and create an equilibrium expression from the “E.” term. HA + H 2 O ↔ H 3 O + + A - I. C. E x +x x+x Since pKa = The value of Ka = = 1.51 x so: Try dropping the -x term. If the value of x comes out to less than 5% of dropping the term is justified. x = 2.75 x so dropping the term was valid. since x = [H 3 O + ], pH = -log(2.75 x ) = 3.56

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8 8 Ammonia, has a K b of 1.8 x Determine the pH of a 0.10 M solution of ammonia. Hint: First write out the equilibrium expression of the base in water. NH 3 + H 2 O ↔ NH OH -

9 9 Ammonia, has a Kb of 1.8 x Determine the pH of a 0.10 M solution of ammonia.. Now make a I.C.E. Chart and fill in the values NH 3 + H 2 O ↔ NH OH - I. C. E x +x 0.10-x+x

10 10 Ammonia, has a Kb of 1.8 x Determine the pH of a 0.10 M solution of ammonia. Create an equilibrium expression from the “E.” term. NH 3 + H 2 O ↔ NH OH - I. C. E x +x 0.10-x+x so: Try dropping the -x term. If the value of x comes out to less than 5% of 0.10 dropping the term is justified. x = 1.34 x so dropping the term was valid. since x = [OH - ], pOH = -log(1.34 x ) = 2.87 and pH = 14 – 2.87 = 11.13

11 11 A weak base, B, has a pK b of Determine the pH of a M solution of B. Hint: First write out the equilibrium expression of the base in water. B + H 2 O ↔ BH + + OH -

12 12 A weak base, B, has a pK b of Determine the pH of a M solution of B. Now make a I.C.E. Chart and fill in the values B + H 2 O ↔ BH + + OH - I. C. E x +x x+x

13 13 A weak base, B, has a pK b of Determine the pH of a M solution of B. Determine the K b and create an equilibrium expression from the “E.” term. B + H 2 O ↔ BH + + OH - I. C. E x +x x+x Since pK b = The value of K b = = 1.51 x so: Try dropping the -x term. If the value of x comes out to less than 5% of dropping the term is justified. x = 2.75 x so dropping the term was valid. since x = [OH - ], pOH = -log(2.75 x ) = 3.56 and pH = 14 – 3.56 = 10.44

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16 16 Trimethlyamine, has a K b of 6.4 x Determine the pH of a 0.10 M solution of trimethylamine. (CH 3 ) 3 N Hint: First write out the equilibrium expression of the base in water. (CH 3 ) 3 N + H 2 O ↔ (CH 3 ) 3 NH + + OH -

17 17 Trimethlyamine, has a K b of 6.4 x Determine the pH of a 0.10 M solution of trimethylamine. (CH 3 ) 3 N Now make a I.C.E. Chart and fill in the values (CH 3 ) 3 N + H 2 O ↔ (CH 3 ) 3 NH + + OH - I. C. E x +x 0.10-x+x

18 18 Trimethlyamine, has a K b of 6.4 x Determine the pH of a 0.10 M solution of trimethylamine. (CH 3 ) 3 N Create an equilibrium expression from the “E.” term. (CH 3 ) 3 N + H 2 O ↔ (CH 3 ) 3 NH + + OH - I. C. E x +x 0.10-x+x so: Try dropping the -x term. If the value of x comes out to less than 5% of 0.10 dropping the term is justified. x = 2.53 x so dropping the term was valid. since x = [OH - ], pOH = -log(2.53 x ) = 2.60 and pH = 14 – 2.60 = 11.40

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