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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 acid in water. CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -

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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. 0.10 00 -x +x 0.10-x+x 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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3 Acetic acid, has a K a of 1.7 x 10 -5. 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. 0.10 00 -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 10 -3 so dropping the term was valid. since x = [H 3 O + ], pH = -log(1.30 x 10 -3 ) = 2.88 and pH = 14 – 2.87 = 11.12

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4 A weak acid, HA, has a pK a of 5.82. Determine the pH of a 0.050 M solution of HA. Hint: First write out the equilibrium expression of the acid in water. HA + H 2 O ↔ H 3 O + + A -

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5 A weak acid, HA, has a pK a of 5.82. Determine the pH of a 0.050 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. 0.050 00 -x +x 0.050-x+x

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6 A weak acid, HA, has a pK a of 5.82. Determine the pH of a 0.050 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. 0.050 00 -x +x 0.050-x+x Since pKa = 5.82. The value of Ka = 10 -5.82 = 1.51 x 10 -6 so: Try dropping the -x term. If the value of x comes out to less than 5% of 0.050 dropping the term is justified. x = 2.75 x 10 -4 so dropping the term was valid. since x = [H 3 O + ], pH = -log(2.75 x 10 -4 ) = 3.56

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8 Ammonia, has a K b of 1.8 x 10 -5. 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 4 + + OH -

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9 Ammonia, has a Kb of 1.8 x 10 -5. 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 4 + + OH - I. C. E. 0.10 00 -x +x 0.10-x+x

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10 Ammonia, has a Kb of 1.8 x 10 -5. 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 4 + + OH - I. C. E. 0.10 00 -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 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

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11 A weak base, B, has a pK b of 5.82. Determine the pH of a 0.050 M solution of B. Hint: First write out the equilibrium expression of the base in water. B + H 2 O ↔ BH + + OH -

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12 A weak base, B, has a pK b of 5.82. Determine the pH of a 0.050 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. 0.050 00 -x +x 0.050-x+x

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13 A weak base, B, has a pK b of 5.82. Determine the pH of a 0.050 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. 0.050 00 -x +x 0.050-x+x Since pK b = 5.82. The value of K b = 10 -5.82 = 1.51 x 10 -6 so: Try dropping the -x term. If the value of x comes out to less than 5% of 0.050 dropping the term is justified. 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

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16 Trimethlyamine, has a K b of 6.4 x 10 -5. 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 -

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17 Trimethlyamine, has a K b of 6.4 x 10-5. 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. 0.10 0 0 -x +x 0.10-x+x

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18 Trimethlyamine, has a K b of 6.4 x 10-5. 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. 0.10 00 -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 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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