83 Marks:1 What is the pH of an aqueous solution made by combining 39.61 mL of a
ID: 1035934 • Letter: 8
Question
83 Marks:1 What is the pH of an aqueous solution made by combining 39.61 mL of a 0.4590 M sodium acetate with 44.27 mL of a 0.3816 M solution of acetic acid to which 2.742 mL of a 0.0643 M solution of NaOH was added? Answer: 84 Marks: 1 What is the pH of an aqueous solution made by combining 40.76 mL of a 0.3965 M sodium acetate with 39.69 mL of a 0.3907 M solution of acetic acid? Answer: 85 Marks: 1 What is the pH of an aqueous solution made by combining 14.88 mL of a 0.1062 M hydrochloric acid with 39.38 mL of a 0.3615 M solution of ammonia? Answer: 86 Marks: 1 In a titration of 38.06 mL of 0.3840 M ammonia with 0.4636 M aqueous nitric acid, what is the pH of the solution when 5.00 mL of the acid have been added? Answer: 87 Marks: 1 In a titration of 38.23 mL of 0.3183 M ammonia with 0.4364 M aquecus nitric acid, what is the pH of the solution when 38.23 mL 10.00 mL of the acid have been added? Answer: In a titration of 45.58 mL of 0.3468 M ammonia with 0.3468 M aqueous nitric acid, what is the pH of the solution when 45.58 mL of the acid have been added? Marks: 1 Answer:Explanation / Answer
1) initial pH
Using Hendersen-Haselbalck equation,
pH = pKa + log(NaC2H3O2]/[C2H3O2]
feeding the given values,
pH = 4.74 + log(0.4590 M x 39.61 ml/0.3816 M x 44.27 ml)
= 4.772
after addition of NaOH = 0.0643 M x 2.742 ml = 0.18 mmoles
pH = 4.74 + log[(0.4590 M x 39.61 ml + 0.18)/(0.3816 M x 44.27 ml- 0.18)]
= 4.781
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2. pH
Using Hendersen-Haselbalck equation,
pH = pKa + log(NaC2H3O2]/[C2H3O2]
feeding the given values,
pH = 4.74 + log(0.3965 M x 40.76 ml/0.3907 M x 39.69 ml)
= 4.760
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3. moles NH3 = 0.3615 M x 39.38 ml = 14.235 mmol
moles HCl = 0.1062 M x 14.88 ml = 1.580 mmol
moles NH4+ formed = 1.580 mmol
moles NH3 remianed = 12.656 mmol
Using Hendersen-Haselbalck equation,
pH = pKa + log(NH3/NH4+)
feeding the given values,
pH = 9.25 + log(12.656/1.580)
= 10.154
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4. moles NH3 = 0.3840 M x 38.06 ml = 14.615 mmol
moles HNO3 = 0.4636 M x 5 ml = 2.318 mmol
moles NH4+ formed = 2.318 mmol
moles NH3 remained = 12.297 mmol
Using Hendersen-Haselbalck equation,
pH = pKa + log(NH3/NH4+)
feeding the given values,
pH = 9.25 + log(12.297/2.318)
= 9.975
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5. moles NH3 = 0.3183 M x 38.23 ml = 12.17 mmol
moles HNO3 = 0.4364 M x 48.23 ml = 21.05 mmol
excess [H+] = 8.88 mmol/86.46 ml = 0.103 M
pH = -log(0.103) = 0.99
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6. Equivalence point
[NH4+] formed = 0.3468 M x 45.58 ml/91.16 ml = 0.1734 M
NH4+ + H2O <==> NH3 + H3O+
let x amount hydrolyzed
Ka = Kw/Kb = [NH3][H3O+]/[NH4+]
1 x 10^-14/1.8 x 10^-5 = x^2/0.1734
x = [H3O+] = 9.81 x 10^-6 M
pH = -log[H3O+] = 5.01
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