Explanation Formula used: p H = − log [ H + ] Moles = V 1 M 1 =V 2 M 2 Calculation: Concentration of HC 2 H 3 O 2 is 0.105 M a) Initial pH H C 2 H 3 O 2 = C 2 H 3 O 2 − ( H + ) Let x amount has dissociated K a = [ C 2 H 3 C O 2 − ] [ H + ] [ H C 2 H 3 O 2 ] 1.8 × 10 − 5 = X 2 / 0.105 x = [ H + ] = 1.37 × 10 − 3 M p H = − log [ H + ] = 2.86 b) The volume of added base required to reach the equivalencepoint’s ( 0.105 M × 20 M L ) / 0.125 M = 16.8 m l Of NaOH S o, the volume ofadded base required to reach equivalence point is 16.8ml c) The pH after 5 ml of NaOH added Moles of acid = 0.105 M × 20 ml = 2.1 mmol. Moles of base added = 0.125 M × 5 ml =0.625 mmol. Excess [ H C 2 H 3 O 2 ] = 1.475 mmol/25 ml = 0.059 M Formed [ C 2 H 3 O 2 N a ] = 0.625 mmol/25 ml = 0.025 p H = p K a + log ( b a s e / a c i d ) = 4...

Chemistry: Structure and Properties Custom Edition for Rutgers University General Chemistry

15th Edition

ISBN: 9781269935678

Author: Nivaldo J. Tro

Publisher: Pearson Education

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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Chapter 18, Problem 69E

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To determine:

The pH at 0 mL, 5 mL, after adding 5 mL, 5 mL beyond equivalent point, equivalent point, one half equivalent point and volume of base required to reach equivalent point.

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Chapter 18, Problem 68E

Chapter 18, Problem 70E

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A 50.0 mL solution of 0.156 M KOH is titrated with 0.312 M HCI. Calculate the pH of the solution after the addition of each of the given amounts of HCl. 0.00 mL pH = 13.2 5.00 mL pH = 12.5 mL pH = 12.84 20.0 mL pH = 12.33 24.0 mL pH = 11.63 25.0 mL pH = 7.0 26.0 mL pH = 2.39 29.0 mL pH =

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Chapter 18 Solutions

Chemistry: Structure and Properties Custom Edition for Rutgers University General Chemistry

Ch. 18 - A buffer is 0.100 M in NH4CI and 0.100 M in NH3....Ch. 18 - What is the pH of a buffer that is 0.120 M in...Ch. 18 - Prob. 3SAQ Ch. 18 - Prob. 4SAQ Ch. 18 - Prob. 5SAQ Ch. 18 - Prob. 6SAQ Ch. 18 - Prob. 7SAQ Ch. 18 - A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN)...Ch. 18 - Prob. 9SAQ Ch. 18 - Prob. 10SAQ

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The pH at 0 mL, 5 mL, after adding 5 mL, 5 mL beyond equivalent point, equivalent point, one half equivalent point and volume of base required to reach equivalent point.

Solution Summary: The author calculates the pH of HC 2 H 3 O 2 and the volume of base required to reach the equivalence point.

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To determine: The pH at 0 mL, 5 mL, after adding 5 mL, 5 mL beyond equivalent point, equivalent point, one half equivalent point and volume of base required to reach equivalent point.

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Chapter 18 Solutions

To determine:

The pH at 0 mL, 5 mL, after adding 5 mL, 5 mL beyond equivalent point, equivalent point, one half equivalent point and volume of base required to reach equivalent point.