16. Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a 0.1 M NaOHsolution (given that pka of formic acid = 3.74).A. 4.74B. 8.74C. 8.37D. 6.0617. When a 0.20 M solution of acetic acid is neutralized with 0.20 M NaOH in 0.50 L of water, the pH of the.resulting solution will be (given that pKa for CH3COOH = 4.74)A. 12.67B. 7.87C. 8.87D. 7.0018. Calculate the equivalent weight of Na2CO3 when it is titrated against HCI in the presence ofphenolphthalein.A. 106B. 53C. 26.5D. 212

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Answered: 16. Calculate the pH at the equivalence…

16. Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a 0.1 M NaOH solution (given that pka of formic acid = 3.74). A. 4.74 B. 8.74 C. 8.37 D. 6.06

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter14: Acid- Base Equilibria

Section: Chapter Questions

Problem 118CP: Malonic acid (HO2CCH2CO2H) is a diprotic acid. In the titration of malonic acid w ith NaOH,...

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Concept explainers

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.

Question

16. Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a 0.1 M NaOH
solution (given that pka of formic acid = 3.74).
A. 4.74
B. 8.74
C. 8.37
D. 6.06
17. When a 0.20 M solution of acetic acid is neutralized with 0.20 M NaOH in 0.50 L of water, the pH of the.
resulting solution will be (given that pKa for CH3COOH = 4.74)
A. 12.67
B. 7.87
C. 8.87
D. 7.00
18. Calculate the equivalent weight of Na2CO3 when it is titrated against HCI in the presence of
phenolphthalein.
A. 106
B. 53
C. 26.5
D. 212

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