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A 43.8 ml sample of a 0.200 M solution of NaCN is titrated by 0.130 M HCl. K, for CN is 2.0 x 105. Calculate the pH of the solution: a. prior to the start of the titration pH- b. after the addition of 47.2 ml. of 0.130 M HCl pH = c. at the equivalence point pH = d. after the addition of 87.6 mL of 0.130 M HCI pH =

A 43.8 ml sample of a 0.200 M solution of NaCN is titrated by 0.130 M HCl. K, for CN is 2.0 x 105. Calculate the pH of the solution: a. prior to the start of the titration pH- b. after the addition of 47.2 ml. of 0.130 M HCl pH = c. at the equivalence point pH = d. after the addition of 87.6 mL of 0.130 M HCI pH =

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.91QP: A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. Kb for CN is 2.0 105....
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A 43.8 ml sample of a 0.200 M solution of NaCN is titrated by 0.130 M HCl. K, for CN is 2.0 x 10-5. Calculate the pH of the solution: a. prior to the start of the titration pH = b. after the addition of 47.2 ml. of 0.130 M HCI pH = c. at the equivalence point pH = d. after the addition of 87.6 mL of 0.130 M HCI pH =
Transcribed Image Text:A 43.8 ml sample of a 0.200 M solution of NaCN is titrated by 0.130 M HCl. K, for CN is 2.0 x 10-5. Calculate the pH of the solution: a. prior to the start of the titration pH = b. after the addition of 47.2 ml. of 0.130 M HCI pH = c. at the equivalence point pH = d. after the addition of 87.6 mL of 0.130 M HCI pH =
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