A 51.5 mL sample of a 0.330 M solution of NaCN is titrated by 0.120 M HCl. K₂ for CN is 2.0 × 10-5. Calculate the pH of the solution: a. prior to the start of the titration pH = b. after the addition of 99.1 mL of 0.120 M HCI pH = c. at the equivalence point pH =

A 51.5 mL sample of a 0.330 M solution of NaCN is titrated by 0.120 M HCl. K₂ for CN is 2.0 × 10-5. Calculate the pH of the solution: a. prior to the start of the titration pH = b. after the addition of 99.1 mL of 0.120 M HCI pH = c. at the equivalence point pH =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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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

A 51.5 mL sample of a 0.330 M solution of NaCN is titrated by 0.120
M HCl. K for CN¯ is 2.0 × 10-5. Calculate the pH of the solution:
a. prior to the start of the titration
pH
b. after the addition of 99.1 mL of 0.120 M HCl
=
pH =
c. at the equivalence point
pH =
d. after the addition of 211 mL of 0.120 M HC1
pH
=

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