A 34.2 mL sample of a 0.110 M solution of NaCN is titrated by 0.170 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 17.7 mL of 0.170 M HCI pH = c. at the equivalence point pH = d. after the addition of 29.7 mL of 0.170 M HCI pH =

A 34.2 mL sample of a 0.110 M solution of NaCN is titrated by 0.170 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 17.7 mL of 0.170 M HCI pH = c. at the equivalence point pH = d. after the addition of 29.7 mL of 0.170 M HCI 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

Hh.129.

A 34.2 mL sample of a 0.110 M solution of NaCN is
titrated by 0.170 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 17.7 mL of 0.170 M HCI
pH =
c. at the equivalence point
pH H
d. after the addition of 29.7 mL of 0.170 M HCI
pH =

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