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 =

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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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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 =
Transcribed Image Text: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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