A 33.8 mL sample of a 0.400 M solution of NaCN is titrated by 0.160 M HCI. Kb for CN is 2.0 x 10-. Calculate the pH of the solution: a. prior to the start of the titration pH %3D b. after the addition of 50.7 mL of 0.160 M HCI pH = c. at the equivalence point pH = d. after the addition of 122 mL of 0.160 M HOCi 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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A 33.8 mL sample of a 0.400 M solution of NaCN is titrated by 0.160 M HCI. K for CN is
2.0 x 10-6. Calculate the pH of the solution:
a. prior to the start of the titration
pH =
%3D
b. after the addition of 50.7 mL of 0.160 M HCl
pH =
%3D
C. at the equivalence point
pH
%3D
d. after the addition of 122 mL of 0.160 M HCl
pH =
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Transcribed Image Text:A 33.8 mL sample of a 0.400 M solution of NaCN is titrated by 0.160 M HCI. K for CN is 2.0 x 10-6. Calculate the pH of the solution: a. prior to the start of the titration pH = %3D b. after the addition of 50.7 mL of 0.160 M HCl pH = %3D C. at the equivalence point pH %3D d. after the addition of 122 mL of 0.160 M HCl pH = Submit Answer Retry Entire Group 9 more group attempts remaining
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