36.4 mL sample of a 0.110 M solution of NaCN is titrated by 0.160 M HCl. K, for prior to the start of the titration after the addition of 10.0 mL of 0.160 M HC1 at the equivalence point after the addition of 30 8 ml of 0 160 M HCI

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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 36.4 mL sample of a 0.110 M solution of NaCN is titrated by 0.160 M HCI. 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 10.0 mL of 0.160 M HCI
pH
c. at the equivalence point
pH
d. after the addition of 30.8 mL of 0.160 M HCI
pH
Transcribed Image Text:A 36.4 mL sample of a 0.110 M solution of NaCN is titrated by 0.160 M HCI. 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 10.0 mL of 0.160 M HCI pH c. at the equivalence point pH d. after the addition of 30.8 mL of 0.160 M HCI pH
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