A 44.4 mL sample of a 0.300 M solution of NaCN is titrated by 0.100 M HC1, K₁ 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 93.2 mL of 0.100 M HCl PH = c. at the equivalence point pH d. after the addition of 173 mL of 0.100 M HCl pH =

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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.144QP
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A 44.4 mL sample of a 0.300 M solution of
NaCN is titrated by 0.100 M
HC1,
K₁ 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 93.2 mL of 0.100 M HCl
pH =
c. at the equivalence point
pH:
d. after the addition of 173 mL of 0.100 M HCl
pH:
Transcribed Image Text:A 44.4 mL sample of a 0.300 M solution of NaCN is titrated by 0.100 M HC1, K₁ 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 93.2 mL of 0.100 M HCl pH = c. at the equivalence point pH: d. after the addition of 173 mL of 0.100 M HCl pH:
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