3. A 50.0-mL sample containing HCN was titrated with 0.220 M KOH and found to have an equivalence point volume of 28.4 ml. Calculate the molarity of HCN in the original sample prior to titration.

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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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3. A 50.0-mL sample containing HCN was titrated with 0.220 M KOH and found to have
an equivalence point volume of 28.4 ml. Calculate the molarity of HCN in the
original sample prior to titration.
4. A 100.0-mL solution consisting of 0.14 M NaC2H30z and 0.16 M HC2H302 was
titrated with 24 mL of 0.25 M KOH. What is the pH of the mixture at equilibrium?
K, for acetic acid=1.8x10-5.
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Transcribed Image Text:3. A 50.0-mL sample containing HCN was titrated with 0.220 M KOH and found to have an equivalence point volume of 28.4 ml. Calculate the molarity of HCN in the original sample prior to titration. 4. A 100.0-mL solution consisting of 0.14 M NaC2H30z and 0.16 M HC2H302 was titrated with 24 mL of 0.25 M KOH. What is the pH of the mixture at equilibrium? K, for acetic acid=1.8x10-5. Next »
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