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.
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.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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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