A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH. What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.a) 7.00 b) 8.76 c) 9.31 d) 11.07
A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH. What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.a) 7.00 b) 8.76 c) 9.31 d) 11.07
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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A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH. What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.
a) 7.00 b) 8.76 c) 9.31 d) 11.07
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Im confused as to how you got .002 moles for NaOH and CN-, I know there are .002 moles of HCN, but how did you get that for NaOH and CN-?
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