A 75.0 mL sample of 0.05 M HCN (pKa = 9.21) is titrated with 0.5 M NAOH. What is [H3O*] in the solution after 3.0 mL of 0.5 M NaOH has been added? O 4.1 x 10-10 M O 1.5 x 10-9 M O 5.2 x 10-13 M O 1.0 x 10-7 M O 9.2 x 10-10 M

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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 75.0 mL sample of 0.05 M HCN (pKa = 9.21) is titrated with 0.5 M
NAOH. What is [H3O*] in the solution after 3.0 mL of 0.5 M NaOH has
been added?
O 4.1 x 10-10 M
O 1.5 x 10-9 M
O 5.2 x 10-13 M
O 1.0 x 10-7 M
O 9.2 x 10-10 M
Transcribed Image Text:A 75.0 mL sample of 0.05 M HCN (pKa = 9.21) is titrated with 0.5 M NAOH. What is [H3O*] in the solution after 3.0 mL of 0.5 M NaOH has been added? O 4.1 x 10-10 M O 1.5 x 10-9 M O 5.2 x 10-13 M O 1.0 x 10-7 M O 9.2 x 10-10 M
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