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Author Question: A 75.0-mL sample of 0.0790 M HCN (Ka = 6.2 1010) is titrated with 0.790 M NaOH. What volume of ... (Read 292 times)

viki

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on: Mar 21, 2021

Question 1

A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 × 10–10) is titrated with 0.272 M NaOH. What is the [H+] in the solution after 3.0 mL of 0.272 M NaOH have been added?

◦ 4.5 × 10–6M
◦ 1.0 × 10–7M
◦ 3.6 M
◦ 2.2 × 10–9M
◦ None of these are correct.

Question 2

A 75.0-mL sample of 0.0790 M HCN (Ka = 6.2 × 10–10) is titrated with 0.790 M NaOH. What volume of 0.790 M NaOH is required to reach the stoichiometric point?

◦ 750. mL
◦ 7.50 mL
◦ 3.75 mL
◦ 75.0 mL
◦ Cannot determine without knowing the pH at the stoichiometric point.
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Marked as best answer by viki on Mar 21, 2021

Leostella20

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Reply #1 on: Mar 21, 2021
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AAMendoza

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Reply #2 on: Nov 16, 2021
Thank you


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