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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 226 times)

viki

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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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