A 75.0-mL sample of 0.0790 M HCN (Ka = 6.2 1010) is titrated ...

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Question 1

A 75.0-mL sample of 0.0500 M HCN (K_a = 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 (K_a = 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.

[Leostella20]

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[AAMendoza]

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