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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 282 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
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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AAMendoza

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Reply #2 on: Nov 16, 2021
Thank you


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