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Author Question: A 100.0-mL sample of 0.583 M H2A (diprotic acid) is titrated with 0.200 M NaOH. After 125.0 mL of ... (Read 56 times)

SGallaher96

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on: Mar 21, 2021

Question 1

A 100.0-mL sample of 0.2 M (CH3)3N (Kb = 5.30 × 10–5) is titrated with 0.2 M HCl. What is the pH at the equivalence point?

◦ 2.6
◦ 8.6
◦ 10.7
◦ 5.4
◦ 7.0

Question 2

A 100.0-mL sample of 0.583 M H2A (diprotic acid) is titrated with 0.200 M NaOH. After 125.0 mL of 0.200 M NaOH has been added, the pH of the solution is 4.50. Calculate Ka1 for H2A.

◦ 4.2 × 10–10
◦ 4.5
◦ 2.4 × 10–5
◦ 0.65
◦ None of these are correct.
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Marked as best answer by SGallaher96 on Mar 21, 2021

isabelt_18

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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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SGallaher96

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Reply 2 on: Mar 21, 2021
Thanks for the timely response, appreciate it

xoxo123

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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