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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 40 times)

SGallaher96

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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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SGallaher96

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Reply 2 on: Mar 21, 2021
Great answer, keep it coming :)


deja

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Reply 3 on: Yesterday
YES! Correct, THANKS for helping me on my review

 

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