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Author Question: A titration of 200.0 mL of 1.00 M H2A was done with 1.37 M NaOH. For the diprotic acid H2A, Ka1 = ... (Read 67 times)

jerry coleman

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on: Mar 21, 2021

Question 1

A titration of 200.0 mL of 1.00 M H2A was done with 1.08 M NaOH. For the diprotic acid H2A, Ka1 = 2.5 × 10–5, Ka2 = 3.1 × 10–9. Calculate the pH after 600.0 mL of 1.08 M NaOH have been added.

◦ 13.491
◦ 0.509
◦ 13.748
◦ 0.252
◦ 9.491

Question 2

A titration of 200.0 mL of 1.00 M H2A was done with 1.37 M NaOH. For the diprotic acid H2A, Ka1 = 2.5 × 10–5, Ka2 = 3.1 × 10–9. Calculate the pH after 100.0 mL of 1.37 M NaOH have been added.

◦ 9.06
◦ 8.85
◦ 5.15
◦ 9.88
◦ 4.94
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Marked as best answer by jerry coleman on Mar 21, 2021

lolol

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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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jerry coleman

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

aruss1303

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Reply 3 on: Yesterday
Wow, this really help

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