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Author Question: Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine, C2H5NH2, by 1.0 M ... (Read 74 times)

Davideckstein7

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on: Mar 21, 2021

Question 1

Calculate the [H+] in a solution that is 0.20 M in NaF and 0.25 M in HF. (Ka = 7.2 × 10–4)

◦ 7.2 × 10–4M
◦ 1.2 M
◦ 9.0 × 10–4M
◦ 0.20 M
◦ 5.8 × 10–4M

Question 2

Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine, C2H5NH2, by 1.0 M perchloric acid, HClO4. (pKb for C2H5NH2 = 3.25)

◦ 6.05
◦ 2.24
◦ 5.53
◦ 2.09
◦ 5.38
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Marked as best answer by Davideckstein7 on Mar 21, 2021

GCabra

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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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Davideckstein7

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

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Reply 3 on: Yesterday
Wow, this really help

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