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Author Question: Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.8 1010) with 0.500 M HCl. For ... (Read 255 times)

rl

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on: Mar 21, 2021

Question 1

Consider the titration of 100.0 mL of 0.100 M H2A (Ka1 = 1.50 × 10–4; Ka2 = 1.00 × 10–8). How many milliliters of 0.100 M NaOH must be added to reach a pH of 5.000?

◦ 41.9 mL
◦ 93.8 mL
◦ 100. mL
◦ 200. mL
◦ 60.0 mL

Question 2

Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.8 × 10–10) with 0.500 M HCl. For calculating the volume of HCl required to reach a pH of 8.0, which of the following expressions is correct? (x = volume in mL of HCl required to reach a pH of 8.0)

◦ = [aniline]
◦ [H+] = x
◦ = [aniline]
◦ = [aniline]
◦ None of these are correct.
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Marked as best answer by rl on Mar 21, 2021

manuelcastillo

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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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kr15t3sh

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Reply #2 on: Mar 29, 2021
Thank you


Shaza Mamdouh

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Reply #3 on: Jun 23, 2021
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