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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 268 times)

rl

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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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