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Author Question: Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of ... (Read 37 times)

warrenjean01

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on: Nov 5, 2023

Question 1

Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka (HCN) = 4.9 × 10–10).
◦ 9.31
◦ 9.18
◦ 9.52
◦ 11.63
◦ 12.59

Question 2

Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline (C6H5NH2). Kb (C6H5NH2) = 3.8 x 10-10

◦ 1.75
◦ 0.81
◦ 4.64
◦ 4.19
◦ 9.09
Chemistry ¦ Chang, Goldsby ¦ 11th Edition
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Marked as best answer by warrenjean01 on Nov 5, 2023

studyforce.com

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Reply #1 on: Nov 5, 2023
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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warrenjean01

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Reply 2 on: Nov 5, 2023
Wow, this really help

laurnthompson

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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