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Author Question: Calculate the molar solubility of AgCl in a 0.15 M solution of NH3(aq). (Ksp (AgCl) = 1.6 x ... (Read 58 times)

byomi

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on: Nov 5, 2023

Question 1

Calculate the molar solubility of AgBr  in a 0.25M  solution of NH3(aq) (Ksp (AgBr) = 7.7 x 10-13  ; Kf  (Ag(NH3)2+) = 1.5 x 107.
◦ 8.8 x 10-7 M
◦ 3.4 x 10-3 M
◦ 8.4 x 10-4 M
◦ 2.5 x 10-1 M
◦ 9.7 x 102 M

Question 2

Calculate the molar solubility of AgCl in a 0.15 M solution of NH3(aq).
         (Ksp (AgCl) = 1.6 x 10-10; Kf (Ag(NH3)2+) = 1.5 x 107)
◦ 6.7 x 10-3 M
◦ 1.3 x 10-5 M
◦ 3.9 x 103 M
◦ 7.5 x 10-2 M
◦ 3.3 x 10-5 M
Chemistry ¦ Chang, Goldsby ¦ 11th Edition
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Marked as best answer by byomi on Nov 5, 2023

athenaheim

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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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byomi

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Reply 2 on: Nov 5, 2023
Excellent

DylanD1323

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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