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Author Question: Addition of 0.0125 mol KOH to 150 mL of a 0.150 M formic acid/0.100 M sodium formate buffer results ... (Read 398 times)

haleyc112

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on: Feb 15, 2020

Question 1

The pH of a 0.150 M formic acid/0.250 M sodium formate buffer = ________? The Ka of formic acid is 1.8 × 10-4.

Question 2

Addition of 0.0125 mol KOH to 150 mL of a 0.150 M formic acid/0.100 M sodium formate buffer results in a pH = ________.  The Ka of formic acid is 1.8 × 10-4.
Chemistry ¦ McMurry, Fay, Robinson ¦ 7th Edition
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Marked as best answer by haleyc112 on Feb 15, 2020

Madisongo23

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Reply #1 on: Feb 15, 2020
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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Eric Flores

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Reply #2 on: Mar 24, 2023
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