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Author Question: A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium formate in water ... (Read 173 times)

Pineapplelove6

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on: Nov 18, 2019
A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium formate in water sufficient to yield 1.00 L of solution.The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the ________ present in the buffer solution. The Ka of formic acid is 1.8 × 10-4.
◦ H3O+
◦ H2O
◦ formic acid
◦ formate
◦ This is a buffer solution: the pH does not change upon addition of acid or base.
Chemistry: A Molecular Approach (Canadian) ¦ Tro, Fridgen, Shaw ¦ 2nd Edition
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Marked as best answer by Pineapplelove6 on Nov 18, 2019

wfdfwc23

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Reply #1 on: Nov 18, 2019
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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Pineapplelove6

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Reply 2 on: Nov 18, 2019
:D TYSM

parker125

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Reply 3 on: Yesterday
Gracias!

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