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Author Question: A 25.0 mL sample of 0.150 mol L-1 formic acid is titrated with a 0.150 mol L-1 NaOH solution. What ... (Read 59 times)

joblessjake

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on: Nov 18, 2019

Question 1

How many millilitres of 0.120 mol L-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L-1 acetic acid to the equivalence point? Acetic acid is monoprotic. The Ka of acetic acid is 1.8 × 10-5.
◦ 60.1
◦ 50.0
◦ 2.82
◦ 41.6
◦ 1.77

Question 2

A 25.0 mL sample of 0.150 mol L-1 formic acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH at the equivalence point? The Ka of formic acid is 1.8 × 10-4.
◦ 10.26
◦ 8.31
◦ 5.69
◦ 7.00
◦ 11.38
Chemistry: A Molecular Approach (Canadian) ¦ Tro, Fridgen, Shaw ¦ 2nd Edition
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Marked as best answer by joblessjake on Nov 18, 2019

IRincones

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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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joblessjake

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Reply 2 on: Nov 18, 2019
Wow, this really help

dyrone

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

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