Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Lindsay on October 28, 2008, 03:15:14 PM
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Hello,
Problem 1
Assuming the density of a 5% acetic acid solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH.
Answer,
I don't really know how to solve problem 1 or 2, but this is my approach in solving them;
1.0 L * 0.05 * 1000g/1L = 50 g
V1=M2V2/M1
0.10 M * 0.025 L/50g *60.05 =
V1 = 0.003 L = 3 mL
Problem 2
A 31.43 mL VOLUME OF 0.108 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 4.441 g sample of vinegar. Calculate the percent acetic acid in the vinegar.
moles of solute = M * V
0.108 M NaOH * 0.03143 = 0.00339 moles of solute NaOH
moles of acetic acid = moles of NaOH
weight of acid = 60.05 g/mol * 0.00339 = 0.204 grams
0.204 g / 4.441 g * 100 = 4.60 %
Please advise.
Thanks,
Lindsay
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V1 = 0.003 L = 3 mL
No idea if it was mastership or sheer luck, but 3 mL it is.
0.204 g / 4.441 g * 100 = 4.60 %
And again :)