[Lindsay]

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

[Borek]

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