Asked by Mark
Vinegar is a dilute solution of acetic acid. In the titration of 5.00 mL of vinegar, 39.75 mL of 0.137 M sodium hydroxide solution was required to neutralize the vinegar to a phenolphthalein end point. Calculate each of the following of the vinegar.
(a) the molarity : I got 1.09 and its correct so im thinking I have to use this for the next part which is
b) the percent by mass (density of solution = 1.007 g/mL)
Thanks:)
(a) the molarity : I got 1.09 and its correct so im thinking I have to use this for the next part which is
b) the percent by mass (density of solution = 1.007 g/mL)
Thanks:)
Answers
Answered by
DrBob222
You CAN use the answer from part A but it isn't necessary.
moles NaOH = M x L
moles vinegar = moles NaOH
moles = grams vinegar/molar mass vinegar and solve for grams.
Then % acetic acid = (mass acetic acid/mass sample)*100 = ??
You will need to use the density information to convert 5 mL to grams to use for mass of the sample. Vinegar is usually about 4-6% acetic acid.
moles NaOH = M x L
moles vinegar = moles NaOH
moles = grams vinegar/molar mass vinegar and solve for grams.
Then % acetic acid = (mass acetic acid/mass sample)*100 = ??
You will need to use the density information to convert 5 mL to grams to use for mass of the sample. Vinegar is usually about 4-6% acetic acid.
Answered by
Mark
Thanks!
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