# During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations.

Sol: CH3COOH + NaOH <---> NaCH3COO +
H20

Convert: 18.62 mL = .01862 L

moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration.
.00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present.

The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required.

During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations.

Sol: CH3COOH + NaOH <---> NaCH3COO +
H20
This step is OK.
Convert: 18.62 mL = .01862 L

moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration.
.00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present. OK to here EXCEPT that since I assume the molarity is really 0.0950 M (and you just omitted the last zero but I may be wrong), then you have three figures and the 0.11 should be carried out to 0.106 g. From here you did not convert grams to percent correctly. That is
%CH3COOH= ([g CH3COOH]/g sample) x 100 or % = (0.106/2.578) x 100 = ??

The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required.

One other point is that since the mL is to four places and the mass of the sample is to four places, I can't understand why the molarity is to only two places. If it really is 0.09500 then you should have four places in the grams CH3COOH, which I believe would be 0.1062. BUT if the 0.095 is correct, then your 0.11 is proper. I hope this helps.

if i have 100% acetic acid bottel how i can prepare 1ml 10mM of it??

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1. I do believe this is the right annswer (.11g Acetic Acid present). When you you use the above equation for the mass percent of acetic acid (g CH3COOH/g sample) you end up with 4.3% Acetic Acid. That is a normal concentration of acetic acid [I looked at other sites for concentrations and in fruits acetic acid conc usually varies from 3 to 5.5%]

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2. u fail

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