Ask A Scientist© Chemistry Archive

name       Liam
status     student
age        16

Question -  In a titration experiment
I know that there is between 4-5% ethanoic(acetic) acid in white vinegar
but I don't known how I can arrive at an exact figure for this. How do I
do this using a 0.100 mol solution of hydrochloric acid in conjunction
with the use of phenophthalein as an indicator. Also what would the
relevent equation/s be for this test?
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Standardize the titrating base, assume it's KOH, with the 0.100 molar HCl.
The applicable equation is:

    Vol.[HCl] *Mol.[HCl] = Vol.[KOH] * Mol.[KOH]

where "Vol." = volume(ml) and "Mol." = molarity (mols/cc)
Plug in the known quantities and solve for Mol.[KOH], which is the only
unknown.

Use KOH instead of NaOH bacause NaOH quickly absorbs CO2 from the atmosphere
forming NaCO3, which throws the standardization off unless you keep the NaOH
away from the atmosphere. Titrate to the same "pinkness" in the
standardization and the titration of the vinegar.

Assume for this illustration that it turns out that Mol.[KOH] = 0.215. Now
titrate a known volume of the acetic acid sample, Vol.[HOAc]. The equation
above again applies:

    Vol.[KOH] *Mol.[KOH] = Vol.[HOAc] * Mol.[HOAc]

and solve for the only unknown quantity, Mol.[HOAc]. Let's say it takes 46.5
ml of standardized KOH to titrate 20.0 ml of the unknown acetic acid to the
same "pinkness".

So: (46.5) * (0.215) = (20.0) * (Mol.[HOAc])

Solving gives (Mol.(HOAc)) = 0.50 [mol./liter]. Since the molecular weight
of acetic acid is 60.0 gm / mol: (0.50 [mol HOAc / liter]) / (60.0 gm HOAc)
/ (mol HOAc) or
30 gm. HOAc / liter, or 3 gm. HOAc / 100 ml., which is a 3% solution.

V. Calder
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The short answer is, you can't. You also need some base,
like KOH. You can use the base to make up a basic
solution. Then you use your 0.100 M HCl solution to
standardize the base solution (which does not have a
known concentration, because most solid bases like KOH
absorb water from the air and therefore cannot be accurately
weighed) via titration, using the phenopthalein as an
indicator. Now that you have a basic solution of
known concentration, you can use it to titrate the vinegar
and determine its acid concentration.

The relevant equation is: for a monoprotic acid of type HX,
reaction with a strong Bronsted-Lowry base (which forms
OH-(aq)) occurs via a neutralization reaction to form H2O;

HX(aq) + OH- (aq) ---> H2O(l) + X- (aq)

Hope this helps,
prof. topper
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Because vinegar is dilute acetic acid, titration with standard base, such as
0.1N NaOH will yield the activity of the acid.  A few drops of methyl orange
indicator (0.1% aqueous solution, look for the color change from red through
orange to yellow) will more accurately mark the equivalence point.  Can also
find equivalence point by plotting pH vs. mL titrant and finding equivalence
by geometry or by first or second derivative.  The equation to find the
percentage of vinegar is:
{(mL of NaOH soln.)x(N of NaOH soln)x(gramEqWt of acetic acid[60.05])x
100%}/{(grams of vinegar sample)x(1000 mL/L)} = % vinegar
for 0.1N NaOH titrant this is simply
{[(mLs titrant)x(.6005)]/[sample size (in grams)]} = % vinegar
Tim Spry
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I don't see how you could titrate the acetic acid in vinegar using just a
standard HCl solution with phenolphthalein indicator.  Phenolphthalein turns
red above about pH 10;  mixtures of HCl and acetic acid will ALWAYS be well
below pH 10.

If you have an alkaline solution of known concentration, say a 0.1 M NaOH
solution, then you can do the titration.  You add a small amount of
phenolphthalein to the acetic acid, and then titrate in the base until you
just begin to get a stable color change.  Then you calculate how many moles
of NaOH you used to neutralize the acetic acid, which must be equal to the
number of moles of acetic acid that were present, according to the equation

HOAc + NaOH --> NaOAc + H2O.

Richard E. Barrans Jr., Ph.D.
Assistant Director
PG Research Foundation, Darien, Illinois
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