Assistance please -- Conversion between Percent Acidity and pH
grist_4_...@juno.com
post your responses here, it would be appreciated....
My son has an experiment that requires five solutions of
varying pH's: 3.5, 4.0. 4.5, 5.0, 5.6
It's Sunday and the science supply store is closed. Besides,
to purchase the litmus paper that would permit precise
measurment of pH over this range would require buying
two products at a cost of over $20.
I have proposed a Plan B, but I don't know how to pull it
off: Dilute standard white vinegar -- already at 5% acidity
-- to achieve the varying pH solutions. But I don't know how
to do the conversion.
Do you know the answer to any of the following:
1). What is the formula for converting between percent
acidity and pH?
2). What is the pH of pure ascetic acid?
3). Are these assumptions correct?
* Vinegar @ 5 percent acidity equals 1 part vinegar and
19 parts water.
* Equal parts of acetic acid and water have the same, or
nearly the same, weight.
* A 5 percent acidity solution of vinegar could be made
into a 2.5 percent acidity solution by adding an additional
20 parts of water (e.g. 1 part acetic acid to 39 parts water).
* Reducing the percent acidity by half does not reduce the pH
by half since the latter is measured logarithmically.
Any and all assistance is appreciated. If you can include a
web site or other source citation, that would be appreciated
as well.
--
JR
I cannot currently access my email account, so if you could
post your responses here, it would be appreciated....
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Mark Tarka
>I cannot currently access my email account, so if you could
>post your responses here, it would be appreciated....
>
>
>My son has an experiment that requires five solutions of
>varying pH's: 3.5, 4.0. 4.5, 5.0, 5.6
>
>It's Sunday and the science supply store is closed. Besides,
>to purchase the litmus paper that would permit precise
>measurment of pH over this range would require buying
>two products at a cost of over $20.
You must have been in touch with the science supply store earlier.
Why didn't you go to the instructor who's responsible for the
assignment of the project and discuss your theory and perhaps get
any supplies you need, gratis?
>I have proposed a Plan B, but I don't know how to pull it
>off: Dilute standard white vinegar -- already at 5% acidity
>-- to achieve the varying pH solutions. But I don't know how
>to do the conversion.
That is, I believe, 5% acetic acid in water, a weight volume
measurement. Assuming 100 grams of solution, 5 grams comprise the
acid (at 60 grams/mole) and 95 grams comprise the water (which is
equal to 95 mL since water has a density of 1 g/mL).
Convert the number of grams of acetic acid into moles of acetic acid
using the molecular weight of the acid, divide that by the 0.095
liters of water and you have the molarity of the acid solution.
Call that value M.
The pH of a solution can be calculated, if you know its hydrogen ion
concentration ([H+]).
Given that you know the molarity of the acetic acid (abbreviated
"HOAc"), then find the Ka for the acid (or the pKa which is simply
equal to the -log(Ka).
Now one merely sets up a table:
[HOAc] [H+] [OAc-]
Initially M 0 0
Finally M-x x x
Here, x is the amount of acetic acid that dissociates into protons
and the acetate anions. Your son does know about dissociation of
acids, yes?
Solve the equation
Ka = [H+][OAc-]/[HOAc] = x^2/M
This is the simplified form (using M and not M-x), but should be
sufficient.
Then
x = square_root(M times Ka)
Just algebra.
Now x = [H+], and you can calculate the pH thus:
pH = -log[H+]
Now you know the pH of your vinegar solution (about 2.4 according to
my calculation).
Knowing the starting concentration (molarity, M) you can calculate
what dilution is required to get a certain pH. You want pH=3.5;
this corresponds to a final concentration of H+ of 3.162e-4 molar.
Your initial concentration is what you calculated earlier (M).
Let's say you want to make up 20 ml of a solution that has a pH of
3.5 or a [H+] of 3.162e-4 molar. Looking at the equation above (Ka
= [H+][OAc]/[HOAc]) you can solve for the molarity of the acetic
acid required to get that first dilution:
[HOAc] = [H+][OAc-]/Ka = (3.162e-4)squared/Ka
Remember, the amount of H+ equals the amount of OAc-, and you've
still got to look-up the value of Ka for acetic acid.
Anyway...call this recent result C2 (concentration of HOAc in the
first dilution). The value M that you found earlier is C1 (the
concentration in the vinegar). We've arbitrarily selected 20 mL as
the volume of the first dilution, and we call that V2. So we need
to find how much of the original solution (the vinegar) we need to
use (called V1) and dilute that amount to the final volume with
tap-water. Mathematically, this is
C1 V1 = C2 V2
You may have or can certainly purchase measuring cups that have mL
increments. Otherwise use fluid ounces for your volumes.
Not so hard to do after 12 years of post-secondary education. If
you want more help with the problem send two upcoming Superbowl
tickets (seats on the 50-yard line) to....
Mark (WARNING...check this freely given information with a
knowledgeable individual before performing any act related to this
discussion in the privacy of your own home. Vinegar contains acid,
and acids can cause injury or even death. Accuracy of the above
information is not guaranteed. Not responsible for typographical
errors. Offer not valid in the 48 contiguous states or in Hawaii or
Alaska.)
Uncle Al
>
> I cannot currently access my email account, so if you could
> post your responses here, it would be appreciated....
>
> My son has an experiment that requires five solutions of
> varying pH's: 3.5, 4.0. 4.5, 5.0, 5.6
>
Mixing up a solution of acid or base to get a solution pH isn't much of
a solution because the slightest system input, like CO2 from the air,
will move the pH. You need a buffer system - a alst of a weak acid and
the acid, with relative concentrations setting the pH. Try lookig in
your offspring's textbook or, here's a bright idea, have HIM crack the
book.
In any case, HOAc is a weak acid, which horribly complicates matters re
concentration vs pKa. It would be much more simple and elegant to
aspirate gastric juice (0.1M HCl) and serially dilute that. And... you
don't need a store to get at it.
--
Uncle Al Schwartz
Uncl...@ix.netcom.com ("zero" before @)
http://uncleal.within.net/
http://pw2.netcom.com/~uncleal0/uncleal.htm
http://www.ultra.net.au/~wisby/uncleal.htm
http://www.guyy.demon.co.uk/uncleal/uncleal.htm
(Toxic URLs! Unsafe for children, Democrats, and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Barry Gow
>I cannot currently access my email account, so if you could
>post your responses here, it would be appreciated....
>My son has an experiment that requires five solutions of
>varying pH's: 3.5, 4.0. 4.5, 5.0, 5.6
>It's Sunday and the science supply store is closed. Besides,
>to purchase the litmus paper that would permit precise
>measurment of pH over this range would require buying
>two products at a cost of over $20.
>I have proposed a Plan B, but I don't know how to pull it
>off: Dilute standard white vinegar -- already at 5% acidity
>-- to achieve the varying pH solutions. But I don't know how
>to do the conversion.
>Do you know the answer to any of the following:
>1). What is the formula for converting between percent
> acidity and pH?
>2). What is the pH of pure ascetic acid?
>3). Are these assumptions correct?
> * Vinegar @ 5 percent acidity equals 1 part vinegar and
> 19 parts water.
> * Equal parts of acetic acid and water have the same, or
> nearly the same, weight.
> * A 5 percent acidity solution of vinegar could be made
> into a 2.5 percent acidity solution by adding an additional
> 20 parts of water (e.g. 1 part acetic acid to 39 parts water).
> * Reducing the percent acidity by half does not reduce the pH
> by half since the latter is measured logarithmically.
>Any and all assistance is appreciated. If you can include a
>web site or other source citation, that would be appreciated
>as well.
There is no simple relation between pH and acidity. Wine is a mixture
of weak acids and any particular wine will be different from another.
`
>--
>JR
There is no simple relationship between pH and acidity. pH meters have become
relatively inexpensive in recent years. I would not be without one. Nor
would I be without a means of measuring TA i.e. 0.1 molar NaOH and a graduated
pipette.