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Hi David -

A quick way to determine the pH would be to purchase some pH-paper from a local pharmacy or drugstore. Many sell "urine dipsticks" that measure pH over a range from 5 - 8, some will also go from pH 1 - 12. Ask the pharmacist on site if you can't find them in the store. This is the quickest, and experimentally, easiest way to measure the pH of vinegar and any dilutions you make.

1. Start with the standard equilibrium reaction indicating the deprotonation of acetic acid, and protonation of acetate ion (CH3COO-).

CH3COOH <-> CH3COO- + H+2. The

Ka = [CH3COO-][H+] products ------------- ----------- [CH3COOH] reactantswhere [things in brackets] = molarity, or concentration of each component.

We thus have Ka = [concentation of CH3COO-] times [concentration of H+] divided by [concentration of CH3COOH].

You can look up the Ka in reference texts such as a __CRC Handbook of Physics &
Chemistry__, or online.
Online sites will tell you the Ka for acetic acid = 1.8 X 10^{-5}.
This value is in *scientific notation*, but it means the same thing as
**0.000018**.

3. Calculate the expected concentration of acetic acid added to a 5% solution of
vinegar. Check the product label on the vinegar, most store-bought brands are
3-5% acetic acid __by
weight__. To convert this number to a concentration (or molarity), assume for
every 100mL of a 5% solution, you have 5g acetic acid + 95g of water. Alter as
needed for the concentration of the vinegar you're using.

Use the ratio:

5g acetic acid = Xg acetic acid ---------------- ----------------- 95g water (1000-X)g water Cross-multiply to get: 5000 - 5X = 95X Next, add 5X to both sides 5000 = 100X Divide both sides by 100 50 = X.. so 50g of acetic acid in 1000mL of water. Conveniently, the density of water is 1g/mL, and we'll assume in your solution of vinegar the density remains approximately 1g/mL. In 1L of this solution, you thus have 50g of acetic acid.

4. Look up the molecular weight of acetic acid CH3COOH, in which you have 2 carbons, 2 oxygens and 4 hydrogens. A Periodic Table comes in handy:

Carbon mass = 2 X 12 a.m.u. Oxygen mass = 2 X 16 a.m.u. Hydrogen mass = 4 X 1 a.m.u.~60 a.m.u. per molecule of acetic acid or 60grams/mole of acetic acid. In your 5% solution of vinegar, you have:

(50g in vinegar) / (60g/mole) = 0.83M solution of acetic acid.5. However, when the acetic acid is added to the water, some of it

Remember that you start with 0.83M of acetic acid. The amount that
*dissociates* is "X", an unknown amount. When CH3COOH dissociates, it
thus produces X amount of H+ AND X amount of CH3COO-

The amount of *undissociated* acetic acid is then (0.83 - X), or what
remains.

Ka = 1.8*10^{-5} or 0.000018.

Plug these values into your equilibrium equation:

Ka = [CH3COO-][H+] ------------- [CH3COOH] 0.000018 = X*X / (0.83-X) multiply both sides by (0.83-X) (0.83-X) * 0.000018 = X6. Now you need to remember (or learn!) about quadratic equations. A quadratic equation has the format:^{2}distribute 0.000018 on the left side 0.0000149 - 0.000018X = X^{2}subtract 0.000018X from both sides 0.0000149 = X^{2}+ 0.000018X make this a quadratic equation, set equal to 0 on one side by subtracting 0.0000149 from both sides 0 = X^{2}+ 0.000018X - 0.0000149

0 = aXFrom the equation we created using acetic acid, we get the following values for a, b, and c:^{2}+ bX + c

0 = 1XAlgebra give us the quadratic formula. Substituting values of a, b, and c will let you solve for the values of X that fulfill the equation^{2}+ 0.000018X - 0.0000149 a b c a = 1 b = 0.000018 c =-0.0000149

X = -b +/- SQUAREROOT(bYou can also use one of the available online quadratic calculators to determine the value of X. Plug in the values for a, b, and c.^{2}- 4ac) ------------------------------ 2a

The one above gives X = +/-0.00385 => positive 0.00385 and -0.00385.

Chemically, we are interested in the positive value 0.00385.

7. Remember that we defined X as the amount of acetic acid that dissociated into H+ and CH3COO-. Knowing the value of X means we can finally determine the pH of a 5% solution of acetic acid!

[H+] = 0.00385MTo get to pH, understand that the "p" is really the Greek letter "rho." Chemically, rho means "the negative log

pH = -log(0.00385)Plug 0.00385 into a calculator and take the log, the change the sign in front of the value you get:

pH = -log(0.00385) pH = -(-2.41) two minuses cancel to a +After all that, the pH of a 5% solution of acetic acid = 2.41pH = 2.41

8. You can plug in different values for your concentration of acetic acid to see how that impacts the pH. Let's say you dilute your vinegar 10-fold. Take one volume of vinegar and add 9 volumes of distilled water. Your concentration of acetic acid is now diluted 10-fold, from 0.83M to 0.083M. Enter the 0.083M into your equation with unknowns X and (0.083-X).

Ka = [CH3COO-][H+] ------------- [CH3COOH] 0.000018 = X*X / (0.083-X) multiply both sides by (0.083-X) (0.083-X) * 0.000018 = XThe key difference is the value of c is 1/10th of what it was in the first euqation we solved: 0.00000149.^{2}distribute 0.000018 among terms 0.00000149 - 0.000018X = X^{2}subtract 0.000018X from both sides 0.00000149 = X^{2}+ 0.000018X make this a quadratic equation, set equal to 0 on one side by subtracting 0.00000149 from both sides 0 = X^{2}+ 0.000018X - 0.00000149 a=2 b=0.000018 c=-0.00000149

Enter these values into the quadratic calculator to get: X = 0.0012

pH = -log[0.0012]You can keep making calculations for different dilutions of your starting solution of vinegar. What happens if you make a 20X, 50X or 100X dilution?pH = 2.91

You could also calculate values with a more complicated system, such as adding a base to acetic acid to raise the solution a given pH, including ones above pH=7. However, I'll leave those problems for when you take your first course in inorganic chemistry.

Hope this helps, and good luck with your project.

-ln

One of the many 46XX at Harvard who fears neither math nor science..

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