### Re: Making solutions with known pH levels

Date: Sun Jun 5 18:54:15 2005
Posted By: Lynn Bry, MD/PhD, Dept. Pathology, Brigham & Women's Hospital, Harvard Medical School
Area of science: Chemistry
ID: 1117153179.Ch
Message:

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.

### Somewhat more complicated method using acid/base calculations:

Another way is to use equilibria reactions to determine values chemically. I don't know how much chemistry and algebra you've had, so some of it might be a little beyond what you're used to seeing in class. Fear not! Ask your parents or teacher at school to help you through the following if you want to calculate actual values (and keep a keen eye on these approaches when you take chemistry and algebra!).

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

```             CH3COOH <-> CH3COO- + H+
```
2. The equlibrium constant (Ka) for this dissociation reaction is then Ka = products/reactants. This value represents the ratio of products of a reaction to the starting reactants when the reaction is at equilibrium. At equilibrium, the rate of reactions forward (->) equals the rate of reactions in reverse (<-).
```        Ka = [CH3COO-][H+]       products
-------------      -----------
[CH3COOH]         reactants

```
where [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 dissociates into CH3COO- + H+. The amount of H+ determines the pH/acidity of the solution. How can you find the amount that remains whole, and the amount that dissociates?

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 = X2    distribute 0.000018 on the left side

0.0000149 - 0.000018X = X2  subtract 0.000018X from both sides

0.0000149 = X2 + 0.000018X  make this a quadratic equation,
set equal to 0 on one side by subtracting
0.0000149 from both sides

0 = X2 + 0.000018X - 0.0000149
```
6. Now you need to remember (or learn!) about quadratic equations. A quadratic equation has the format:
```    0 = aX2 + bX + c
```
From the equation we created using acetic acid, we get the following values for a, b, and c:
```0 = 1X2 + 0.000018X - 0.0000149
a          b         c

a = 1
b = 0.000018
c =-0.0000149
```
Algebra 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 0 = X2 + 0.000018X - 0.0000149.
```             X = -b +/- SQUAREROOT(b2 - 4ac)
------------------------------
2a
```
You 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.

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+] = 0.00385M
```
To get to pH, understand that the "p" is really the Greek letter "rho." Chemically, rho means "the negative log10 of the following concentration," or..
```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 +

pH = 2.41
```
After all that, the pH of a 5% solution of acetic acid = 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 = X2    distribute 0.000018 among terms

0.00000149 - 0.000018X = X2  subtract 0.000018X from both sides

0.00000149 = X2 + 0.000018X  make this a quadratic equation,
set equal to 0 on one side by subtracting
0.00000149 from both sides

0 = X2 + 0.000018X - 0.00000149

a=2    b=0.000018     c=-0.00000149
```
The key difference is the value of c is 1/10th of what it was in the first euqation we solved: 0.00000149.

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

```pH = -log[0.0012]
pH = 2.91
```
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?

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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