**James Richard Fromm**

There are several methods used to prepare aqueous buffer solutions, but the basic
principle of all of them is the same: **the solution must contain both the acid and
the base of a conjugate pair, in reasonably similar and in reasonably high concentrations**.
Only then can the buffer solution properly perform its task of maintaining solution pH
when small amounts of strong acid or strong base are added or formed by other chemical
reactions. The preparation of buffer solutions is shown in the following examples.

Example. Let us work out the experimental directions for making up one liter of an
aqueous buffer solution of pH 7.0. First, we select an acid-base conjugate pair such that
its p*K*_{a} is about 7.0. For the phosphoric acid system:

*K*_{1} = [H_{3}O^{+}][H_{2}PO_{4}^{-}]/[H_{3}PO_{4}]
= 7.11 x 10^{-3}

*K*_{2} = [H_{3}O^{+}][HPO_{4}^{2-}]/[H_{2}PO_{4}^{-}]
= 6.23 x 10^{-8}

The value of *K*_{2} is reasonably close to 1 x 10^{-7}. When pH =
7.0, [H_{3}O^{+}] = 1.0 x 10^{-7}, so using

*K>*_{2} (thereby ignoring *K*_{1} and *K*_{3},
which is approximate but legitimate - one can, in dealing with polyprotic acids, generally
treat the different constants as independent unless two of them have very similar values):

6.23 x 10^{-8} = [1 x 10^{-7}][HPO_{4}^{2-}]/[H_{2}PO_{4}^{-}]

[HPO_{4}^{2-}]/[H_{2}PO_{4}^{-}]
= 0.623, reasonably close to 1.

Now let us set the concentration of one component at an arbitrary but reasonably high
value, such as [H_{2}PO_{4}^{-}] = 1 x 10^{-2} molar. Then
0.623 = [HPO_{4}^{2-}]/1 x 10^{-2}, and [HPO_{4}^{2-}]
= 6.23 x 10^{-3} molar.

We now need one liter of a solution which is **both** 6.23 x 10^{-3}
molar in HPO_{4}^{2-} and 1.0 x 10^{-2} molar in H_{2}PO_{4}^{-}.
This requires 1.0 L x 6.23 x 10^{-3} mol/L HPO_{4}^{2-} = 6.23 x
10^{-3} mol HPO_{4}^{2-} and 1.0 L x 1.0 x 10^{-2} mol/L H_{2}PO_{4}^{-}
= 1.0 x 10^{-2} mol H_{2}PO_{4}^{-}. Since it is
physically impossible to add single ions to solutions, we have to add compounds. Choosing
the sodium salts because they are reasonably easily available specifies Na_{2}HPO_{4}
(molar mass = 142 g/mol) and NaH_{2}PO_{4} (molar mass = 120 g/mol).

6.23 x 10^{-3} mol Na_{2}HPO_{4} x 142 g/mol Na_{2}HPO_{4}
= 0.885 g Na_{2}HPO_{4}

1.0 x 10^{-2} mol NaH_{2}PO_{4} x 120 g/mol NaH_{2}PO_{4}
= 1.20 g NaH_{2}PO_{4}

The overall instructions for the preparation of one liter of aqueous buffer of pH 7.00
are: "Weigh out 1.20 g NaH_{2}PO_{4} solid and 0.885 g Na_{2}HPO_{4}
solid and dilute with water to one liter in a volumetric flask to get a buffer of pH
7.0."

Copyright 1997 James R. Fromm