How to Make a Phosphate Buffer

Useful for Biological Applications at Near-Neutral pH

Chemistry set
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In chemistry, a buffer solution serves to maintain a stable pH when a small amount of acid or base is introduced into a solution. A phosphate buffer solution is especially useful for biological applications, which are especially sensitive to pH changes, since it is possible to prepare a solution near any of three pH levels.

The three pKa values for phosphoric acid (from the CRC Handbook of Chemistry and Physics) are 2.16, 7.21 and 12.32. Monosodium phosphate and its conjugate base, disodium phosphate are usually used to generate buffers of pH values around 7, for biological applications, as shown here.

  • Note: Remember that pKa is not easily measured to an exact value. Slightly different values might be available in the literature from different sources.

Making this buffer is a bit more complicated than making TAE and TBE buffers, but it is not difficult and should take only about 10 minutes.

Materials You Will Need:

  • Monosodium phosphate
  • Disodium phosphate.
  • Phosphoric acid or sodium hydroxide (NaOH).
  • pH meter and probe
  • Volumetric flask
  • Graduated cylinders
  • Beakers
  • Stir bars
  • Stirring hotplate

Step 1. Decide on the Buffer Properties

Before making a buffer, you should first know what molarity you want it to be, what volume to make and what the desired pH is. Most buffers work best at concentrations between 0.1 M and 10 M. The pH should be within 1 pH unit of the acid/conjugate base pKa. For simplicity, this sample calculation will create 1 liter of buffer.

2. Determine the Ratio of Acid to Base

Use the Henderson-Hasselbalch equation (below) to determine what ratio of acid to base is required to make a buffer of the desired pH. Use the pKa value nearest your desired pH and the ratio will refer to the acid-base conjugate pair that corresponds to that pKa.

HH Equation: pH = pKa + log ([Base]/[Acid])

For a buffer of pH 6.9, [Base]/[Acid] = 0.4898

Substitute for [Acid] and Solve for [Base]

The desired molarity of the buffer is the sum of [Acid] + [Base].

For a 1 M buffer, [Base] + [Acid] = 1 and [Base] = 1 - [Acid]

By substituting this into the ratio equation, from step 2, you get:

[Acid] = 0.6712 moles/L

Solve for [Acid]

Using the equation: [Base] = 1 - [Acid], you can calculate that:

[Base] = 0.3288 moles/L

3. Mix the Acid and Conjugate Base

After you've used the Henderson-Hasselbalch equation to calculate what ratio of acid to base is required for your buffer, prepare just under 1 liter of solution using the correct amounts of monosodium phosphate and disodium phosphate.

4. Check the pH

Use a pH probe to confirm that the correct pH for the buffer is reached. Adjust slightly as necessary, using phosphoric acid or sodium hydroxide (NaOH).

5. Correct the Volume

Once the desired pH is reached, bring the volume of buffer to 1 liter. Then dilute the buffer as desired. This same buffer can be diluted to create buffers of 0.5 M, 0.1 M, 0.05 M or anything in between.

Here are two examples of how a phosphate buffer can be calculated.

Example #1

In this example, the procedure requires a 0.1 M Na-phosphate buffer, pH 7.6.7.6.

Using the Henderson-Hasselbalch equation (pH = pKa + log ([salt]/[acid]), the salt is Na2HPO4 and the acid is NaHzPO4. The buffer will be most effective at its pKa— the point where [salt] = [acid]. In this example, it's evident that if the [salt] > [acid], the pH will then be greater than the pKa; conversely, if [salt] < [acid], the pH will be less than the pKa. It follows, then, that if we make up a solution of acid (NaH2PO4) its pH will be less than the pKa, and therefore less than the pH required for the solution to act as an effective buffer.

It's necessary, then, to titrate it with a base with a pH closer to the pKa. NaOH offers a suitable base, since it maintains sodium as the cation:

NaH2PO4 + NaOH--+ Na2HPO4 + H20.

Once titrated to the correct pH, the solution may be diluted to the volume that gives the required molarity. The HendersonHasselbalch equation states that the pH is determined by the ratio of salt to acid, rather than their absolute concentrations. There are three implications:

  • The only by-product in this reaction is water.
  • The molarity of the buffer is governed by the mass of the NaH2PO4 acid, which is weighed out, and the final volume of the solution is made up.
  • The concentration of NaOH has no relevance, so any concentration can be used.
  • The reaction requires only a simple calculation of molarity and a single weighing—only one solution needs to be mixed, and because all the weighed material is used in the buffer, nothing is wasted.

Make sure to not weigh out the "salt" (NazHPO4) in the first instance, since this produces an unwanted by-product. In a salt solution, the pH will exceed the pKa, requiring titration with an acid to lower the pH. If HC1 is used, the reaction will be:

Na2HPO4 + HC1--+ NaH2PO4 + NaC1,

which produces NaC1 of an indeterminate concentration—not wanted in the buffer.

In some situations, it's necessary to have a gradient of [NaC1] superimposed on the buffer—such as in an ion-exchange ionic-strength gradient elution, for example. When this is the case, two buffers are required for the two chambers of the gradient generator—the starting buffer (the equilibration buffer without added NaC1, or with the starting concentration of NaC1) and the finishing buffer, which is identical to the starting buffer but which also contains the finishing concentration of NaC1. When making up the finishing buffer, make sure to account for common ion effects caused by the sodium ion.

Example #2

In this example, the requirement is for an ionic-strength gradient finishing buffer, 0.1 M Na-phosphate buffer, pH 7.6, containing 1.0 M NaCl.

Here, the NaC1 is weighed out and made up together with NaHEPO4—because the common ion effects are accommodated by the titration, it avoids complex calculations. For 1 liter of buffer, dissolve NaH2PO4.2H20 (15.60 g) and NaC1 (58.44 g) in about 950 ml of distilled H20, titrated to pH 7.6 using a fairly concentrated NaOH solution, made up to 1 liter.