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 the process is not difficult and should take only about 10 minutes.

Materials

To make your phosphate buffer, you'll need the following materials:

  • 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 creates 1 liter of buffer.

Step 2. Determine the Ratio of Acid to Base

Use the Henderson-Hasselbalch (HH) 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; the ratio refers 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

Step 3. Mix the Acid and Conjugate Base

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

Step 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).

Step 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, as described by Clive Dennison, Department of Biochemistry at the University of Natal, South Africa.

Example No. 1

The requirement is for a 0.1 M Na-phosphate buffer, pH 7.6.

In the Henderson-Hasselbalch equation, pH = pKa + log ([salt] / [acid]), the salt is Na2HPO4 and the acid is NaHzPO4. A buffer is most effective at its pKa, which is the point where [salt] = [acid]. From the equation it is clear that if the [salt] > [acid], the pH will be greater than the pKa, and if [salt] < [acid], the pH will be less than the pKa. Therefore, if we were to make up a solution of the acid NaH2PO4, its pH will be less than the pKa, and therefore will also be less than the pH at which the solution will function as a buffer.

To make a buffer from this solution, it will be necessary to titrate it with a base, to a pH closer to the pKa. NaOH is a suitable base because it maintains sodium as the cation:

NaH2PO4 + NaOH--+ Na2HPO4 + H20.

Once the solution has been titrated to the correct pH, it may be diluted (at least over a small range, so that deviation from ideal behavior is small) to the volume that will give the desired molarity. The HH equation states that the ratio of salt to acid, rather than their absolute concentrations, determines the pH. Note that:

  • In this reaction the only by-product is water.
  • The molarity of the buffer is determined by the mass of the acid, NaH2PO4, which is weighed out, and the final volume to which the solution is made up. (For this example 15.60 g of the dihydrate would be required per liter of final solution.)
  • The concentration of the NaOH is of no concern, so any arbitrary concentration can be used. It should, of course, be concentrated enough to effect the required pH change in the available volume.
  • The reaction implies that only a simple calculation of molarity and a single weighing is required: only one solution needs to be made up, and all of the material weighed out is used in the buffer—that is, there is no waste.

    Note that it is not correct to weigh out the "salt" (Na2HPO4) in the first instance, as this gives an unwanted by-product. If a solution of the salt is made up, its pH will be above the pKa and it will require titration with an acid to lower the pH. If HC1 is used, the reaction will be:

    Na2HPO4 + HC1--+ NaH2PO4 + NaC1,

    yielding NaC1, of an indeterminate concentration, which is not wanted in the buffer. Sometimes—for example, in an ion exchange ionic-strength gradient elution—it is required to have a gradient of, say, [NaC1] superimposed on the buffer. Two buffers are then required, for the two chambers of the gradient generator: the starting buffer (that is, the equilibration buffer, without added NaC1, or with the starting concentration of NaC1) and the finishing buffer, which is the same as the starting buffer but which additionally contains the finishing concentration of NaC1.

    In making up the finishing buffer, common ion effects (due to the sodium ion) must be taken into account.

    Example as noted in the journal Biochemical Education 16(4), 1988.

    Example No. 2

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

    In this case, the NaC1 is weighed out and made up together with the NaHEPO4; common ion effects are accounted for in the titration, and complex calculations are thus avoided. For 1 litre of buffer, NaH2PO4.2H20 (15.60 g) and NaC1 (58.44 g) are dissolved in about 950 ml of distilled H20, titrated to pH 7.6 with a fairly concentrated NaOH solution (but of arbitrary concentration) and made up to 1 liter. 

    Example as noted in the journal Biochemical Education 16(4), 1988.