What is the set of chemical equations that describe the buffering action of phosphate buffered saline (PBS)? Calculate theoretically the pH of phosphate buffered saline.

Here’s what these equations are.

A phosphate buffered saline (PBS) buffer usually contains the following species

  • Sodium chloride, ##NaCl##;
  • Potassium chloride, ##KCl##;
  • Disodium phosphate, ##Na_2HPO_4##;
  • Monopotassium phosphate, ##KH_2PO_4##.

The two species thata give PBS its buffer capacity are the hydrogen phosphate, ##HPO_4^(2-)##, and dihydrogen phosphate, ##H_2PO_4^(-)## ions.

An is established between these two ions in solution, with dihydrogen phosphate acting as an acid, i.e. donating a proton, and hydrogen phosphate acting as a base, i.e. accepting a proton.

##underbrace(H_2PO_((aq))^(-))_(color(blue)(“acid”)) + H_2O_((l)) rightleftharpoons underbrace(HPO_(4(aq))^(2-))_(color(green)(“conj base”)) + H_3O_((l))^(+)## ##” “color(red)((1))##

When a strong acid is added to the buffer, the excess hydronium ions will be consumed by the hydrogen phosphate ion

##H_3O_((aq))^(+) + HPO_(4(aq))^(2-) -> H_2PO_(4(aq))^(-) + H_2O_((l))##

The strong acid will thus be converted to a weak acid. Likewise, when a strong base is added, the excess hydroxide ions will be consumed by the dihydrogen phosphate ion.

##OH_((aq))^(-) + H_2PO_(4(aq))^(-) -> HPO_(4(aq))^(2-) + H_2O_((l))##

The strongbase will thus be converted to a weak base.

In relation to equation ##color(red)((1))##, you can say that

  • Excess hydronium ions will shift the equilibrium to the left;
  • Excess hydroxide ions will shift the equilibrium to the right.

In order to calculate the of a PBS buffer, you can use the Hendeson-Hasselbalch equation

##pH_”sol” = pK_a + log(([“conj base”])/([“weak acid”]))##

In your case, you have

##pH_”sol” = pK_a + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))##

To get the ##pK_a##, you need the value of the acid dissociation constant, ##K_a##, for dihydrogen phosphate.

##K_a = 6.23 * 10^(-8)##

By definition, ##pK_a## is equal to

##pK_a = -log(K_a) = -log(6.23 * 10^(-8)) = 7.21##

The H-H equation becomes

##pH_”sol” = 7.21 + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))##

If the buffer contains equal concentrations of hydrogen phosphate and dihydrogen phosphate, then the pH of the solution will be equal to the ##pK_a##.

##pH_”sol” = 7.21 + underbrace(log(([HPO_4^(2-)])/([H_2PO_4^(-)])))_(color(blue)(“=0”)) = 7.21##

Usually, 1X PBS buffers have a pH of about 7.4. A bigger concentration of hydrogen phosphate is used, which will determine the pH to be bigger than ##pK_a##.

A common way to prepare 1X PBS buffers is to use

##[HPO_4^(2-)] = “10 mM”## ##[H_2PO_4^(-)] = “1.8 mM”##

This will give you

##pH_”sol” = 7.21 + log((10cancel(“mM”))/(1.8cancel(“mM”))) = 7.95##

You’d then use hydrochloric acid to adjust the pH to 7.4.

Read more on that here:

http://cshprotocols.cshlp.org/content/2006/1/pdb.rec8247