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QUESTION

# 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