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.

Answer 1

Here's what these equations are.

Typically, a phosphate buffered saline (PBS) buffer comprises the subsequent species:

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

Between these two ions in solution, an equilibrium reaction is created, with hydrogen phosphate functioning as a base and dihydrogen phosphate acting as an acid, or giving up 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))#

The hydrogen phosphate ion will absorb the excess hydronium ions when a strong acid is added to the buffer.

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

This will turn the strong acid into a weak acid. Similarly, the dihydrogen phosphate ion will eat up the excess hydroxide ions in the presence of a strong base.

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

Thus, the strong base will change into a weak base.

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

The pH of a PBS buffer can be determined using the Hendeson-Hasselbalch equation.

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

As for you, you've

#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 formula turns into

#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#.

One typical method for making 1X PBS buffers is to use

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

This will provide you with

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

After that, the pH would be adjusted to 7.4 using hydrochloric acid.

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Answer 2

Axel Alvarado

The chemical equations describing the buffering action of PBS are:

[ \text{H}_2\text{PO}_4^- + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{HPO}_4^{2-} ] [ \text{HPO}_4^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{PO}_4^{3-} ]

The theoretical pH of PBS can be calculated using the Henderson-Hasselbalch equation:

[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

For PBS, the relevant pKa values are approximately 2.1 for ( \text{H}_2\text{PO}_4^- ) and 7.2 for ( \text{HPO}_4^{2-} ). Given that PBS typically contains equimolar concentrations of ( \text{H}_2\text{PO}_4^- ) and ( \text{HPO}_4^{2-} ), the concentration ratio is 1:1. Plugging the values into the Henderson-Hasselbalch equation:

[ \text{pH} = 7.2 + \log(1) = 7.2 ]

Therefore, the theoretical pH of phosphate buffered saline (PBS) is 7.2.

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