# 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.

Jun 26, 2015

Here's what these equations are.

#### Explanation:

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

• Sodium chloride, $N a C l$;
• Potassium chloride, $K C l$;
• Disodium phosphate, $N {a}_{2} H P {O}_{4}$;
• Monopotassium phosphate, $K {H}_{2} P {O}_{4}$.

The two species thata give PBS its buffer capacity are the hydrogen phosphate, $H P {O}_{4}^{2 -}$, and dihydrogen phosphate, ${H}_{2} P {O}_{4}^{-}$ ions.

An equilibrium reaction 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.

${\underbrace{{H}_{2} P {O}_{\left(a q\right)}^{-}}}_{\textcolor{b l u e}{\text{acid")) + H_2O_((l)) rightleftharpoons underbrace(HPO_(4(aq))^(2-))_(color(green)("conj base}}} + {H}_{3} {O}_{\left(l\right)}^{+}$ $\text{ } \textcolor{red}{\left(1\right)}$

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

${H}_{3} {O}_{\left(a q\right)}^{+} + H P {O}_{4 \left(a q\right)}^{2 -} \to {H}_{2} P {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$

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.

$O {H}_{\left(a q\right)}^{-} + {H}_{2} P {O}_{4 \left(a q\right)}^{-} \to H P {O}_{4 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)}$

The strongbase will thus be converted to a weak base.

In relation to equation $\textcolor{red}{\left(1\right)}$, 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 pH of a PBS buffer, you can use the Hendeson-Hasselbalch equation

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

$p {H}_{\text{sol}} = p {K}_{a} + \log \left(\frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]}\right)$

To get the $p {K}_{a}$, you need the value of the acid dissociation constant, ${K}_{a}$, for dihydrogen phosphate.

${K}_{a} = 6.23 \cdot {10}^{- 8}$

By definition, $p {K}_{a}$ is equal to

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(6.23 \cdot {10}^{- 8}\right) = 7.21$

The H-H equation becomes

$p {H}_{\text{sol}} = 7.21 + \log \left(\frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]}\right)$

If the buffer contains equal concentrations of hydrogen phosphate and dihydrogen phosphate, then the pH of the solution will be equal to the $p {K}_{a}$.

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 $p {K}_{a}$.

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

$\left[H P {O}_{4}^{2 -}\right] = \text{10 mM}$
$\left[{H}_{2} P {O}_{4}^{-}\right] = \text{1.8 mM}$

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.