Question

In some natural systems the pH must be maintained within very narrow limits. For example, in human blood the pH must remain close to 7.4 or cell deterioration occurs. Blood contains several weak acid/conjugate base equilibria called buffers which control the pH. One weak acid present in blood is the dihydrogen phosphate ion, H2PO4-1 for which the equilibrium in aqueous solution is:

H2PO4-1(aq) <---------> HPO4-2(aq) + H+1(aq)

0.50 mol H2PO4-1 and 0.50 mol HPO4-2 are in equilibrium in 1L of aqueous solution, the pH of the solution is 7.2. If 0.01 mol of HCl is added to the 1L solution, assuming that all the added H+ ions are used up in the equilibrium shift, calculate the concentration of H2PO4-1(aq) and HPO4-2(aq) and hence the pH of this new solution.

Answer 1

First we write down the Henderson Hasselbach equation for this particular reaction, as shown below:

pH = pKa + log([conjugate base]/[acid])

In this reaction, the acid is H2PO₄^-, which gives the conjugate base HPO₄^2- upon dissociation.

So, we re-write the above equation as:

pH = pKa + log([HPO₄^2-]/[H2PO₄^-])

Since it is given that both acid and conjugate base have same number of moles, equal to 0.5, in 1 L solution, so we have:

[HPO₄^2-] = [H2PO₄^-] = moles/volume = 0.5/1 = 0.5 M

So,

pH = pKa + log(0.5/0.5) = pKa + log(1) = pKa

So,

pH = pKa = 7.2

Thus for this acid dissociation reaction, pKa = 7.2

Now, when we add 0.01 mol of HCl, we essentially add 0.01 mol of H+, which causes the reaction to shift in the reverse direction. So total moles of acid increase by 0.01 and total moles of conjugate base decrease by 0.01, following thid addition.

Re-writing the equation as:

pH = pKa + log((0.5+0.01)/(0.5-0.01))

So,

pH = 7.2 + log((0.5+0.01)/(0.5-0.01)) = 7.217

Thus we see only a slight increase in the pH owing to the buffering action of the acid-conjugate base pair system.

Final conc of acid and base are:

[Acid] = [H2PO4-] = (0.5+0.01)1 = 0.51 M

[Conjugate Base] = [HPO4^2-] = (0.5-0.01) = 0.49 M