Buffer exercise answer 1

100 cm³ of pH 7.0 buffer needs to be made from some 0.1 mol dm^-3 potassium dihydrogenphosphate (KH₂PO₄) solution and 0.1 M dipotassium hydrogenphosphate (K₂HPO₄) solution. The following equilibrium is set up:

H₂PO₄^-(aq) HPO₄²^-(aq) + H⁺(aq)

and the K_a has a value of 1.39 × 10^-7 mol dm^-3. Calculate the appropriate volumes of the solutions which need to be used.

pK_a = - log₁₀K_a = - log₁₀(1.39 × 10^-7) = 6.86

We need to use the following rearranged equation

This gives 6.86 - 7.0 = - 0.14 = log₁₀([HA]/[A^-]

[HA] = [H₂PO₄^-] and [A^-] = [HPO₄²^-]

log₁₀([H₂PO₄^-]/[HPO₄²^-]) = - 0.14

We remove the log with the 10^x (inv log) function

So [H₂PO₄^-]/[HPO₄²^-] = 10^-^0.14 = 0.72

As the two solutions have the same concentration, we could achieve this ratio with 0.72 cm³ of H₂PO₄^- and 1 cm³ of HPO₄^2-. This would only have a total volume of 1.72 cm³ and we are looking for 100 cm³.

We need 100/1.72 = 58.1 lots of 1.72 cm³

So we need 58.1 × 1 = 58.1 cm³ of HPO₄^2-

and 58.1 × 0.72 = 41.9 cm³ of H₂PO₄^-