Question

Phosphoric acid is a triprotic acid with the following pKa values:

pKa1=2.148

pKa2=7.198

pKa3=12.375

You wish to prepare 1.000 L of a 0.0100 M phosphate buffer at pH 7.580. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?

Mass NaH2PO4=? Mass Na2HPO4=?

Answer 1

SOLUTION

Our buffer will undergo following dissociation reaction

H₂PO₄^- <---------> HPO₄^2- pKa2=7.198 (Bz choosen pH is close to pKa2)

pH = pKa2 + log[HPO₄^2- ] / [H₂PO₄^-]

log[HPO₃^2- ] / [H₂PO3^-] = pH - pKa2 = 7.580 - 7.198 = 0.382

[HPO₃^2- ] / [H₂PO3^-] = antilog(0.382) = 2.4

[HPO₄^2- ] = [H₂PO₄^-] X 2.4 =======1)

Further Molarity of solutio = 0.0100M

It implies [HPO₄^2- ] + [H₂PO₄^-] = 0.0100M ====2)

Using eqn 1) in eqn 2)

[H₂PO₄^-] X 2.4 + [H₂PO₄^-] = 0.0100M

3.4[H₂PO3^-] = 0.0100M

[H₂PO3^-] = 0.0100M / 3.4 = 0.00294M

Hence from eqn 2) [HPO₄^2- ] = 0.0100M - [H₂PO₄^-] = 0.0100 - 0.00294 = 0.00706M

Mass of Na₂HPO₄^2- required = No. of moles X Molar mass of Na₂HPO₄ ( Because HPO₄^2- is obtained from Na₂HPO₄)

Mass of [Na₂HPO₄^2- ] = 0.00706 X 141.95 ( bz solution in 1L in volume hence Molarity = No. of moles) = 1.002g

Similarly Mass of NaH₂PO₄^- = No. of moles X Molar mass of NaH₂PO_{4 =} 0.00294 X 119.95 = 0.352g