Question

A phosphate buffer contains 0.075M K₂HPO₄ and 0.125M KH₂PO₄. pKa values for H₃PO₄ are 2, 6.8, and 12.2. Calculate the concentrations of:

1.) H₂PO₄^- concentration

2.) H⁺ concentration

Answer 1

H3PO4 ---------> H+ + H2PO4-       pKa1 = 2

H2PO4- ---------> H+ + HPO4^2-      pKa = 6.8

HPO4^2- ----------> H+ + PO4^3-      pKa = 12.2

(1)

[HPO4^2-] = 0.075 M   and [H2PO4-] = 0.125 M

So Answer for 1st part is [H2PO4-] = 0.125 M

Now in order to calculate [H+] we have calculate each [H+] and add them to get total [H+]

pKa1 = 2 , Ka1 = 10^-2 = 0.01 ( Since pKa = -log[Ka] )

pKa2 = 6.8 , Ka2 = 10^-6.8 = 1.58 x 10^-7

pKa3 = 12.2 , Ka3 = 6.31 x 10^-13

For 1st step ionisation

[H+] = [H2PO4-] = 0.125 M

For 2nd part

Ka2 = [HPO4^2-]*[H+] / [H2PO4-]

[H+] = Ka2 * [H2PO4-] / [HPO4^2-] = 1.58 x 10^-7 * 0.125 / 0.075 = 2.63 x 10^-7 M

For 3rd part Ka3 <<< Ka2

So ,

[H+] will be same for part 2 as it will not dissociation is negligible

[H+] = 2.63 x 10^-7

So,

Total [H+] = 0.01 + 2 * ( 2.63 x 10^-7)

[H+] = 0.01 M