Question

Determine the volume of a solution of 1 M HCl that must be added to adjust the pH from 9 to 4 in 100 mL of a 100 mM solution of phosphoric acid. Phosphoric acid has pKs of 2.15, 6.82, and 12.38. For this problem, first estimate the amount of HCl needed based on your knowledge of titration, then accurately determine the correct amount using the Henderson-Hasselbalch equation.

Answer 1

H₃PO₄(aq) + H₂O(l) <==>H₂PO₄^-(aq) + H₃O⁺(aq)

H₂PO₄^-(aq) + H₂O(l) <==>HPO₄^2-(aq) + H₃O⁺(aq)

HPO₄^2-(aq)+ H₂O(l) <==>PO₄^3-(aq) + H₃O⁺(aq)

we have, PKa1=2.15 ; Pka2 =6.82 ; PKa3 = 12.38

pH = 9.0 , it is after Pka2 , thus based on Pka values we can deduce that :

Principal phosphate species in system : HPO₄^2-(aq)

100 mM = 0.1 M

on adding HCl, we will Have,

HPO₄^2-(aq) + H₃O⁺(aq) <==> H₂PO₄^-(aq) + 2 H₂O(l)

we will have ~ 1:1 titration ,

thus V(HCl) ~ 0.1 M *100 mL / 1.0 M ~ 10 mL

pH = pKa1 + log {[H₂PO₄^-]/[H₃PO₄]}

4.0 = 2.15 + log {[H₂PO₄^-]/[H₃PO₄]}

log {[H₂PO₄^-]/[H₃PO₄]} = 1.85

[H₂PO₄^-]/[H₃PO₄] = 70.8

[H₂PO₄^-] + [H₃PO₄] = 0.1 M

thus, [H₂PO₄^-] = 0.0986 M; [H₃PO₄] = 0.0014 M

Thus , amount of HCl needed :

V(HCl) = 1.014*0.1 M *100 mL / 1.0 M = 10.14 mL