Question

Why does concentration of OH get cut in half and then a third at the second and third equivalence points? If the titration of H₂PO₄⁻ in a urine sample was continued until all of the acid in the solution was neutralized.

2 Equivalents of NaOH are needed to neutralize H2PO4- [PO4-31 = [OH) Equivalence point 3 Buffer Region 3 [HPO4-21 = { [0h) HPO4-2 + OH- = P04-3 + H20 - - - - - - - Equivalence point 2 Buffer Region 2 pH [H2PO4+] = [OH). H2PO4 + OH- = HPO4-2+ H20 - - - - - - -- - Equivalence point 1 Buffer Region 1 H3PO4 + OH = H2PO4 + H2O Volume of OH added (mL

Answer 1

At second equivalent point :

H₃PO₄ + 2 OH^- $\rightarrow$ HPO₄^2- + 2 H₂O

That means when twice the number of moles of OH^- are added to the phosphoric acid , to reach second equivalent point . Hence the HPO₄^2- moles are formed is half the OH^- moles used . Hence [HPO₄^2-] = (1/2 )× [OH^-]

At third equivalent point :

H₃PO₄ + 3 OH^- $\rightarrow$ PO₄^3- + 3 H₂O

That means when thrice the number of moles of OH^- are added to the phosphoric acid , to reach third equivalent point . Hence the PO₄^3- moles are formed is one third the OH^- moles used . Hence [PO₄^3-] = (1/3)× [OH^-]