Question

Recall that, in general, for the reaction H2O + HX --> H3O+ + X-

pH= pK - log [HX]\[X-]

Using this relation, write pH expressions corresponding to pK1, pK2, and pK3 for H3PO4.

Answer 1

First, write the reactions for the H₃PO₄ hydrolisis:

H₃PO₄ + H₂O -----------> H₂PO₄^- + H₃O⁺   Ka1

H₂PO₄^- + H₂O ----------> HPO₄^2- + H₃O⁺   Ka2

HPO₄^2- + H₂O ----------> PO₄^3- + H₃O⁺ Ka3

For the first reaction:

pH₁ = -logKa1 + log [H₂PO₄^-]/[H₃PO₄] = pKa1 + log[H₂PO₄^-]/[H₃PO₄]

For the second reaction:

pH₂ = -logKa2 + log[HPO₄^2-]/[H₂PO₄^-] = pKa2 + log[HPO₄^2-]/[H₂PO₄^-]

For the third reaction:

pH₃ = -logKa3 + log[PO₄^3-]/[HPO₄^2-​] = pKa3 + log[PO₄^3-]/[HPO₄^2-​]

Hope this helps