Question

What is the equilibrium concentration of PO4 in a 0.511 M solution of H3PO4 (Ka1 = 7.5E-3, ka2= 6.2E-8, ka3= 4.8E-13) I troes using the 5% rule but when i checked my work i was at 11%. I dont know how to solve this if i cant use the 5% rule. Help please.

Answer 1

First, We will use the relation, pKa1 = [H2PO4-][H+] / [H3PO4]

The important equilibrium reaction is:

H3PO4 + H2O <=> H3O^+ + H2PO4^-

Thus Ka1 = [H3O^+][H2PO4^-]/[H3PO4] = 7.5 x 10^-3

Let [H3PO4] = 0.511 - x; [H3O^+] = x; [H2PO4^-] = x. This is the condition at equilibrium.

(x)(x)/(0.511 - x) = 7.5 x 10^-3 assuming x to be small we have,

7.5 x 10^-3 = x^2 / 0.511

x = 0.062 = [H3O+] = [H2PO4-]

Remaining [H3PO3] = 0.511 - 0.062 = 0.449

lets consider the second equilibrium using data from the 1st.

H2PO4^- + H2O <=> H3O^+ + HPO4^2-

thus, Ka2 = 6.2 x 10^-8 = [H3O^+][HPO4^2-]/[H2PO4^-]

Let [H2PO4^-] = 0.062 = [H3O^+];

and, [HPO4^2-] = x

6.2 x 10^-8 = (x)(0.062)/(0.062) Rearrange this into the classic quadratic equation.

x = 6.2 x 10^-8 M = [HPO4^2-]

Let us consider the third disscociation,

Ka3 = 4.8 x 10^-13 = [H3O+][PO4^3-]/[HPO4^2-]

Let, [PO4^2-] = x = [H3O+]

We get by substituting the values,

4.8 x 10^-13 = (x)(0.062)/(6.2 x 10^-8)

x = 4.8 x 10^-19

Thus, PO4^2- = 4.8 x 10^-19