Question

For a 0.110 M H3PO4 solution, find the equilibrium concentrations of H3PO4, H2PO4-, HPO4^2-, PO4^3-, and H3O+.

For H3PO4: Ka1= 7.5x10^-3, Ka2= 6.2x10^-8, Ka3= 4.2x10^-13

Answer 1

For the first dissociation of the acid

H3PO4 ---- > H2PO₄^-(aq) + H₃O⁺ ; Ka1 = 7.5x10^-3

0.110M, ------- 0 ---------------- 0

(0.110 - x)M, x M, ----------- x M

Ka1 = 7.5*10^-3 = [H2PO₄^-(aq)]*[H₃O⁺] / [H3PO4] = x² / (0.110 - x)

=> x = 0.0252 M

Hence [H3PO4] = (0.110 - x) M = (0.110 - 0.0252 ) = 0.0848 M (answer)

After first dissociation, [H2PO₄^-(aq)] = [H₃O⁺] = x =  0.0252 M

For the second dissociation of the acid

H2PO₄^-(aq) --- >HPO₄^2-(aq)  + H₃O⁺ ; Ka2 = 6.2x10^-8

0.0252 M, ---------- 0 ---------------- 0.0252 M

(0.0252 - y)M, y M, ------------ (0.0252 + y) M

Ka2 = 6.2x10^-8  = [HPO₄^2-(aq]*[H₃O⁺] / [H2PO₄^-(aq)] = y * (0.0252 + y)  / (0.0252 - y)

=> y² + 0.0252y - 1.5624*10^-9 = 0

=> y = 6.20x10^-8 M

Hence [H2PO₄^-(aq)] = (0.0252 - y)M 0.0252 M (answer)

[HPO₄^2-(aq)] = y = 6.20x10^-8 M

Since the third dissociation is very weak, we can neglect the H3O+ produced due to third dissociation. Hence

[H3O+] =  (0.0252 + y) M 0.0252 M (answer)

For the third dissociation of the acid

HPO₄^2-(aq) --- >PO₄^3-(aq)  + H₃O⁺ ; Ka3 = 4.2x10^-13

6.20x10^-8 M, --- 0 ----------- 0.0252 M

(6.20x10^-8 - z)M, z M, ------ 0.0252 M

Ka3 = 4.2x10^-13  = [PO₄^3-(aq)]*[H₃O⁺] / [HPO₄^2-(aq)] = z* 0.0252 / (6.20x10^-8 - z)

=> z = 1.03x10^-18 M

Hence [PO₄^3-(aq)] = z = 1.03x10^-18 M (answer)

=> y = 6.20x10^-8 M

Hence [H2PO₄^-(aq)] = (0.0252 - y)M 0.0252 M (answer)

[HPO₄^2-(aq)] = y = 6.20x10^-8 M