Question

what is the ph of a solution that is .900 M k2Po4. ka 1= 7.6*10^-3, ka2=6.2*10^-8, ka3=4.2*10^-13

Answer 1

​​​​​​Lets look at the reaction :-

HPO4(2-) +H2O <===> PO4(3-) + H3O(+).

K⁺ ion is a spectator ion so we need not bother about it, sink nce there is only one disassociation, and if you look above, the Ka for this = Ka³ = 4.2 × 10^-13

​​​​So,
Ka = [PO₄(3-)] [H3O(+)] / [HPO4(2-)] = 4.2×10^-13

We can establish from the equation that [PO3(3-)] = [H3O(+)]
and we can also establish that [H3O+] = [H+] as the number of protons attached to water to formed H3O+ is the same as the number of H3O+ molecules themselves

Hence Ka = [H+]^2 / [HPO4(2-)] = 4.2×10^-13
pKa = -lg(Ka) = lg[HPO4(2-)] - lg[H+]^2 = - lg(4.2×10^-13)
pH = - lg[H+]
- lg [H+]^2 = - 2 lg[H+]
2pH = - lg (4.2×10^-13) - lg [HPO4(2-)]
2pH = - lg (4.2×10^-13) - lg (0.900)
pH = 6.212