Acids and Bases in Water Question...Determining Equations?

Question

Question: Write the Ka expression for each of the following in water: (a) H2PO4-, (b) H3PO2, (c) HSO4-

My Approach:

(a) H2PO4 - + OH- <----> H2O + HPO4 2- >>> ka = [HPO4]/[OH-][H2PO4]

(b) H3PO2 + OH- <---> H2O + H2PO2 - >>> ka = [H2PO2]/ [OH][H3PO2]

(c) HSO4 - + H2O <----> H3O+ + SO4 2- >>> ka = [H3O][SO4]/ [HSO4]

I know I'm probably not right, but I cannot see it any other way. Can someone help me correct my problems either by explaining or showing me the right answers? Thanks!

Answer 1

There souldn't be any reference to OH- in a Ka expression. Your third one is correct, except that you must still put the ion charges in.

Why not:

(H2PO4)- + H2O ----> (HPO4)2- + H3O+

Ka = [(HPO4)2-] x [H3O+]/[(HPO4)2-]

Missing out the water itself in the normal way.

Answer 2

(a) this acid is H3PO4 so it has three acidic hydrogen:
First equilibrium constant:
Ka1=[H2PO4-][H3O+]/[H3PO4]; pKa1=2.15
Second equilibrium constant:
Ka2=[HPO4 (2-)][H3O+]/[H2PO4-]; pKa2=7.21
Third equilibrium constant:
Ka3=[PO4 (3-)][H3O+]/[H2PO4(2-)]; pKa3=12.26
all tha acids react with water in the formation of H3O+
Ka1>>Ka2>Ka3

(b)this acid has just one acidic Hydrogen So;
Ka=[H2PO2 (-)][H3O+]/[H3PO2];

(c)this acid is sulfuric acid with formulation of H2SO4
this acid also react with water in two steps and have two equilibrium constant:
Ka1=[HSO4(-)][H3O+]/[H2SO4];
Ka2=[SO4(2-)][H3O+]/[HSO4(-)];
Ka1>>Ka2;

Answer 3

The equilibrium consistent for acids constantly contains the concentrations of the products divided by using the concentration of the reactant. H3PO4 ---> H+ + H2PO4- Ka = [H+][H2PO4-] / [H3PO4] HClO2 ----> H+ + ClO2- Ka = [H+][ClO2-] / [HClO2] CH3COOH ----> CH3COO- + H+ Ka = [CH3COO-][H+] / [CH3COOH] HCO3- ----> H+ + CO32- Ka = [H+][CO32-] / [HCO3-] HSO4- ----> H+ + SO42- Ka = [H+][SO42-] / [HSO4-]