And #K_a=([H_3O^+][A^-])/([HA])#
If #K_a# is LARGE, then we gots a strong acid...if #K_a# is small, then we got a weak acid, and the equilibrium LIES to the LEFT as written.
For #a.# we gots a diprotic acid.....
#H_3PO_4(aq) + 2H_2O(l) rightleftharpoonsHPO_4^(-) +2H_3O^+#
#K_a=([HPO_4^(2-)][H_3O^+]^2)/([H_3PO_4])#
For #b.# we gots a monoprotic weak acid.....
#HClO_2(aq) + H_2O(l) rightleftharpoonsClO_2^(-) +H_3O^+#
#K_a=([HClO_2^(-)][H_3O^+])/([HClO_2])#
For #c.# we gots a monoprotic weak acid.....
#HOAc(aq) + H_2O(l) rightleftharpoonsAcO^(-) +H_3O^+#
#K_a=??#. Have bash at the remaining expressions yourself.