How do you write the balanced acid equation and the dissociation expression Ka for following compounds in water? a) #H_3PO_4# b) #HClO_2# c) #CH_3COOH# d) #HCO_3^-# e) #HSO_4^-#

1 Answer
Sep 15, 2017

Answer:

#HA(aq) + H_2O(l) rightleftharpoons H_3O^+ + A^-#

Explanation:

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.