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^-#

Question

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^-#

Ellie Andrews · Accepted Answer

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

Cameron Andrews · Answer

undefined a) $H_3PO_4 \rightarrow 3H^+ + PO_4^{3-}$
   $K_a = \frac{[H^+]^3}{[PO_4^{3-}]} $

b) $HClO_2 \rightarrow H^+ + ClO_2^-$
   $K_a = \frac{[H^+][ClO_2^-]}{[HClO_2]}$

c) $CH_3COOH \rightarrow H^+ + CH_3COO^-$
   $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$

d) $HCO_3^- \rightarrow H^+ + CO_3^{2-}$
   $K_a = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$

e) $HSO_4^- \rightarrow H^+ + SO_4^{2-}$
   $K_a = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

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^-#