#NH_4^+# is a weak acid, which would give us an idea if not about the actual value, then at least about its dissociation constant's dimension. Knowing that the ammonium ion (#NH_4^+#) is a weak acid, we should expect #K_a# to be less than #1#.
That is, a very small quantity of products ( ammonia and hydronium cation ) will be formed, the reaction favoring the reactans to a large degree.
#NH_4^+ (aq) + H_2O(l) <=> NH_3 (aq) + H_3^+O (aq)#
#[K_a]# is defined as
#K_a = ([NH_3] * [H_3^+O])/([NH_4^+])#, and can be seen as a measure of an acid's strenght in solution. The best way to go about actually solving a problem is by using the ICE method (I've linked a wikipedia article on this).
#NH_4^+# has an actual dissociation constant (I've provided an acid strenght table) of #K_a = 5.8 * 10^(-10)#.
