#H_3C-C(=O)OH +H_2O rightleftharpoonsH_3C-C(=O)O^(-) + H_3O^+#
#K_a# #=# #([H_3O^+][H_3C-C(=O)O^(-)])/([H_3C-C(=O)OH])#
Now it is a given that #[H_3O^+]# #=# #0.0019*mol*L^-1#. Given the stoichiometry of the equation, #[HO^-]# #=# #0.0019*mol*L^-1# as well, and #[H_3C-C(=O)OH]=(0.200-0.0019)*mol*L^-1#.
Thus #K_a# #=# #(0.0019)^2/(0.200-0.0019)# #=# #1.82xx10^(-5)#.