H_3C-C(=O)OH +H_2O rightleftharpoonsH_3C-C(=O)O^(-) + H_3O^+H3C−C(=O)OH+H2O⇌H3C−C(=O)O−+H3O+
K_aKa == ([H_3O^+][H_3C-C(=O)O^(-)])/([H_3C-C(=O)OH])[H3O+][H3C−C(=O)O−][H3C−C(=O)OH]
Now it is a given that [H_3O^+][H3O+] == 0.0019*mol*L^-10.0019⋅mol⋅L−1. Given the stoichiometry of the equation, [HO^-][HO−] == 0.0019*mol*L^-10.0019⋅mol⋅L−1 as well, and [H_3C-C(=O)OH]=(0.200-0.0019)*mol*L^-1[H3C−C(=O)OH]=(0.200−0.0019)⋅mol⋅L−1.
Thus K_aKa == (0.0019)^2/(0.200-0.0019)(0.0019)20.200−0.0019 == 1.82xx10^(-5)1.82×10−5.