Given that the reaction is between Bismuth(III) hypochlorite and acetic acid.
Molecular equation for the reaction can be represented as:
[tex]Bi(OCl)_{3}(aq)+3CH_{3}COOH(aq)-->(CH_{3}COO)_{3} Bi(aq)+3HOCl(aq)[/tex]
Total ionic equation for this reaction is: Bismuth(III) hypochlorite ionizes completely in solution where acetic acid being a weak acid partially ionizes in solution. On the product side, bismuth acetate being a salt completely ionizes to give 3 acetate ions and 1 bismuth ion. The other product is hypochlorous acid which ionizes partially in solution being a weak acid.
[tex]Bi^{3+}(aq)+3OCl^{-}(aq)+3CH_{3}COOH(aq)-->3CH_{3}COO^{-}(aq)+Bi^{3+}(aq)+HOCl(aq)[/tex]
Therefore, there are 4 free ions ([tex]3CH_{3}COO^{-} and one Bi^{3+}[/tex]) on the product side of the reaction.
