Question

How many free ions are there on the products side of the total ionic equations that results from the reaction between bismuth 3 and hypochlorite and acetic acid?

Answer 1

Given that the reaction is between Bismuth(III) hypochlorite and acetic acid.

Molecular equation for the reaction can be represented as:

$Bi(OCl)_{3}(aq)+3CH_{3}COOH(aq)-->(CH_{3}COO)_{3} Bi(aq)+3HOCl(aq)$

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.

$Bi^{3+}(aq)+3OCl^{-}(aq)+3CH_{3}COOH(aq)-->3CH_{3}COO^{-}(aq)+Bi^{3+}(aq)+HOCl(aq)$

Therefore, there are 4 free ions ($3CH_{3}COO^{-} and one Bi^{3+}$) on the product side of the reaction.