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.