Question

Write a net ionic equation for the reaction with \(\mathrm{OH}^{-}\) by which (a) \(\mathrm{Sb}^{3+}\) forms a precipitate. (b) antimony(III) hydroxide dissolves when more OH \(^{-}\) is added. (c) \(\mathrm{Sb}^{3+}\) forms a complex ion.

Short answer

Question: Write the net ionic equations for the following reactions involving antimony(III) cations and hydroxide anions: (a) Formation of a precipitate of antimony(III) hydroxide. (b) Dissolution of antimony(III) hydroxide precipitate when more hydroxide ions are added. (c) Formation of a complex antimony(III) ion with hydroxide ions. Answer: (a) \(\mathrm{Sb^{3+}(aq) + 3OH^{-}(aq) -> Sb(OH)3(s)}\) (b) \(\mathrm{Sb(OH)3(s) + OH^{-}(aq) -> Sb(OH)4^{-}(aq)}\) (c) \(\mathrm{Sb^{3+}(aq) + 4OH^{-}(aq) -> Sb(OH)4^{-}(aq)}\)

Step-by-step solution

Write a balanced molecular equation

We will first write the balanced molecular equation for the reaction: \(\mathrm{Sb^{3+}(aq) + 3OH^{-}(aq) -> Sb(OH)3(s)}\)

Write the net ionic equation

Now, we'll eliminate the spectator ions and write the net ionic equation for the precipitation reaction: \(\mathrm{Sb^{3+}(aq) + 3OH^{-}(aq) -> Sb(OH)3(s)}\) In this case, there are no spectator ions, so the net ionic equation is the same as the molecular equation. (b) Dissolution of antimony(III) hydroxide precipitate when more hydroxide ions are added.

Write a balanced molecular equation

Write the balanced molecular equation for the dissolution of the antimony(III) hydroxide precipitate: \(\mathrm{Sb(OH)3(s) + OH^{-}(aq) -> Sb(OH)4^{-}(aq)}\)

Write the net ionic equation

Now, we'll eliminate the spectator ions and write the net ionic equation for the dissolution process: \(\mathrm{Sb(OH)3(s) + OH^{-}(aq) -> Sb(OH)4^{-}(aq)}\) In this case, there are no spectator ions, so the net ionic equation is the same as the molecular equation. (c) Formation of a complex antimony(III) ion with hydroxide ions.

Write a balanced molecular equation

Write the balanced molecular equation for the formation of a complex antimony(III) ion with hydroxide ions: \(\mathrm{Sb^{3+}(aq) + 4OH^{-}(aq) -> Sb(OH)4^{-}(aq)}\)

Write the net ionic equation

Now, we'll eliminate the spectator ions and write the net ionic equation for the complex ion formation: \(\mathrm{Sb^{3+}(aq) + 4OH^{-}(aq) -> Sb(OH)4^{-}(aq)}\) In this case, there are no spectator ions, so the net ionic equation is the same as the molecular equation.