Question

Write a net ionic equation for the reaction between aqueous solutions of (a) sodium acetate \(\left(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) and nitric acid. (b) hydrobromic acid and strontium hydroxide. (c) hypochlorous acid and sodium cyanide. (d) sodium hydroxide and nitrous acid.

Short answer

The net ionic equations for the given reactions are as follows: (a) sodium acetate and nitric acid: $$ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}_{(aq)} + \mathrm{H}^{+}_{(aq)} \rightarrow \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)}. $$ (b) hydrobromic acid and strontium hydroxide: $$ 2 \mathrm{H}^{+}_{(aq)} + 2 \mathrm{OH}^{-}_{(aq)} \rightarrow 2 \mathrm{H}_2 \mathrm{O_{(l)}}. $$ (c) hypochlorous acid and sodium cyanide: $$ \mathrm{H}^{+}_{(aq)} + \mathrm{CN}^{-}_{(aq)} \rightarrow \mathrm{HCN_{(aq)}}. $$ (d) sodium hydroxide and nitrous acid: $$ \mathrm{OH}^{-}_{(aq)} + \mathrm{H}^{+}_{(aq)} \rightarrow \mathrm{H}_2 \mathrm{O_{(l)}}. $$

Step-by-step solution

Write the balanced chemical equation.

The reaction between sodium acetate \((\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)})\) and nitric acid \((\mathrm{HNO}_{3 (aq)})\) forms sodium nitrate \((\mathrm{NaNO}_{3 (aq)})\) and acetic acid (In this case, it remains as \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)}\)). The balanced chemical equation is: $$ \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)} + \mathrm{HNO}_{3 (aq)} \rightarrow \mathrm{NaNO}_{3 (aq)} + \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)}. $$

Break up the aqueous species into their respective ions.

Splitting the aqueous species into ions, we get: $$ \mathrm{Na}^{+}_{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-_{(aq)} + \mathrm{H}^{+}_{(aq)} + \mathrm{NO}_{3}^{-}_{(aq)} \rightarrow \mathrm{Na}^{+}_{(aq)} + \mathrm{NO}_{3}^{-}_{(aq)} + \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)}. $$

Remove spectator ions.

Spectator ions are ions that do not change or react during the reaction. In this case, the sodium ions \((\mathrm{Na}^+_{(aq)})\) and the nitrate ions \((\mathrm{NO}_{3}^{-}_{(aq)})\) can be removed, as they do not change during the reaction.

Write the net ionic equation.

The net ionic equation for the reaction between sodium acetate and nitric acid is: $$ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-_{(aq)} + \mathrm{H}^{+}_{(aq)} \rightarrow \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2 (aq)}. $$ (b) hydrobromic acid and strontium hydroxide

Write the balanced chemical equation.

The reaction between hydrobromic acid \((\mathrm{HBr_{(aq)}})\) and strontium hydroxide \((\mathrm{Sr(OH)}_{2 (aq)})\) forms strontium bromide \((\mathrm{SrBr}_{2 (aq)})\) and water \((\mathrm{H}_2 \mathrm{O_{(l)}})\). The balanced chemical equation is: $$ 2 \mathrm{HBr_{(aq)}} + \mathrm{Sr(OH)}_{2 (aq)} \rightarrow \mathrm{SrBr}_{2 (aq)} + 2 \mathrm{H}_2 \mathrm{O_{(l)}}. $$

Break up the aqueous species into their respective ions.

Splitting the aqueous species into ions, we get: $$ 2 \mathrm{H}^{+}_{(aq)} + 2 \mathrm{Br}^{-}_{(aq)} + \mathrm{Sr}^{2+}_{(aq)} + 2 \mathrm{OH}^{-}_{(aq)} \rightarrow \mathrm{Sr}^{2+}_{(aq)} + 2 \mathrm{Br}^{-}_{(aq)} + 2 \mathrm{H}_2 \mathrm{O_{(l)}}. $$

Remove spectator ions.

In this case, the spectator ions are strontium ions \((\mathrm{Sr}^{2+}_{(aq)})\) and bromide ions \((\mathrm{Br}^{-}_{(aq)})\).

Write the net ionic equation.

The net ionic equation for the reaction between hydrobromic acid and strontium hydroxide is: $$ 2 \mathrm{H}^{+}_{(aq)} + 2 \mathrm{OH}^{-}_{(aq)} \rightarrow 2 \mathrm{H}_2 \mathrm{O_{(l)}}. $$ (c) hypochlorous acid and sodium cyanide

Write the balanced chemical equation.

The reaction between hypochlorous acid \((\mathrm{HClO_{(aq)}})\) and sodium cyanide \((\mathrm{NaCN_{(aq)}})\) forms hydrogen cyanide \((\mathrm{HCN_{(aq)}})\) and sodium hypochlorite \((\mathrm{NaClO_{(aq)}})\). The balanced chemical equation is: $$ \mathrm{HClO_{(aq)}} + \mathrm{NaCN_{(aq)}} \rightarrow \mathrm{HCN_{(aq)}} + \mathrm{NaClO_{(aq)}}. $$

Break up the aqueous species into their ...

Splitting the aqueous species into ions, we get: $$ \mathrm{H}^{+}_{(aq)} + \mathrm{ClO}^{-}_{(aq)} + \mathrm{Na}^{+}_{(aq)} + \mathrm{CN}^{-}_{(aq)} \rightarrow \mathrm{HCN_{(aq)}} + \mathrm{Na}^{+}_{(aq)} + \mathrm{ClO}^{-}_{(aq)}. $$

Remove spectator ions.

The spectator ions are sodium ions \((\mathrm{Na}^{+}_{(aq)})\) and hypochlorite ions \((\mathrm{ClO}^{-}_{(aq)})\).

Write the net ionic equation.

The net ionic equation for the reaction between hypochlorous acid and sodium cyanide is: $$ \mathrm{H}^{+}_{(aq)} + \mathrm{CN}^{-}_{(aq)} \rightarrow \mathrm{HCN_{(aq)}}. $$ (d) sodium hydroxide and nitrous acid

Write the balanced chemical equation.

The reaction between sodium hydroxide \((\mathrm{NaOH_{(aq)}})\) and nitrous acid \((\mathrm{HNO}_{2 (aq)})\) forms sodium nitrite \((\mathrm{NaNO}_{2 (aq)})\) and water \((\mathrm{H}_2 \mathrm{O_{(l)}})\). The balanced chemical equation is: $$ \mathrm{NaOH_{(aq)}} + \mathrm{HNO}_{2 (aq)} \rightarrow \mathrm{NaNO}_{2 (aq)} + \mathrm{H}_2 \mathrm{O_{(l)}}. $$

Break up the aqueous species into their respective ions.

Splitting the aqueous species into ions, we get: $$ \mathrm{Na}^{+}_{(aq)} + \mathrm{OH}^{-}_{(aq)} + \mathrm{H}^{+}_{(aq)} + \mathrm{NO}_{2}^{-}_{(aq)} \rightarrow \mathrm{Na}^{+}_{(aq)} + \mathrm{NO}_{2}^{-}_{(aq)} + \mathrm{H}_2 \mathrm{O_{(l)}}. $$

Remove spectator ions.

In this case, the spectator ions are sodium ions \((\mathrm{Na}^{+}_{(aq)})\) and nitrite ions \((\mathrm{NO}_{2}^{-}_{(aq)})\).

Write the net ionic equation.

The net ionic equation for the reaction between sodium hydroxide and nitrous acid is: $$ \mathrm{OH}^{-}_{(aq)} + \mathrm{H}^{+}_{(aq)} \rightarrow \mathrm{H}_2 \mathrm{O_{(l)}}. $$