Question

Write a net ionic equation for the reaction with ammonia by which (a) \(\mathrm{Cu}(\mathrm{OH})_{2}\) dissolves. (b) \(\mathrm{Cd}^{2+}\) forms a complex ion. (c) \(\mathrm{Pb}^{2+}\) forms a precipitate.

Short answer

Question: Write the net ionic equations for the reaction of ammonia with (a) Copper(II) hydroxide, (b) Cadmium ion, and (c) Lead ion. Answer: (a) \(\mathrm{Cu}(\mathrm{OH})_{2} + 4\mathrm{NH}_{3} \Rightarrow [\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+} + 2\mathrm{OH}^{-}\); (b) \(\mathrm{Cd}^{2+} + 4\mathrm{NH}_{3} \Rightarrow [\mathrm{Cd}(\mathrm{NH}_{3})_{4}]^{2+}\); (c) \(\mathrm{Pb}^{2+} + 2\mathrm{NH}_{3} + 2\mathrm{H}_{2}\mathrm{O} \Rightarrow \mathrm{Pb}(\mathrm{OH})_{2}\)

Step-by-step solution

Identify the reactions with ammonia

For each reaction, we have to understand how ammonia reacts with the respective substance/ion, and then write down the balanced chemical equations. (a) Copper(II) hydroxide, Cu(OH)₂, dissolves in the presence of ammonia to form the tetraamine copper(II) complex ion: \(\mathrm{Cu}(\mathrm{OH})_{2} + 4\mathrm{NH}_{3} \rightarrow [\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+} + 2\mathrm{OH}^{-}\) (b) Cadmium ion and ammonia form the complex ion: \(\mathrm{Cd}^{2+} + 4\mathrm{NH}_{3} \rightarrow [\mathrm{Cd}(\mathrm{NH}_{3})_{4}]^{2+}\) (c) Lead ion reacts with ammonia to form lead(II) hydroxide precipitate: \(\mathrm{Pb}^{2+} + 2\mathrm{NH}_{3} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Pb}(\mathrm{OH})_{2} + 2\mathrm{NH}_{4}^{+}\)

Write the net ionic equations

Now we can write the net ionic equations by leaving out the spectator ions and listing only the reacting species: (a) \(\mathrm{Cu}(\mathrm{OH})_{2} + 4\mathrm{NH}_{3} \Rightarrow [\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+} + 2\mathrm{OH}^{-}\) In this case, there are no spectator ions, so the net ionic equation is the same as the balanced chemical equation. (b) For the second reaction: \(\mathrm{Cd}^{2+} + 4\mathrm{NH}_{3} \Rightarrow [\mathrm{Cd}(\mathrm{NH}_{3})_{4}]^{2+}\) Again, there are no spectator ions, so the net ionic equation is the same as the balanced chemical equation. (c) Finally, for the third reaction, there is one spectator ion, NH₄⁺: \(\mathrm{Pb}^{2+} + 2\mathrm{NH}_{3} + 2\mathrm{H}_{2}\mathrm{O} \Rightarrow \mathrm{Pb}(\mathrm{OH})_{2} + 2\mathrm{NH}_{4}^{+}\) The net ionic equation is: \(\mathrm{Pb}^{2+} + 2\mathrm{NH}_{3} + 2\mathrm{H}_{2}\mathrm{O} \Rightarrow \mathrm{Pb}(\mathrm{OH})_{2}\)