Question

Three solutions are mixed together to form a single solution. One contains \(0.2 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\), the second contains \(0.1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S}\), and the third contains \(0.1 \mathrm{~mol}\) \(\mathrm{CaCl}_{2}\). (a) Write the net ionic equations for the precipitation reaction or reactions that occur. (b) What are the spectator ions in the solution?

Short answer

Three precipitation reactions occur when the solutions are mixed: 1. \(Pb^{2+}(aq) + S^{2-}(aq) \rightarrow PbS(s)\) 2. \(Pb^{2+}(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s)\) 3. \(Ca^{2+}(aq) + S^{2-}(aq) \rightarrow CaS(s)\) The spectator ions in the solution are \(Na^+\) and \(CH_3COO^-\).

Step-by-step solution

1. \(Pb^{2+}\) and \(S^{2-}\)

These ions could form \(PbS\), which is insoluble according to rule 4. The net ionic equation is: \[Pb^{2+}(aq) + S^{2-}(aq) \rightarrow PbS(s)\]

2. \(Pb^{2+}\) and \(Cl^-\)

These ions could form \(PbCl_2\), which is insoluble according to rule 3. The net ionic equation is: \[Pb^{2+}(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s)\]

3. \(Ca^{2+}\) and \(S^{2-}\)

These ions could form \(CaS\), which is insoluble according to rule 4. The net ionic equation is: \[Ca^{2+}(aq) + S^{2-}(aq) \rightarrow CaS(s)\] There are no more possible precipitation reactions between the given ions. #b) Spectator ions in the solution#

Spectator ions

The spectator ions are the ions that do not participate in any precipitation reactions. In this case, we can see from the net ionic equations that the two spectator ions are: 1. Sodium ion (\(Na^{+}\)) 2. Acetate ion (\(CH_3COO^{-}\))