Question 5.2: Using Solubility Guidelines to Predict Precipitation Reactio......

Analyze

All the compounds shown in (a), (b), and (c) are soluble and they provide ions in solution. By using the solubility guidelines in Table 5.1, determine whether the positive ions from one compound combine with the negative ions of the other to form soluble or insoluble compounds. If only soluble compounds are formed, then all ions remain in solution (no reaction). If an insoluble compound is formed, then the insoluble compound precipitates from the solution. The net ionic equation for the precipitation reaction is obtained by eliminating the spectator ions from the full ionic equation.

Solve

For each of (a), (b) and (c), apply the strategy described above.

(a) In aqueous solution, we get Na^{+}  and  OH^{−} from NaOH and Mg^{2+}  and  Cl^{−}  from  MgCl_{2}. The combination of Na^{+}  and  Cl^{−} gives NaCl, a soluble compound; thus, the Na^{+}  and  Cl^{−} ions remain in solution.

However, the Mg^{2+}  and  OH^{−} ions combine to produce Mg(OH)_{2}, an insoluble compound. The full ionic equation is

2 Na^{+}(aq) + 2  OH^{−}(aq) + Mg^{2+}(aq) + 2  Cl^{−}(aq) → Mg(OH)_{2}(s) + 2  Na^{+}(aq) + 2  Cl^{−}(aq)

With the elimination of spectator ions, we obtain

2 OH^{−}(aq) + Mg^{2+}(aq) → Mg(OH)_{2}(s)

(b) In aqueous solution, we get Ba^{2+}  and  S^{2−} from BaS and Cu^{2+}  and  SO_{4}^{2−}  from  CuSO_{4}. The Ba^{2+}  and  SO_{4}^{2−} ions combine to form BaSO_{4}, an insoluble compound, and the Cu^{2+}  and  S^{2−} ions combine to form CuS, also an insoluble compound. The full ionic equation is

Ba^{2+}(aq) + S^{2−}(aq) + Cu^{2+}(aq) + SO_{4}^{2−}(aq) → BaSO_{4}(s) + CuS(s)

The equation above is also the net ionic equation because there are no spectator ions.

(c) We get NH_{4}^{+}, SO_{4}^{2−}, Zn^{2+},  and  Cl^{−} ions in solution. Because all possible combinations of positive and negative ions lead to water soluble compounds, all of the ions remain in solution. No reaction occurs.

Assess

Problems of this type can also be solved by using a diagrammatic approach, which is illustrated for part (a).
As you gain experience, you should be able to go directly to a net ionic equation without first having to write an ionic equation that includes spectator ions.