Question 7.FP.1: An unidentified solid is dissolved in water to produce a cle...
An unidentified solid is dissolved in water to produce a clear, colorless solution, and this solution conducts electricity. A second solution is added, and a precipitate forms. Once the precipitate has settled, the liquid above it does not conduct electricity perceptibly. What does this experiment tell you about the bonding in the initial solid and in the precipitate, and how does it tell you that?
Strategy
This question is another example of conceptual problem solving. We have seen three types of chemical bonding in this chapter, and from the observable behavior of this experiment, we need to infer which one is important in the unknown solid. The other piece of information we need to include involves what allows a solution to conduct electricity, as discussed in Chapter 3.
For any substance to conduct electricity, it must allow a charge to move. The charged particles in solution are usually ions, so this flow of charge arises from the movement of ions. This strongly suggests that the original solid was ionically bonded. How does this explain the observations? When the ionic solid is dissolved, its constituent ions are dispersed in solution, causing the solution to conduct electricity. The addition of the second solution forms a precipitate. This tells us a couple of things. First, because it was formed from two solutions, the precipitate must be an insoluble ionic compound. Moreover, the constituent ions in the precipitate must have come from both solutions. If the anion came from the first solution, the cation must have come from the second. We can also note that because the final solution no longer conducts electricity, it must not contain significant concentrations of any ions. This points toward a few possibilities. Each of the individual solutions must have contained both anions and cations to be electrically neutral. When the two solutions were mixed, these ions must have recombined to form two products: the cation from the first solution combining with the anion from the second, and vice versa. So we have two products to account for. One possibility is that two different precipitates may have formed, removing all of the ions. Alternatively, the second product could be water, formed from the neutralization reaction between an acid and a base.