The exposed electrodes of a light bulb are placed in a solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in an electrical circuit such that the light bulb is glowing. You add a dilute salt solution, and the bulb dims. Which of the following could be the salt in the solution? a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) c. $\mathrm{K}_{2} \mathrm{SO}_{4}$ b. \(\mathrm{NaNO}_{3}\) d. \(\mathrm{Ca}(\mathrm{NO}_{3})_{2}\) Justify your choices. For those you did not choose, explain why they are incorrect

The exposed electrodes of a light bulb are placed in a solution of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which is a strong electrolyte and dissociates completely into ions (\(2\,\mathrm{H}^{+}\) and \(1\,\mathrm{SO}_{4}^{-2}\)) in water. The bulb is glowing due to the flow of current through the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution, which is acting as a conductor.

When a dilute salt solution is added, a change in conductivity occurs, which in turn causes a change in the brightness of the light bulb. The significant effect is caused by the presence of ions from the salt solution, which may interact with the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) ions already present in the solution. If a salt solution forms a precipitate with the sulfate ion (\(\mathrm{SO}_{4}^{-2}\)), then the concentration of ions available to conduct electricity would decrease, causing the bulb to dim.

Now let's examine each salt option and determine if it forms a precipitate with sulfate ions or not: a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) - Barium nitrate is soluble in water and dissociates into \(\mathrm{Ba}^{2+}\) and \(2\,\mathrm{NO}_{3}^{-}\) ions. However, \(\mathrm{Ba}^{2+}\) ions have a high affinity for sulfate ions and readily form a precipitate of barium sulfate (\(\mathrm{BaSO}_4\)). This would lead to a decrease in the ions concentration and hence, the bulb would dim. b. \(\mathrm{NaNO}_3\) - Sodium nitrate is soluble in water and dissociates into \(\mathrm{Na}^{+}\) and \(\mathrm{NO}_{3}^{-}\) ions. These ions don't form any precipitate with sulfate ions. Thus, the bulb's brightness will not be affected. c. \(\mathrm{K}_{2}\mathrm{SO}_{4}\) - Potassium sulfate is soluble in water and dissociates into \(2\,\mathrm{K}^{+}\) and \(\mathrm{SO}_{4}^{-2}\) ions. These ions don't form any precipitate with sulfate ions. Thus, the bulb's brightness will not be affected. d. \(\mathrm{Ca}(\mathrm{NO}_{3})_{2}\) - Calcium nitrate is soluble in water and dissociates into \(\mathrm{Ca}^{2+}\) and \(2\,\mathrm{NO}_{3}^{-}\) ions. \(\mathrm{Ca}^{2+}\) ions interact with sulfate ions to form a precipitate of calcium sulfate (\(\mathrm{CaSO}_4\)). This would lead to a decrease in the ions concentration and hence, the bulb would dim.

The correct salt solutions that could dim the light bulb are \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) (option a) and \(\mathrm{Ca}(\mathrm{NO}_{3})_{2}\) (option d) because they form precipitates with sulfate ions, reducing the concentration of ions available to conduct electricity. The other options, \(\mathrm{NaNO}_3\) (option b) and \(\mathrm{K}_{2}\mathrm{SO}_{4}\) (option c), do not form any precipitates, and thus, they are incorrect choices for this question.