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. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) b. \(\mathrm{NaNO}_{3}\) c. \(\mathrm{K}_{2} \mathrm{SO}_{4}\) d. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) Justify your choices. For those you did not choose, explain why they are incorrect.

All salts in the options are soluble in water, so they will dissociate into their respective cations and anions when added to the solution. In the process, they can either increase or decrease the conductivity of the solution, based on the ions they release. 2. Determine the effect of each salt on the conductivity

For the salt to decrease the conductivity of the solution, it should either: a) react with the \(\mathrm{H}^{+}\) or \(\mathrm{SO}_4^{2-}\) ions and form an insoluble compound, effectively reducing their concentration, or b) reduce the number of ions available to carry electric charge. 3. Analyze each salt and eliminate incorrect options

a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\): This salt dissociates into \(\mathrm{Ba}^{2+}\) and \(\mathrm{NO}_3^-\) ions. \(\mathrm{Ba}^{2+}\) will react with \(\mathrm{SO}_4^{2-}\) ions to form \(\mathrm{BaSO}_{4}\), which is an insoluble compound. Thus, it reduces the number of \(\mathrm{SO}_4^{2-}\) ions carrying the charge, decreasing the conductivity of the solution. b. \(\mathrm{NaNO}_{3}\): This salt dissociates into \(\mathrm{Na}^{+}\) and \(\mathrm{NO}_3^{-}\) ions. These ions will not react with the \(\mathrm{H}^{+}\) or \(\mathrm{SO}_{4}^{2-}\) ions and won't form any insoluble compound, nor reduce the number of ions available to carry the electric charge. So, it will increase the conductivity of the solution. c. \(\mathrm{K}_{2} \mathrm{SO}_{4}\): This salt dissociates into \(\mathrm{K}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) ions. These ions will not react with the \(\mathrm{H}^{+}\) or \(\mathrm{SO}_{4}^{2-}\) ions and won't form any insoluble compound, nor reduce the number of ions available to carry the electric charge. So, it will increase the conductivity of the solution. d. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\): This salt dissociates into \(\mathrm{Ca}^{2+}\) and \(\mathrm{NO}_3^-\) ions. These ions will not react with the \(\mathrm{H}^{+}\) or \(\mathrm{SO}_{4}^{2-}\) ions and won't form any insoluble compound, nor reduce the number of ions available to carry the electric charge. So, it will increase the conductivity of the solution. 4. Choose the correct answer and justify

From the above analysis, only \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) would cause the light bulb to dim. Therefore, the answer is option a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\). It is the correct choice because it reacts with the \(\mathrm{SO}_{4}^{2-}\) ions to form an insoluble compound, effectively reducing their concentration and the conductivity of the solution. The other options will not cause the light bulb to dim because they will not form insoluble compounds nor reduce the number of ions carrying electric charge.