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. \(\operatorname{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) Justify your choices. For those you did not choose, explain why they are incorrect.

First, we need to list the ions that each given salt dissociates into when added to the solution: a. Ba(NO3)2 -> Ba²⁺ + 2NO₃⁻ b. NaNO3 -> Na⁺ + NO₃⁻ c. K2SO4 -> 2K⁺ + SO₄²⁻ d. Ca(NO3)2 -> Ca²⁺ + 2NO₃⁻

Now, let's analyze how the ions from each salt will interact with the H2SO4 solution: a. The Ba²⁺ ion could form a precipitate with the sulfate ions already present in the H2SO4 solution: Ba²⁺ + SO₄²⁻ -> BaSO4(s). The formation of a precipitate would decrease the overall number of free ions in the solution, reducing its conductivity. b. The addition of Na⁺ and NO₃⁻ would simply increase the number of free ions in the solution, potentially increasing its conductivity. Therefore, NaNO3 would not cause the light bulb to dim. c. Similar to NaNO3, K2SO4 would add more K⁺ and SO₄²⁻ ions, possibly increasing the conductivity of the solution. It would not cause the light bulb to dim. d. The addition of Ca²⁺ and NO₃⁻ from Ca(NO3)2 would increase the number of free ions in the solution, potentially increasing its conductivity. It would not cause the light bulb to dim.

Based on the analysis, the correct salt that could cause the light bulb to dim is: a. Ba(NO3)2 The Ba²⁺ ions can form a precipitate when reacting with the sulfate ions in the H₂SO₄ solution, reducing the overall number of free ions and consequently reducing the conductivity of the solution.