Question

The exposed electrodes of a light bulb are placed in a solution of${\mathrm{H}}_2{\mathrm{SO}}_4$ in an electrical circuitsuch 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.

Short answer

$\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2$and$\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ arethe salt in the solution.

Step-by-step solution

 Step 1: Definition

The capacity of an aqueous solution to conduct an electric current is measured by conductivity.

Explanation for incorrect

The solution may easilydissolve into${\mathrm{H}}^{+}$ and${\mathrm{HSO}}_4^{-}$ ions. so, the solution of${\mathrm{H}}_2{\mathrm{SO}}_4$conducts electricity. When${\mathrm{NaNO}}_3$ is introduced to a solution,it separates into${\mathrm{Na}}^{+}$ and${\mathrm{NO}}_3^{-}$ ions.

In the same way, the salt ${\mathrm{K}}_2{\mathrm{SO}}_4$, will dissociate into ${\mathrm{K}}^{+}$ and ${\mathrm{SO}}_4^{-}$other ions added to the solution. Because these salts add more ions to the solution, the solution's conductivity improves.

Explanation for the correct

When$\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2$ is introduced to a solution, it dissociates into ${\mathrm{Ba}}^{2+}$and${\mathrm{NO}}_3^{-}$ions, In solution, the ionlocalid="1650356829785" ${\mathrm{Ba}}^{2+}$reacts with${\mathrm{SO}}_4^{2-}$ to generate insoluble ${\mathrm{BaSO}}_4$. When$\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$is introduced to a solution, it dissociates into ${\mathrm{Ca}}^{2+}$and ${\mathrm{NO}}_3^{-}$ions, which are then released into the solution. In the solution, the ion${\mathrm{Ca}}^{2+}$will react with${\mathrm{HSO}}_4^{-}$to generate insoluble ${\mathrm{CaSO}}_4$.

The production of insoluble species will account for the total reduction in the amount of ions in the solution, even though these species will give twice as much as ions in the solution. As a result, conductivity decreases. As a result, the light dims. As a result, the potential salts in the solution are $\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2$and$\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ .