You have exposed electrodes of a light bulb in a solution of H2SO4 such that the light bulb is on. You add a dilute solution and the bulb grows dim. Which of the following could be in the solution? a. Ba(OH)2

b. NaNO3

c. K2SO4

d. Cu(NO3)2

e. none of these

The answer is B but I don’t understand why please explain :))

Edit: found the answer in case someone needs it
A solution of H2SO4 is conductive because of the H+ and HSO4- ions.
The addition of sodium nitrate or potassium sulfate will increase the number of ions in solution which will increase the conductivity of the solution.
When Ca3+ or Ba2+ (in the form of calcium nitrate or barium nitrate) is added, it will react with HSO4- to make insoluble CaSO4 or BaSO4.
The catch is that the solution of calcium or barium nitrate will introduce twice as many nitrate ions as sulfate ions that are removed from solution. Plus, the removal of sulfate ions will leave H+ in solution.
Therefore, I don't believe that there will be a decrease of conductivity of the solution as calcium or barium sulfate is formed.
The addition of barium oxide or barium hydroxide to a solution of sulfuric acid would cause an overall decrease of number of ions in solution and cause a dimming of the light bulb. This is because it will not add additional ions to the solution.
BaO + H+ + HSO4- --> BaSO4)s) + H2O