Question

the exposed electrodes of a light bulb are placed in a solution of h₂so₄ 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?
o nacl
o caso₄
o bacl₂
o cacl₂
o none of the above

Explanation:

Step1: Analyze the reaction of each salt with \(H_2SO_4\)

• For \(NaCl\): \(NaCl + H_2SO_4\) (dilute) does not react significantly to form a precipitate. The solution still has free - moving ions (\(Na^+\), \(Cl^-\), \(H^+\), \(SO_4^{2 -}\)) and the bulb will remain glowing.
• For \(CaCl_2\): \(CaCl_2+H_2SO_4 = CaSO_4\downarrow+2HCl\). \(CaSO_4\) is slightly soluble. There are still enough ions (\(H^+\), \(Cl^-\)) in the solution for the bulb to glow (though the intensity may change a bit, but not go dim).
• For \(BaCl_2\): \(BaCl_2 + H_2SO_4=BaSO_4\downarrow + 2HCl\). \(BaSO_4\) is highly insoluble. But \(HCl\) is a strong acid and dissociates completely (\(HCl = H^++Cl^-\)). So there are still free - moving ions (\(H^+\), \(Cl^-\)) in the solution and the bulb will remain glowing.
• For \(Cu_2SO_4\): Assuming it is \(CuSO_4\) (copper (II) sulfate), \(CuSO_4\) and \(H_2SO_4\) do not react. The solution has \(Cu^{2 +}\), \(SO_4^{2 -}\), \(H^+\) ions and the bulb will remain glowing.

Since none of the salts will cause the bulb to go dim (by removing almost all free - moving ions), the answer is based on the fact that no reaction (that would remove most ions) occurs in a way that the bulb goes dim for the given salts.

Answer:

none of the above