# Question 947ff

Dec 8, 2016

Here's why that is the case.

#### Explanation:

The idea here is that an aqueous solution, i.e. a solution that has water as the solvent, can only conduct electricity if it contains ions.

Keep in mind that pure water does not conduct electricity very well because it contains very, very low concentrations of hydronium cations, ${\text{H"_3"O}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$

$2 {\text{H"_ 2"O" _ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-} \to$ the auto-ionization of water

Now, sulfuric acid is a strong acid, which means that it dissociates completely in aqueous solution to produce cations and anions.

You can assume that both dissociations are strong, meaning that a solution of sulfuric acid will contain

${\text{H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O" _ ((l)) -> 2"H"_ 3"O"_ ((aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Since you now have a significant amount of hydronium cations and of sulfate anions, ${\text{SO}}_{4}^{2 -}$, floating around, this solution will conduct electricity, which is why the light bulb lights up.

Barium hydroxide, "Ba"("OH")_2#, is a strong base, which means that it dissociates completely to form barium cations and hydroxide anions -- keep in mind that barium hydroxide is not very soluble, but the amount that dissolves does so completely.

${\text{Ba"("OH") _ (2(s)) ->"Ba"_ ((aq))^(2+) + 2"OH}}_{\left(a q\right)}^{-}$

When you add barium hydroxide to the sulfuric acid solution, the hydronium cations and the hydroxide anions will neutralize each other to produce water

${\text{H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O}}_{\left(l\right)}$

This will reduce the number of ions present in solution, which implies that the solution's capacity to conduct electricity will also decrease.

Moreover, barium sulfate, ${\text{BaSO}}_{4}$, is considered insoluble in aqueous solution, which means that some of the barium cations and of the sulfate anions will combine to form a precipitate -- this actually depends on the concentrations of sulfuric acid and barium hydroxide

${\text{Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO}}_{4 \left(s\right)} \downarrow$

This will reduce the number of ions floating around in solution even more, further decreasing its ability to conduct electricity.

So there you have it. The neutralization reaction that occurs between barium hydroxide and sulfuric acid and the possible precipitation reaction that produces barium sulfate will reduce the number of ions present in solution.

This will in turn decrease the solution's ability to conduct electricity, which is why the light dims.