Question in the description

Question

Experimentally, you can monitor conductivity of a solution to prove the presence of ions in solution.

Consider the reaction of sulfuric acid, H2SO4, a strong acid, with barium hydroxide, Ba(OH)2, a strong base. The net ionic reaction being observed in this experiment is:

Ba2+(aq) + 2 OH– (aq) + 2 H+ (aq) + SO4 2–(aq)  BaSO4(s) + 2 H2O(l)

How would you expect the measured conductivity to change as a solution of sulfuric acid is slowly added to a solution of barium hydroxide

How would the conductivity change as the moles of sulfuric acid added exceeds the initial moles of barium hydroxide present?

Answer 1

Ba²⁺(aq) + 2 OH^– (aq) + 2 H⁺ (aq) + SO₄ ^2–(aq) ==> BaSO₄(s) + 2 H₂O(l)

Conductivity depends on the number of ions in solution. In this case, that would be Ba²⁺, OH^- H⁺, and SO₄^2- ions.

At the start of the reaction, you have some Ba²⁺ and some OH^- ions in solution. As you add H₂SO₄, you add some H⁺ and some SO₄^2- ions. This will increase conductivity, but as the BaSO₄ forms, both Ba²⁺ and SO₄^2-ions are removed from solution as a precipitate. Also H₂O is formed and this is a poor conductor of current. So, the conductivity will decrease.

After moles of H₂SO₄ exceeds moles of barium hydroxide, then you have only H₂SO₄ present and since this is a strong acid (for the first ionization), the conductivity will then increase because you have H⁺ and HSO₄^- ions present in solution. This is an electrolyte and will conduct a current.