[pirateboy]

Hello, all. I have a question regarding the dissociation of an element

We recently did a lab experiment where we measured the varying conductivity of water in which we initially added a dilute solution of Ba(OH)₂ and then added increasing amounts of H₂SO₄ and watched as the conductivity dropped, hit the equivalence point and then increased.

The relationship of the volume of H₂SO₄ to conductivity was linear, which I guess makes sense, since presumably the electrons are getting paired ("used up") at the same ratio every time, and conductivity is just where is resitivity. (?)

But my real question is this. I know this reaction is taking place because the Barium ions dissociate once it is the barium hydroxide is dissolved in water. And I know that the barium then bonds to the sulfate in the sulfiric acid. I'm just curious as to why I should know that the Barium would dissociate if someone hadn't told me.

I'd venture this has something to do with the the fact that barium's 6th s-shell is full (has two electrons) thus the Barium is inclined to give away electrons. But why does it ditch the hydroxide as soon as it hits the water?  Are some of the H molecules and the O moleulces forming up to make water together and leaving a few oxygen's floating around as unbonded ions?

Edit: added more descriptive title

[Borek]

Ba(OH)₂ is a strong base, and as all strong bases it is fully dissociated in water.

[DevaDevil]

this is a question about solubilities more than anything.
look up the solubilities of barium hydroxide and barium sulfate.

the OH^- ions in barium hydroxide do not dissociate in any way; when sulfuric acid comes in, a proton from the acid will neutralize this base and form water

the barium remains 2+ throughout, first as soluted ion, then as non-soluble BaSO₄, so do not worry about valence shells.

so initially the conductance goes down because :
neutralization of OH^- ions and crystallizing of barium sulfate (reducing charge carriers in solution)