[Capital]

Please click on the link for the question:

http://i1008.photobucket.com/albums/af204/John132456/Chem.jpg?t=1306647211

The answer is B.

I thought the answer was A because I thought when H2S and NaOH react, it is a double replacement reaction so the equation is

H2S (aq)  + 2NaOH(aq) --> 2H2O (l) and Na2S(aq)

Na2S is aqueous because it is soluble according to my solubility table so the net ionic reaction will simply be H+ + OH-  --> H2O  (everything else is spectator)

Why is this not the case? (The answer is B)

And how do you type subscripts and charges?

[Borek]

What do you know about H₂S strength? Is it fully dissociated in water solutions?

For subscripts and superscripts use [ sub][ /sub] and [ sup][ /sup] tags (no spaces after [). You can enter them directly or use formatting buttons in the advanced editor.

[Vidya]

H2S is not completely dissociated
hence there is an equilibrium  and only Na+ ions are spectators ions

[Capital]

I know H₂S is a weak acid but on my acid-base table, it ionizes to form HS^-. My acid-base table does not say that HS^- is amphiprotic since HS^- is not on the acid side of the table. So according to my acid-base table, H₂S is monoprotic, not diprotic.

Wouldn't the net ionic equation be H₂S + OH^- --> HS^- + H₂O ?

[Borek]

This is tricky and full of handwaving. HS^- is a very weak acid, with K_a2 (second dissociation constant) given different values in different sources. Value that I usually use is 12 - that means in solutions with pH > 12 more than half of the HS^- will be neutralized to S^2-. In 1M solution of strong base S^2- concentration will be about 100 times larger than the concentration of HS^-.