Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the diprotic acid H2SeO4.

Video transcript

welcome back everyone. We need to show the stepwise association of a di protic acid carbonic acid and the corresponding equilibrium equations. So let's begin by showing carbonic acid and how it dissolves in water. So we would have plus H20. Liquid and this is going to be an equilibrium. Now recognize that carbonic acid would be our Bronston Lowry acid and so it's going to be a proton donor. So we would observe our carbonic acid donating a proton to water which would be our bronze, said Lowry base, accepting this proton. And as a result we would form the following product where we have hydro knee um produced as well as our an ion bicarbonate. So that's H C. 03 minus. Now we want to make note of our first equilibrium for this dissociation. So we would say that K A one is equal to our concentration of our products over reactant and this would only be our species. So we would exclude water from this equilibrium equation. So we would have the concentration of our products being our concentration of hydro ni um multiplied by our concentration of bicarbonate, H C. 03, one minus over our concentration of our only Aquarius reactant being our carbonic acid. So our concentration of H two C 03. So now let's move on to our next part of our dissociation which would begin with bicarbonate. So h we have a church C. 03 minus further dissolving in water in a second equilibrium where our bicarbonate acts again as a bronze stead. Lowry acid donating a proton to water being our bronze. Did Lowry base, accepting this proton where we have the formation of again hydro ni um and I'll use the color red here again. So H 30 plus plus our new an ion now being depopulated and a result of C. 032 minus which we should recognize as carbonate and so now we would form a second equilibrium constant, which is going to be again the concentration of our product. So this case the concentration of hydro ni um multiplied by the concentration of carbonate, C. 032 minus, divided by our concentration of our only Aquarius reactant being our bicarbonate. So H C 031 minus. So for our final answers, we've successfully showcased the stepwise association of carbonic acid given by the following two highlighted equations as well as we've shown our equilibrium equations as the following two highlighted equations of K one and K two. So all of the work we've outlined will correspond to choice. Be in the multiple choice as the final answer. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video