# Diprotic acids and weak acids/bases

Hi,

I was reviewing some of my notes and was wondering what the correct steps for this would be (assuming in aqueous solution).
If you have ## H_2CO_3 ⇔ H^+ + HCO_3^- ## finding the final concentration of HCO3- is simple enough.

Now, for ## HCO_3^- ⇔ H^+ + CO_3^{2-} ## this would change the HCO3- concentration and the initial equilibrium from part 1 would have to be recalculated, right? I somewhat understand that this would be a very negligible since Ka values decreases as you go from the diprotic to monoprotic forms. But isn't it necessary to do a recalculation to find the more accurate answer? I find a problem with this....when would we stop recalculating? I understand the effects would be minimal, I'm just having trouble fully verifying this....any clarification would be great!

Also, if you have a weak base and acid, ex. HCO3- again. Although the Ka and Kb values do differ by a few orders of magnitude, how would you calculate the final concentration for the products and reactants of another molecule that has Ka ≈ Kb? Does such a molecule exist?

Sorry for the misguided questions. I am just trying to figure out when we can say a solution is equilibrium since we have all these sub-reactions occurring and it seems there is no point in time where there equilibrium since once you find the equilibrium concentration for one reaction, it changes due to another reaction that the products are a part of. Thanks in advance!

These relationships only apply to systems in equilibrium. You do not keep recalculating but you solve a system of multiple equations and multiple variables simultaneously, assuming that you can get enough equations written down to allow you to solve for all of the unknowns. This usually involves writing out all of the reactions and their corresponding Keq expressions as well as, typically, mass and charge balance statements.

To throw another wrench in your gears, you should be aware that carbonic acid (H2CO3) spontaneously breaks down to carbon dioxide and water in aqueous solution. This gives another equilibrium expression to consider for such a system.

Borek
Mentor
Go through these three - it is just an introduction, but it shows general approach to more and more complicated pH problems (third page describes diprotic acid, but don't go there without reading the earlier ones):

http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation

http://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

(third page assumes acid is dissolved and its total concentration known, as Yanick signaled, in the case of carbonic acid situation is a little bit more complicated).

The approach presented - deriving one equation in one variable (H+) - is not the only one possible. You can also solve system of equations numerically, without any initial preparations, or do some of the steps to convert initial system of equations to an easier to solve one. Which approach works best depends on the problem at hand.

1 person
Go through these three - it is just an introduction, but it shows general approach to more and more complicated pH problems (third page describes diprotic acid, but don't go there without reading the earlier ones):

http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation

http://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

(third page assumes acid is dissolved and its total concentration known, as Yanick signaled, in the case of carbonic acid situation is a little bit more complicated).

The approach presented - deriving one equation in one variable (H+) - is not the only one possible. You can also solve system of equations numerically, without any initial preparations, or do some of the steps to convert initial system of equations to an easier to solve one. Which approach works best depends on the problem at hand.

Wow, exactly what I was looking for. Thank you!