Diprotic acids and weak acids/bases

[MathewsMD]

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 K_a 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 K_a and K_b 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 K_a ≈ K_b? 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!

 

[Yanick]

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]

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.

 

[MathewsMD]

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!

 

[LudusRex]

Hello,

I can understand your confusion and questions regarding diprotic acids and weak acids/bases. Let me try to clarify some of your concerns.

Firstly, for the equilibrium reaction $H_2CO_3 ⇔ H^+ + HCO_3^-$, you are correct in saying that finding the final concentration of HCO3- is simple enough. This is because the equilibrium constant (Ka) for this reaction is relatively large, indicating that the reaction favors the formation of products (H^+ and HCO3^-). Therefore, the concentration of HCO3- will be higher than the concentration of H2CO3 in the solution.

Now, for the reaction $HCO_3^- ⇔ H^+ + CO_3^{2-}$, you are also correct in saying that this will change the HCO3- concentration and the initial equilibrium from part 1 would have to be recalculated. This is because the equilibrium constant (Ka) for this reaction is smaller than the one for the previous reaction, indicating that the reaction favors the formation of reactants (HCO3^- and CO3^2-). However, as you mentioned, the effect on the HCO3- concentration will be minimal since the Ka values decrease as you go from the diprotic to monoprotic forms. This means that the reaction will not significantly shift the equilibrium and the final concentration of HCO3- will still be higher than the concentration of H2CO3.

To answer your question about when we would stop recalculating, the answer is that it depends on the accuracy and precision needed for your specific experiment or application. In most cases, the effects of the subsequent reactions will be negligible and the initial equilibrium concentrations can be used to determine the final concentrations. However, if a high level of accuracy is required, then the subsequent reactions may need to be taken into account.

Moving on to your question about weak bases and acids, such as HCO3-. In this case, the Ka and Kb values do differ by a few orders of magnitude, indicating that the reaction favors the formation of either H^+ or OH^- depending on the conditions. In such cases, the equilibrium concentrations of the products and reactants can be determined using the Henderson-Hasselbalch equation, which takes into account the Ka and Kb values. And yes, there are molecules that have Ka ≈ Kb, such as water