Is NaHCO3 a versatile compound in aqueous solutions?

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SUMMARY

Aqueous solutions of sodium bicarbonate (NaHCO3) can function as both an acid and a base, demonstrating amphiprotic behavior. The acid dissociation reaction is represented as NaHCO3(aq) + H2O(l) → H3O+(aq) + NaCO3-(aq), while the base reaction is NaHCO3(aq) + H2O(l) → OH-(aq) + NaH2CO3+(aq). The equilibrium constants are given as Ka1 = 4.2 x 10^-7 and Ka2 = 4.8 x 10^-11, indicating that Ka1 predominates, suggesting that NaHCO3 primarily acts as an acid in solution. The analysis of Kb values further supports this conclusion, with Kb1 calculated as 2.38 x 10^-8 and Kb2 as 2.08 x 10^-4, confirming that the release of protons is favored over the uptake of protons.

PREREQUISITES
  • Understanding of acid-base equilibria
  • Familiarity with dissociation constants (Ka and Kb)
  • Knowledge of aqueous solution chemistry
  • Ability to perform equilibrium calculations
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  • Study the hydrolysis of Na2CO3 and its relationship to H2CO3
  • Learn about weak acid-base equilibria and their calculations
  • Explore the concept of amphiprotic substances in detail
  • Investigate the implications of pH in amphiprotic salts
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Chemistry students, educators, and professionals interested in acid-base chemistry, particularly those studying weak acids and their behavior in aqueous solutions.

synergix
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Homework Statement



Write two equations that illustrate that an aqeous solution of NaHCO3 can act either as an acid or a base. In pure water show quantitatively which of the two reactions predominate. ka1= 4.2x10-7, Ka2=4.8x10-11


The Attempt at a Solution


as an acid:

NaHCO3(aq) + H2O(l)---> H3O+(aq) + NaCO3-(aq)

Base:

NaHCO3(aq) + H2O(l) ------> OH-(aq) + NaH2CO3+(aq)

The second part is where I am having some trouble. I am not sure if I am doing this correctly so i will try to show and explain what I think will happen as best I can. I think Ka1 will predominate the reason being that kb1= (1x10^-14)/(4.2x10^-7)=2.38x10^-8
so Ka1>Kb1 therefore this reaction will proceed. Meaning most of the sodium bicarbonate will act as an acid and release protons. whereas Kb2=(1x10^-14)/(4.8x10^11)=2.08x10^-4.
so Kb2>>Ka2 so releasing a 2nd proton is not very favorable the species is much more likely to pick up another proton then it is to release another proton.
 
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I guess if I am correct then the fact that Ka1 is quantitatively greater then kb1 is the proof that ka1 will predominate.
 
Your written reactions are good. For which dissocation of bicarbonate is more favorable, check the Na2CO3 hydrolysis steps on the way towards H2CO3, or check the dissocation steps of H2CO3 toward Na2CO3. You know how to find Ka from corresponding Kb, right?
 
How does NaHCO3 become Na2CO3? I am not sure exactly what you mean by checking the Na2CO3 hydrolysis steps on the way towards H2CO3.
 
Synergix, maybe I was sloppy with my terminology or notation. You were on a good track with your idea of comparing the acidic dissociation constant for bicarbonate to the equilibrium constant for HCO3- + H20 <===> H2CO3 + OH-
Are you studying weak acid-base equilibria right now? If so, you will soon know how to evaluate those constants.
 
Ya we have covered that already as part of the last section. I will look at it again maybe I can figure it out thanks.
 
pH of amphiprotic salt

I am not sure you can say which of the reactions dominate. If you will read through the lecture you will find that ratio [H2CO3]/[CO32-] is close to 1, that means both reactions proceed to the same degree.
 

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