Calculate Equilibrium Constant for Polyprotic Acid Reaction

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SUMMARY

The equilibrium constant for the reaction of bicarbonate ions (HCO3-) forming carbonic acid (H2CO3) and carbonate ions (CO32-) in a sodium bicarbonate (NaHCO3) solution is calculated using the relationship between the dissociation constants of the involved species. The correct equilibrium constant is determined to be 1.1 x 10^(-4), derived from the product of the base dissociation constant (Kb) and the second acid dissociation constant (Ka2). The initial miscalculation arose from incorrectly multiplying the equilibrium constants without accounting for the necessary division by the concentrations of hydrogen ions ([H+]) and hydroxide ions ([OH-]).

PREREQUISITES
  • Understanding of acid-base equilibria and polyprotic acids
  • Familiarity with equilibrium constant expressions (Ka and Kb)
  • Knowledge of the dissociation reactions of bicarbonate ions
  • Basic algebra for manipulating equilibrium expressions
NEXT STEPS
  • Study the derivation of equilibrium constants for polyprotic acids
  • Learn how to calculate Kb and Ka values for various acids
  • Explore the concept of ionic product of water (Kw) and its implications
  • Practice calculating equilibrium constants for other acid-base reactions
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Chemistry students, educators, and professionals involved in acid-base chemistry, particularly those focusing on polyprotic acids and their equilibria.

i_love_science
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The principal equilibrium in a solution of NaHCO3 is
HCO3-(aq) + HCO3-(aq) <-> H2CO3(aq) + CO32-(aq)
Calculate the value of the equilibrium constant for this reaction.

My solution:

This overall reaction is the same as the sum of the following reactions:
HCO3-(aq) <-> H2CO3(aq) + OH-(aq)
HCO3-(aq) <-> CO32-(aq) + H+(aq)

The overall equilibrium constant is (Kb)(Ka2) = [10^(-14) / 4.3*10^(-7)]*[4.8*10^(-11)] = 1.1^10^(-18).

The correct answer is 1.1*10^(-4).

Can anyone explain where I went wrong? Thanks.
 
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i_love_science said:
The overall equilibrium constant is (Kb)(Ka2)

Is it?
 
If you multiply the expressions for the equilibrium constants of the two reactions, you will find you have [H+] and [OH-] in the numerator. To get the equilibrium constant for the original reaction, you need to divide by [H+][OH-].
 

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