1. Chemistry
2. Physical Chemistry
3. Chemical Equilibrium
4. 69803

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Calculate the value of the equilibrium constant for each of the following reactions ( we abbreviate citric acid as H3Cit) using the Ka values for citric acid and for carbonic acid:

H3Cit + HCO3- <--> H2Cit- + H2O +CO2
H3Cit + 3HCO3- <--> Cit3- + 3H2O +3CO2
Given Ka1 of H3Cit =8.4x10-4; Ka2 of H3Cit = 1.8x10-5; Ka3 of H3Cit = 4.0x10-6; Ka1 of H2CO3 = 4.3x10-7; Ka2 of H2CO3 = 5.6x10-11

Hint: When reactions are added, the corresponding equilibrium constants are multiplied.

The Ka values are known for the H3Cit and its equilibrium. How/where is the HCO3- Ka value found/calculated to include it into the equation. Also, the same question with the H2O and the CO2. These are not acids so how is this a part of this set up.