Bicarbonate-CO2 Equilibrium

Q.
This question has been submitted in the following subject: science A body of water already at equilbrium contains 10^-5 mol/litter of carbonic acid, H2CO3. a)What is the concentration of bicarbonate, HCO3^-, at pH values of 4, 7, and 10. b)Use the same pH conditions in part a) but in seawater having an ionic strength of 0.5M. (Because pH is properly defined on the basis of hydrogen activity, ionic strength corrections need only be made for HCO3^-.) Not sure what to do. There should be an answer for each pH value.
ANSWER:
(a)
Ki = [H+][HCO3-] / [H2CO3] = 4.3x10^-7 (from a table of Ka values)
For a pH of 4, [H+] = 1x10^-4
[1x10^-4][HCO3-] / [(1x10^-5)] = 4.3x10^-7
[HCO3-] = [H2CO3](Ka) / [H+]
[HCO3-] = (1x10^-5)(4.3x10^-7) / (1x10^-4) = 4.3x10^-8
You find [HCO3-] for other pH’s in a similar way. As the pH rises, the concentration of HCO3- should also rise.
(b)
Reference:
http://www.chembuddy.com/?left=pH-calculation&right=ionic-strength-activity-coefficients
The hydrogen activity coefficient can be calculated by:
Log(f) = [(-.51z^2)*sqrt(I)] / [1 + sqrt(I)] z = ionic charge, I = ionic strength
Log(f) = -0.21125 and f = 0.615
The hydrogen activity is
a = (f)[H+]
For pH = 4,
a = {H+} = (0.615)(1x10^-4) = 6.15x10^-5 <--- H+ activity replacing [H+]
[HCO3-] = [H2CO3](Ka) / [H+]
[HCO3-] = (1x10^-5)(4.3x10^-7) / (6.15x10^-5) = 7.0x10^-8
For other pH’s, do similar calculations with the same activity coefficient.

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