The amount of dissolved CO_2 in the blood is usually given as a volume of gaseou

Question

The amount of dissolved CO_2 in the blood is usually given as a volume of gaseous CO_2 dissolved in a volume of blood. The amount of dissolved CO_2 depends on the partial pressure of CO_2 in the lung (Henry's Law). When the partial pressure of CO_2 is 60 mmHg in the lung, the amount of dissolved CO_2 in the blood is 55ml of CO_2/dL blood. The volume of CO_2 corresponds to its gas form. Using the ideal gas law, calculate the concentration of dissolved CO_2 in the blood in mol/L Using equation (1) from Part I calculate the concentration of H_2CO_3 present in the blood. Do not approximate. Calculate the amount of HCO_3 presetn in the blood for a physiological pH of 7.3.

Explanation / Answer

From Ideal gas law calculate the concentration of dissolved CO2 in the blood is calculated as follows:

PV= nRT

Here P = 60mmHg or 0.07895 atm T= 37 C or 310 K and V= 55 ml or 0.055 L; R= gas constant = 0.0821 L atm / K mol

n = PV/RT

n = 0.07895 atm *0.055 L /0.0821 L atm / K mol*310 K

n= 1.71*10^-4 mol

this number of moles of CO2 is present in 1 dL of blood or 0.1 L

Thus the concentration of CO2 = 1.71*10^-4 mol /0.1 L

= 1.71*10^-3 M

CO2 + H2O --- > H2CO3

Here 1 mole of CO2 is form 1 mol of H2CO3 so concentration of H2CO3 is same as CO2.

CO2(aq) + H2O(l)          <==> H+ + HCO3^-
1.71*10^-3............... ............0 ........0 ............. initial
-x ...................... ...........         +x ..... +x ...........change

1.71*10^-3-x ............... ........... x ........ x ............. equilibrium

Here pH = 7.3

[H+]= 5.01*10^-8

Ka = [H+] [HCO3-] / [CO2]

Here the concentration of [H+] =[HCO3-] = 5.01*10^-8

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