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Blood is buffered with CO(2) " and " HCO...

Blood is buffered with `CO_(2) " and " HCO_(3)^(-)`. What is the ratio of the base concentration to the acid ( i.e., `CO_(2) (aq) " plus " H_(2)CO_(3))` concentration to maintain the pH of blood at 7.4 ? The first dissociation constant of `H_(2)CO_(3) (H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)) " is " 4.2xx10^(-7)` where the `H_(2)CO_(3)` is assumed to include `CO_(2)(aq)` i.e., dissolved `CO_(2)`. [ log 4.2=0.6232]

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The pH of the a solution containing 0.4 M HCO_(3)^(-) is : [K_(a_(1)) (H_(2)CO_(3)) = 4 xx 10^(-7), K_(a_(2)) (HCO_(3)^(-)) = 4 xx 10^(-11)]

The following reaction is known to occur in the body, If CO_(2) escapes from the system CO_(2) + H_(2)O hArr H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)

HCO_(3)^(-) is conjugate base of H_(2)CO_(3) .

The equilibrium constant of given reaciton will be HCO_(3)^(-)+H_(2)OhArrH_(2)CO_(3)+OH^(-)

The equilibrium constant of given reaciton will be HCO_(3)^(-)+H_(2)OhArrH_(2)CO_(3)+OH^(-)

BaCO_(3)darr+CO_(2)+H_(2)O to Ba(HCO_(3))_(2)