The `pH` of blood is `7,4`. If the buffer in blood constitute `CO_(2)` and `HCO_(3)^(Theta)` ions, calculate the ratio of conjugate base of acid `(H_(2)CO_(3))` to maintain the `pH` of blood. Given `K_(1)` of `H_(2)CO_(3) = 4.5 xx 10^(-7)`.

Explanation

`CO_(2)` with `H_(2)O` forms `H_(2)CO_(3)`
`CO_(2)+H_(2)O Leftrightarrow H^+ +HCO_(3)^(-)`
`K_(1)=([H^+] [HCO_(3)^-])/([CO_2])=4.5 xx 10^(-7)`
Now, `pH =-log [H^+]=7.4, [H^+]=4.0 xx 10^(-8)`
Thus, `([HCO_(3)^-])/([CO_2])=(4.5 xx 10^(-7))/(4 xx 10^(-8))=11`