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)`.

Options

A. `11.25`
B. `10.0`
C. `8.5`
D. None

Answer

A

Explanation

`CO_(2) + H_(2)O hArr H_(2)CO_(3) hArr H^(o+) + HCO_(3)^(Theta)`
`K_(1) = ([H^(o+)][HCO_(3)^(Theta)])/([H_(2)CO_(3)]) = ([H^(o+)][HCO_(3)^(Theta)])/([CO_(2)])`
`(K_(1))/([H^(o+)]) = ([HCO_(3)^(Theta)])/([CO_(2)]) = (["Conjugate base"])/(["Acid"])`
`pH = 7.4, log [H^(o+)] =- 7 - 0.4 + 1- 1 = bar(8).6`
`:. [H^(o+)] = "Antilog" (bar(8).6) = 4 xx 10^(-8)`
`:. ([HCO_(3)^(Theta)])/([CO_(2)]) = (K_(1))/([H^(o+)]) = (4.5 xx 10^(-7))/(4xx10^(-8)) = 11.25`