The pH of blood is 7.4. If the buffer in blood is $C O_{2}$ and $H C O_{3}^{-},$ calculate the ratio of conjugate base to acid $(H_{2} C O_{3})$ to maintain the pH of the blood. Given $K_{1}$ of $H_{2} C O_{3} = 4.5 \times 10^{- 7}$

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10.0

8.5

None of these

11.25

answer is C.

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Detailed Solution

$C O_{2} + H_{2} O \overset{}{⇌} H C O_{3}^{-} + H^{+};$

$K_{1} = \frac{[H C O_{3}^{-}] [H^{+}]}{[C O_{2}]} = 4.5 \times 10^{- 7}$

or $\frac{[H C O_{3}^{-}]}{[C O_{2}]} = \frac{4.5 \times 10^{- 7}}{[H^{+}]} = \frac{4.5 \times 10^{- 7}}{4 \times 10^{- 8}} = 11.25$

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