The [HCO_{3}^{-}]/[H_{2}CO_{3}] ratio in the | vedclass

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(A) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$Here,the salt is $[HCO_{3}^{-}]$ and

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$Here,the salt is $[HCO_{3}^{-}]$ and the acid is $[H_{2}CO_{3}]$.Substituting the given values:$7.4 = 6.1 + \log \frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}$Subtracting $6.1$ from both sides:$1.3 = \log \frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}$Taking the antilog of both sides:$\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]} = 10^{1.3} \approx 20$Thus,the ratio is approximately $20$.

The $[HCO_{3}^{-}]/[H_{2}CO_{3}]$ ratio in the buffer present in blood $(pH = 7.4)$ is approximately $(pK_{a} = 6.1$ for $H_{2}CO_{3})$.

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