Respuesta :
Answer: Option (A) is the correct answer.
Explanation:
The given data is as follows.
pH = 7.40, [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex]
[tex]pK_{a}[/tex] = 6.10
We have to find [tex]\frac{[HCO_^{-}{3}]}{[CO_{2}]}[/tex] = ?
According to Henderson-Hasselbalch equation,
pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]
Hence, putting the given values into the above equation as follows.
pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]
or, pH = [tex]pK_{a} + log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]
7.40 = 6.10 + [tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]
[tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = 1.30
[tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = antilog (1.30)
= 20
Since, it is given that [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex].
Therefore, [tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] or [tex]\frac{[HCO^{-}_{3}]}{[CO_{2}]}[/tex] = [tex]\frac{20}{1}[/tex]
Thus, we can conclude that the bicarbonate (HCO3-) : carbon dioxide ratio for a normal blood pH of 7.40 is 20:1.
