Answer:
The ratio, [tex]\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}[/tex], is 10.7
Explanation:
[tex]H_{2}CO_{3}[/tex] is an weak acid and [tex]HCO_{3}^{-}[/tex] is it's conjugate base.
So, according to Henderson-Hasselbalch equation, pH of this buffer system can be represented as-
[tex]pH=pK_{a}(H_{2}CO_{3})+log(\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]})[/tex]
Here pH=7.40, [tex]pK_{a}(H_{2}CO_{3})[/tex]=6.37
So, [tex]7.40=6.37+log(\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]})[/tex]
or, [tex]\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}=10.7[/tex]
So the ratio, [tex]\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}[/tex], is 10.7