Equilibrium Question 809
Question: Assuming that the buffer in the blood is $ CO_2-HCO_3^{-} $ . Calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH of 7.4. $ K_1(H_2CO{ _3})=4.5\times {10^{-7}} $
Options:
A) 11
B) 8
C) 6
D) 14
Show Answer
Answer:
Correct Answer: A
Solution:
$ CO_2 $ with $ H_2O $ forms $ H_2CO_3 $
$ CO_2+H_2O\rightarrow {H^{+}}+HCO_3^{-} $
$ K_1=\frac{[{H^{+}}][HCO_3^{-}]}{[CO_2]}=4.5\times {10^{-7}} $ Again $ pH=-\log [{H^{+}}]=7.4 $
$ \therefore ,[{H^{+}}]=4.0\times {10^{-8}} $
$ \therefore ,\frac{[HCO_3^{-}]}{[CO_2]}=\frac{4.5\times {10^{-7}}}{4\times {10^{-8}}}=11 $
