Equilibrium Question 804
Question: When $ CO_2 $ dissolves in water, the following equilibrium is established $ CO_2+2H_2O\rightarrow H_3{O^{+}}+HCO_3^{-} $ for which the equilibrium constant is $ 3.8\times {10^{-7}} $ and pH = 6.0. The ratio of $ [ HCO_3^{-} ] $ to $ [CO_2] $ would be
Options:
A) $ 3.8\times {10^{-13}} $
B) $ 3.8\times {10^{-1}} $
C) 6.0
D) 13.4
Show Answer
Answer:
Correct Answer: B
Solution:
$ CO_2+2H_2O\rightarrow H_3{O^{+}}+HCO_3^{-}; $
$ K_{c}=\frac{[H_3{O^{+}}][HCO_3^{-}]}{[CO_2]}; $
$ \frac{[HCO_3^{-}]}{[CO_2]}=\frac{3.8\times {10^{-7}}}{{10^{-6}}}=3.8\times {10^{-1}} $
