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}} $



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