When $C O_{2}$ is bubbled in excess of water, the following equilibrium is established.
$C O_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + H C O_{3}^{-},$ $K_{c} = 3.8 \times 10^{- 7}, P^{H} = 6.$

What would be the $\frac{[H C O_{3}^{-}]}{[C O_{2}]}$ ?

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0.0038

0.038

0.38

answer is D.

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Detailed Solution

$C O_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + H C O_{3}^{-}$

$K_{c} = \frac{[H_{3} O^{+}] [H C O_{3}^{-}]}{[C O_{2}]}$

$3.8 \times 10^{- 7} = \frac{10^{- 6} \times [H C O_{3}^{-}]}{[C O_{2}]}$

$\frac{[H C O_{3}^{-}]}{[C O_{2}]} = 3.8 \times 10^{- 1} = 0.38$

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