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Question

When $$CO_{2}$$ dissolves in water the following equilibrium is established, $$CO_{2}+H_{2}O\leftrightharpoons H_{3}O^{+}+HCO_{3}^{-}$$, for which the equilibrium constant $$3.8 \times 10^{-7}$$ and $$p{H} = 6.0$$, then ratio of $$\left [ HCO_{3}^{-} \right ]$$  to $$\left [ CO_{2} \right ]$$ would be :


A
3.8×101
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B
3.8×1013
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C
13.4
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D
6.0
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Solution

The correct option is A $$3.8\times 10^{-1}$$
The expression of $$K_{c}$$ j for the given reaction will be:

$$K_{c}=\cfrac{{[H]^+}{{[HCO}_{3}]^-}}{[CO_2]}$$

$$\therefore{3.8}\times 10^{-7}= \dfrac{(10^{-6})[HCO_{3}^{-}]}{[CO_{2}]}$$
$$\dfrac{[HCO_{3}^{-}]}{[CO_{2}]}= 0.38$$

Hence, option A is correct.

Chemistry

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