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When CO(2) is bubbled in excess of water...

When `CO_(2)` is bubbled in excess of water, the following equilibrium is establisheed.
`CO_(2) + 2H_(2)O hArr H_(3)O^(+) + HCO_(3)^(-)`
`K_(c) = 3.8 xx 10^(-7), p^(H) = 6`
What would be the `[HCO_(3)^(-)]//[CO_(2)]`

A

6

B

`0.0038`

C

`0.038`

D

`0.38`

Text Solution

Verified by Experts

The correct Answer is:
D
[weak acid]= `[CO_(2)]`
[salt] =`[HCO_(3)^(-)]`
`:. P^(H)=P^(ka)+"log"_(10)([HCO_(3)^(-)])/([CO_(2)])`
`6(7-log 3.8)+log([CO_(3)^(-2)])/([CO_(2)])`
`6=3.2+log ([CO_(3)^(-2)])/([CO_(2)]), ([CO_(3)^(-2)])/([CO_(2)])=0.38`
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