When `CO_(2)` dissolves in water, the following equilibrium is established.
`CO_(2)+2H_(2)O hArr H_(3)O^(+) +HCO_(3)^(-)`
for which the equilibrium constant is `3.8 xx 10^(-6)` and pH 6.0. What would be the ratio of concentration of bicarbonate ion to carbon dioxide `i.e.[HCO_(3)^(-)]//[CO_(2)]`

Options

A. `3.8 xx 10^(12)`
B. 3.8
C. 6
D. 13.4

Answer

B

Explanation

`K=([H_(3)O^(+)][HCO_(3)^(-)])/([CO_(2)][H_(2)O]^(2))`. As `pH =6.0 `
`:. [H_(3)O^(+)]=10^(-6)=([H_(3)O^(+)][HCO_(3)^(-)])/([CO_(2)])`
`(H_(2)O` is in excess , `:. `its conc. Remains constant )
`([HCO_(3)^(-)])/([CO_(2)])=(K)/([H_(2)O^(+)])=(3.8 xx 10^(-6))/(10^(-6))=3.8`