1.

`pH=7.40, K_(1) "of" H_(2)CO_(3)=4.5xx10^(-7)`. What will be the ratio of `[HCO_(3)^(-)]` to `[H_(2)CO_(3)]` ?

Answer» `H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)`
`K_(1) = ([H^(+)][HCO_(3)^(-)])/([H_(2)CO_(3)])`
`:. ([HCO_(3)^(-)])/([H_(2)CO_(3)])=(K_(1))/([H^(+)])`
pH = 7.40 means - log `[H^(+)] = 7.4`
or `log [H^(+)] = - 7.4 bar(8) . 6 or [H^(+)] = 3.981 xx 10^(-8)`
`:. ([HCO_(3)^(-)])/([H_(2)CO_(3)]) = (4.5xx10^(-7))/(3.981xx10^(-8))=11.3`


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