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)] ?

pH=7.40, K_(1) "of" H_(2)CO_(3)=4.5xx10^(-7). What will be the ratio o

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» SOLUTION :`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`