Blood is a buffer of H_(2)CO_(3) and[HCO_(3)^(-)]with pH = 7.40. Given K_(1)of H_(2)CO_(3) = 4.5xx10^(-7). What will be the ratio of [HCO_(3)^(-)] to [H_(2)CO_(3)] in the blood ?

Blood is a buffer of H_(2)CO_(3) and[HCO_(3)^(-)]with pH = 7.40. Given

1.	Blood is a buffer of H_(2)CO_(3) and[HCO_(3)^(-)]with pH = 7.40. Given K_(1)of H_(2)CO_(3) = 4.5xx10^(-7). What will be the ratio of [HCO_(3)^(-)] to [H_(2)CO_(3)] in the blood ?
Answer» SOLUTION :`H_(2)CO_(3)hArr H^(+)+HCO_(3)^(-), K_(1)=([H^(+)][HCO_(3)^(-)])/([H_(2)CO_(3)]) or ([HCO_(3)^(-)])/([H_(2)CO_(3)])=(K_(1))/([H^(+)])` `PH = 7.40` means - log `[H^(+)] = 7.4` or log `[H^(+)] = -7.4= bar(8).6or [H^(+)]=3.981 xx 10^(-8)` `([HCO_(3)^(-1)])/([H_(2)CO_(3)])=(4.5xx10^(-7))/(3.981xx10^(-8))=11.3`.