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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 ? |
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Answer» `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).6 or [H^(+)]=3.981 xx 10^(-8)` `([HCO_(3)^(-1)])/([H_(2)CO_(3)])=(4.5xx10^(-7))/(3.981xx10^(-8))=11.3`. |
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