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The `pH` of blood is 7.4 . What is the ratio of `[(HPO_(4)^(2-))/(H_(2)PO_(4)^(-))]` in the blood. `pK_(a)(H_(2)PO_(4)^(-))=7.1`A. `2:1`B. `1:2`C. `3:1`D. `1:3` |
Answer» Correct Answer - 1 `{:H_(2)PO_(4)^(-),hArr,HPO_(4)^(2-),+,H^(+):}` `8xx10^(-8)=((HPO_(4)^(2-))/(H_(2)PO_(4)^(2-)))xx4xx10^(-8)" "rArr" "((HPO_(4)^(2-))/(H_(2)PO_(4)^(2-)))=(2)/(1)` |
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