• # question_answer Blood is a buffer of ${{H}_{2}}C{{O}_{3}}$ and $[HCO_{3}^{-}]$with pH = 7.40. Given ${{K}_{1}}$ of ${{H}_{2}}C{{O}_{3}}=4.5\,\times {{10}^{-7}}$. What will be the ratio of $[HCO_{3}^{-}]$ to $[{{H}_{2}}C{{O}_{3}}]$ in the blood?

${{H}_{2}}C{{O}_{3}}\rightleftharpoons {{H}^{+}}\,+HCO_{3}^{-},$  ${{K}_{1}}=\,\frac{[{{H}^{+}}]\,[HCO_{3}^{-}]}{[{{H}_{2}}C{{O}_{3}}]}$              or     $\frac{[HCO_{3}^{-}]}{[{{H}_{2}}C{{O}_{3}}]}\,=\frac{{{K}_{1}}}{\,[{{H}^{+}}]}$ $pH=7.40$ means $-\log \,[{{H}^{+}}]\,=7.4$ or $\log \,[{{H}^{+}}]\,=-7.4=\overline{8}.6$ or $[{{H}^{+}}]\,=3.981\times {{10}^{-8}}$ $\frac{[HCO_{3}^{-}]}{[{{H}_{2}}C{{O}_{3}}]}\,=\frac{4.5\times {{10}^{-7}}}{3.981\times {{10}^{-8}}}\,=11.3$.