Answer:
\[{{H}_{2}}C{{O}_{3}}\rightleftharpoons {{H}^{+}}\,+HCO_{3}^{-},\] \[{{K}_{1}}=\,\frac{[{{H}^{+}}]\,[HCO_{3}^{-}]}{[{{H}_{2}}C{{O}_{3}}]}\] 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 \[[{{H}^{+}}]\,=3.981\times {{10}^{-8}}\]
\[\frac{[HCO_{3}^{-}]}{[{{H}_{2}}C{{O}_{3}}]}\,=\frac{4.5\times {{10}^{-7}}}{3.981\times {{10}^{-8}}}\,=11.3\].
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