11th Class Chemistry Equilibrium Question Bank 11th CBSE Chemistry Ionic Equilibrium

  • 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?

    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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