Assuming that the buffer in the blood is $\mathrm{CO}_{2}-\mathrm{HCO}_{3}^{-} .$ Calculate the ratio of…

Assuming that the buffer in the blood is $\mathrm{CO}_{2}-\mathrm{HCO}_{3}^{-} .$ Calculate the ratio of conjugate base to acid necessary to maintain blood at its proper $\mathrm{pH}$ of $7.4$ $\mathrm{K}_{1}\left(\mathrm{H}_{2} \mathrm{CO}_{3}ight)=4.5 \times 10^{-7}$
  1. 11
  2. 8
  3. 6
  4. 14

Solution

$\mathrm{CO}_{2}$ with $\mathrm{H}_{2} \mathrm{O}$ forms $\mathrm{H}_{2} \mathrm{CO}_{3}$
$\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} ightleftharpoons \mathrm{H}^{+}+\mathrm{HCO}_{3}^{-}$
$\mathrm{K}_{1}=\frac{\left[\mathrm{H}^{+}ight]\left[\mathrm{HCO}_{3}^{-}ight]}{\left[\mathrm{CO}_{2}ight]}=4.5 \times 10^{-7}$
Again $\mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]=7.4$
$\therefore \quad\left[\mathrm{H}^{+}ight]=4.0 \times 10^{-8}$
$\therefore \quad \frac{\left[\mathrm{HCO}_{3}^{-}ight]}{\left[\mathrm{CO}_{2}ight]}=\frac{4.5 \times 10^{-7}}{4 \times 10^{-8}}=11$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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