Calculate the $\mathrm{pH}$ of a solution containing $0.1 \mathrm{M}$ $\mathrm{HCO}_{3}^{-}$ and $0.2…

Calculate the $\mathrm{pH}$ of a solution containing $0.1 \mathrm{M}$ $\mathrm{HCO}_{3}^{-}$ and $0.2 \mathrm{M} \mathrm{CO}_{3}^{2-}$ $\left[\mathrm{K}_{1}\left(\mathrm{H}_{2} \mathrm{CO}_{3}ight)=4.2 \times 10^{-7} \times 10ight.$ and $\left.\mathrm{K}_{2}\left(\mathrm{HCO}_{3}^{-}ight)=4.8 \times 10^{-11}ight]$
  1. $3.18$
  2. $10.62$
  3. $6.62$
  4. $9.31$

Solution

$\mathrm{HCO}_{3}^{-} ightarrow \mathrm{H}^{+}+\mathrm{CO}_{3}^{2-}$
$\mathrm{K}_{2}=\frac{\left[\mathrm{H}^{+}ight]\left[\mathrm{CO}_{3}^{2-}ight]}{\left[\mathrm{HCO}_{3}^{-}ight]}=4.8 \times 10^{-11}$
$\left[\mathrm{H}^{+}ight]=\frac{4.8 \times 10^{-11}\left[\mathrm{HCO}_{3}^{-}ight]}{\left[\mathrm{CO}_{3}^{2-}ight]}$
$=4.8 \times 10^{-11}(0.1 / 0.2)$
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]$
$=-\log \left(4.8 \times 10^{-11} \times 0.5ight)=10.62$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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