When $\mathrm{CO}_{2}$ dissolves in water, the following equilibrium is established $\mathrm{CO}_{2}+2…

When $\mathrm{CO}_{2}$ dissolves in water, the following equilibrium is established $\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HCO}_{3}^{-}$ for which the equilibrium constant is $3.8 \times 10^{-7}$ and $\mathrm{pH}=6.0$. The ratio of $\left[\mathrm{HCO}_{3}^{-}ight]$ to $\left[\mathrm{CO}_{2}ight]$ would be
  1. $3.8 \times 10^{-13}$
  2. $3.8 \times 10^{-1}$
  3. $6.0$
  4. $13.4$

Solution

\(\begin{aligned} & \mathrm{K}= \frac{\left[\mathrm{H}_3 \mathrm{O}^{+}ight]\left[\mathrm{HCO}_3^{-}ight]}{\left[\mathrm{CO}_2ight]\left[\mathrm{H}_2 \mathrm{O}ight]^2} \mathrm{AspH} \\ &=6.0\left[\mathrm{H}_3 \mathrm{O}ight]^{+}=10^{-6} \end{aligned}\) \(\mathrm{K}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}ight]\left[\mathrm{HCO}_3^{-}ight]}{\left[\mathrm{CO}_2ight]\left[\mathrm{H}_2 \mathrm{O}ight]^2}\) \(\left(\mathrm{H}_2 \mathrm{O}ight.\) is in excess, therefore its conc. remains constant) \(\begin{aligned} & \frac{\left[\mathrm{HCO}_3^{-}ight]}{\left[\mathrm{CO}_2ight]}=\frac{\mathrm{K}}{\left[\mathrm{H}_3 \mathrm{O}^{+}ight]} \\ & =\frac{3.8 \times 10^{-7}}{10^{-6}}=3.8 \times 10^{-1} \end{aligned}\) /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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