When CO_2 dissolves in water, the following&n | vedclass

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Question

$\mathrm{Ka}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]}$

$\frac{\left[\mathrm

Vedclass · Accepted Answer

$3.8 	imes 10^{-1}$

Vedclass · Answer

$\mathrm{Ka}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]\left[\mathrm{HCO}_{3}^{-}ight]}{\left[\mathrm{CO}_{2}ight]}$

$\frac{\left[\mathrm{HCO}_{3}^{-}ight]}{\left[\mathrm{CO}_{2}ight]}=\frac{\mathrm{Ka}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]}$

$=\frac{3.8 	imes 10^{-7}}{10^{-6}}=3.8 	imes 10^{-1}$

When $CO_2$ dissolves in water, the following equilibrium is established

$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $

for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-

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When $CO_2$ dissolves in water, the following equilibrium is established $C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $ for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-