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 :-

  • A

    $3.8 \times 10^{-13}$

  • B

    $3.8 \times 10^{-1}$

  • C

    $6.0$

  • D

    $13.4$

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