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

When CO_2 dissolves in water, the following equilibrium is establ…

MCQ

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 :-

✓

Correct option: B.

$3.8 \times 10^{-1}$

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

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

$=\frac{3.8 \times 10^{-7}}{10^{-6}}=3.8 \times 10^{-1}$

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.