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6-2.Equilibrium-II (Ionic Equilibrium)
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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 :-
A
$3.8 \times 10^{-13}$
B
$3.8 \times 10^{-1}$
C
$6.0$
D
$13.4$
Solution
$\mathrm{Ka}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]}$
$\frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]}=\frac{\mathrm{Ka}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}$
$=\frac{3.8 \times 10^{-7}}{10^{-6}}=3.8 \times 10^{-1}$
Standard 11
Chemistry
