Question

When $C{O_2}$ dissolves in water the following equilibrium is established
$C{O_2} + 2{H_2}O \to {H_3}{O^ + } + HC{O_3}^ -$ ;
For which the equilibrium constant is $3.8 \times {10^{ - 6}}{\text{ and pH = 6}}{\text{.0}}$ . What would be the ratio of concentration of bicarbonate ions to carbon dioxide?

Answer 1

The equilibrium constant is defined as the ratio of units of pressure or concentration of products to reactants for a chemical equation. The equilibrium constant basically determines the forward and backward flow rate ratio in equilibrium.

Complete step by step answer:
Equilibrium constant is a constant defined for the chemical reactions which is basically the ratio of pressure or concentration units of products to reactants when reaction is at equilibrium stage. Only aqueous phase units are included in this term equilibrium constant, solids and excess reagents or products are excluded during the calculation of equilibrium constant.
The reaction; where carbon dioxide dissolves in water is as follows:
$C{O_2} + 2{H_2}O \to {H_3}{O^ + } + HC{O_3}^ -$
The equilibrium constant for this chemical equation is given as $3.8 \times {10^{ - 6}}{\text{ and pH = 6}}{\text{.0}}$.
Therefore the equilibrium constant for this chemical equation will be represented as the ratio of the product of hydronium ion and carbonate ion with carbon dioxide.
hence the equilibrium constant is as follows:
${\text{Kc = }}\dfrac{{{\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{][HC}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{]}}}}{{{\text{[C}}{{\text{O}}_{\text{2}}}{\text{]}}}}$
The concentration of hydronium ion will be calculated from the given pH value.
${\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{] = 1}}{{\text{0}}^{{\text{ - pH}}}}$
${\text{[}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{] = 1}}{{\text{0}}^{{\text{ - 6}}}}$
Now assume the ratio of concentration of bicarbonate ion to carbon dioxide as $x$
Now put all the values in above equilibrium formula and calculate the value of $x$
$3.8 \times {10^{ - 6}}{\text{ = }}{10^{ - 6}} \times x$
$\Rightarrow x = 3.8$

So the ratio of concentration of bicarbonate ions to carbon dioxide is 3.8.

Note:
The concentration of ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$ ion and ${{\text{H}}^{\text{ + }}}$ion are considered to be equal for numerical performed in chemistry, because ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$ ion is also represented as (${{\text{H}}_{\text{2}}}{\text{O + }}{{\text{H}}^{\text{ + }}}$ ) so both of them are same and we can calculate the concentration of hydronium ion in the same way as that of hydrogen ion.