To find the equilibrium constant expression for the given chemical reaction, we need to identify the appropriate form of the equilibrium expression for [tex]\(K_a\)[/tex].
Given the reaction:
[tex]\[ \text{H}_2\text{CO}_3 (aq) + \text{H}_2\text{O} (l) \rightleftarrows \text{H}_3\text{O}^+ (aq) + \text{HCO}_3^- (aq) \][/tex]
The equilibrium constant expression for an acid dissociation reaction can be written generally as:
[tex]\[ K_a = \frac{[\text{Products}]}{[\text{Reactants}]} \][/tex]
When writing the equilibrium expression, we need to remember that the concentration of pure liquids (like water) is typically not included because it remains essentially constant and can be omitted from the equilibrium expression.
Using this principle, we write the equilibrium expression for the given reaction:
[tex]\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \][/tex]
Let's match this expression with the options provided:
A. [tex]\( K_a = \frac{[\text{H}_3\text{O}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \)[/tex]
B. [tex]\( K_a = \frac{[\text{H}_2\text{CO}_3][\text{H}_2\text{O}]}{[\text{H}_3\text{O}^+][\text{HCO}_3^-]} \)[/tex]
C. [tex]\( K_a = \frac{[\text{H}_3\text{O}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3][\text{H}_2\text{O}]} \)[/tex]
D. [tex]\( K_0 = \frac{[\text{H}_2\text{CO}_3]}{[\text{H}_3\text{O}^+][\text{HCO}_3^-]} \)[/tex]
From these options, the correct equilibrium expression corresponds to option A:
[tex]\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \][/tex]
So, the correct answer is:
A. [tex]\(K _{ a }=\frac{\left[ H _3 O ^{+}\right]\left[ HCO _3^{-}\right]}{\left[ H _2 CO _3\right]}\)[/tex].
