What is the pH of a 0.97 M solution of carbonic acid (H2CO3) given that its acid dissociation constant (Ka) is 4.5 x 10^-7?

The first step in calculating the pH of a 0.97 M solution of carbonic acid (H2CO3) is to use the provided acid dissociation constant (Ka). Carbonic acid dissociates in water as follows:

H₂CO₃ ⇌ H⁺ + HCO₃^–

The expression for the acid dissociation constant (Ka) is:

Ka =  rac{[H^+][HCO_3^-]}{[H_2CO_3]} 

Here, we know that:

• [H2CO3] initially = 0.97 M
• Let ‘x’ be the concentration of H⁺ ions produced at equilibrium.

At equilibrium, the concentrations will be:

• [H₂CO₃] = 0.97 – x
• [H⁺] = x
• [HCO₃^–] = x

Substituting these into the Ka expression:

4.5 x 10^-7 =  rac{x imes x}{0.97 – x} 

Since Ka is quite small compared to the concentration of the acid, we can assume that x is much smaller than 0.97. Thus, 0.97 – x is approximately 0.97:

4.5 x 10^-7 =  rac{x^2}{0.97} 

Now, solve for x:

x² = 4.5 x 10^-7 × 0.97

x² = 4.365 x 10^-7

x ≈ 6.6 x 10^-4

Now, since x represents [H⁺], we can calculate the pH:

pH = -log[H⁺]

pH = -log(6.6 x 10^-4)

pH ≈ 3.18

Therefore, the pH of a 0.97 M solution of carbonic acid is approximately 3.18.