Question

The first and second dissociation constants of \( \text{H}_2\text{CO}_3 \) are \( 6.761 \times 10^{-7} \) and \( 4.68 \times 10^{-11} \), respectively. If the concentration of \( \text{H}_2\text{CO}_3 \) is 1 molal and pH = 6, the \( \sum \text{CO}_2 \) in the solution (assuming ideal condition) is \(\underline{\hspace{0cm}}\) molal. (round off to 3 decimal places)

Hint:

When calculating \( \sum \text{CO}_2 \) in a solution, use the dissociation constants and the pH value to find the concentration of each ion, and sum them for the total \( \text{CO}_2 \).

Answer 1

Correct Answer: 1.675

We are given the dissociation constants:
- \( K_1 = 6.761 \times 10^{-7} \),
- \( K_2 = 4.68 \times 10^{-11} \),
and the concentration of \( \text{H}_2\text{CO}_3 \) is 1 molal. The pH is 6, so the concentration of \( \text{H}^+ \) is: \[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-6} \, \text{mol/L}. \] The dissociation reactions for \( \text{H}_2\text{CO}_3 \) are: \[ \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- K_1 = 6.761 \times 10^{-7}, \] \[ \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} K_2 = 4.68 \times 10^{-11}. \] To calculate \( [\text{CO}_2] \), we use the dissociation constant expressions. Assuming ideal conditions, the total concentration of \( \text{CO}_2 \) is the sum of \( [\text{HCO}_3^-] \) and \( [\text{CO}_3^{2-}] \). Substituting the values, the total \( \sum \text{CO}_2 \) is approximately: \[ \boxed{1.675} \, \text{molal}. \]