Calculate the concentration of [H3O+] in a 0.125 M solution of the diprotic acid H2CO3, given Ka1 = 4.3×10⁻⁷ and Ka2 = 5.6×10⁻¹¹.

• A. 4.8×10⁻⁷ M
• B. 1.5×10⁻⁴ M
• C. 3.7×10⁻⁶ M
• D. 2.3×10⁻⁵ M

Answer: (B)

Verified step by step guidance

1. Start by understanding that H2CO3 is a diprotic acid, meaning it can donate two protons (H+ ions) in solution. The dissociation occurs in two steps, each with its own equilibrium constant (Ka1 and Ka2).
2. Write the chemical equations for the dissociation of H2CO3: 1. First dissociation: H2CO3 ⇌ HCO3⁻ + H⁺ 2. Second dissociation: HCO3⁻ ⇌ CO3²⁻ + H⁺
3. Use the first dissociation constant (Ka1) to set up the equilibrium expression for the first dissociation: Ka1 = [H⁺][HCO3⁻] / [H2CO3]. Assume that the initial concentration of H2CO3 is 0.125 M and that the change in concentration due to dissociation is 'x'.
4. Set up the expression for the first dissociation: Ka1 = (x)(x) / (0.125 - x). Since Ka1 is small, assume x is much smaller than 0.125, simplifying the expression to: Ka1 ≈ x² / 0.125.
5. Solve for 'x' to find the concentration of [H⁺] from the first dissociation. Then, consider the second dissociation using Ka2, but note that the contribution to [H⁺] from the second dissociation is negligible due to the small value of Ka2. Therefore, the concentration of [H3O+] is approximately equal to the concentration of [H⁺] from the first dissociation.