Calculate the [H3O+] of a 0.115 M H2CO3 solution, given that Ka1 = 4.3 x 10^-7 and Ka2 = 5.6 x 10^-11.

• A. 1.5 x 10^-7 M
• B. 2.2 x 10^-4 M
• C. 5.6 x 10^-11 M
• D. 3.6 x 10^-5 M

Answer: (B)

Verified step by step guidance

1. Identify that H2CO3 is a diprotic acid, meaning it can donate two protons (H+). It has two dissociation constants: Ka1 for the first dissociation and Ka2 for the second.
2. Write the equilibrium expressions for the dissociation of H2CO3: H2CO3 ⇌ H+ + HCO3- with Ka1 = 4.3 x 10^-7, and HCO3- ⇌ H+ + CO3^2- with Ka2 = 5.6 x 10^-11.
3. Assume that the first dissociation is the primary source of H+ ions because Ka1 is significantly larger than Ka2. Set up the expression for Ka1: Ka1 = [H+][HCO3-] / [H2CO3].
4. Assume that the initial concentration of H2CO3 is 0.115 M and that the change in concentration due to dissociation is x. Therefore, [H+] = x, [HCO3-] = x, and [H2CO3] = 0.115 - x.
5. Substitute these values into the Ka1 expression and solve for x, which represents [H+]. Since Ka1 is small, you can assume x is much smaller than 0.115, simplifying the calculation to Ka1 ≈ x^2 / 0.115.