What is the pH of a 0.090 M H2CO3 solution, given that the first dissociation constant (Ka1) is 4.3×10^-7 and the second dissociation constant (Ka2) is 5.6×10^-11?

• A. pH = 6.10
• B. pH = 3.68
• C. pH = 5.23
• D. pH = 4.52

Answer: (B)

Verified step by step guidance

1. Start by understanding that carbonic acid (H2CO3) is a diprotic acid, meaning it can donate two protons (H+). The dissociation occurs in two steps, each with its own dissociation constant: Ka1 for the first dissociation and Ka2 for the second.
2. For the first dissociation, write the equilibrium expression: H2CO3 ⇌ HCO3^- + H+. The equilibrium constant expression is given by: $K_1=\frac{\mathrm{[HCO3^-]}\times \mathrm{[H+]}}{\mathrm{[H2CO3]}}$.
3. Assume that the initial concentration of H2CO3 is 0.090 M and that the change in concentration due to dissociation is small. Let x be the concentration of H+ ions produced from the first dissociation. Therefore, [HCO3^-] = x and [H+] = x, while [H2CO3] ≈ 0.090 M.
4. Substitute these values into the equilibrium expression for Ka1: $K_1=\frac{x\times x}{0.090}=4.3\times {10}^{-}$. Solve for x, which represents [H+].
5. Calculate the pH using the formula: $\mathrm{pH}=-\log \left(\mathrm{[H+]}\right)$. Since the second dissociation constant (Ka2) is much smaller, its contribution to the [H+] is negligible, and the pH is primarily determined by the first dissociation.