For a tribasic acid, , , and . The value of at equilibrium in an aqueous solution originally having — Ionic Equilibrium Chemistry Question
Question
For a tribasic acid, $\text{H}_3\text{A}$, $K_{a1} = 2 \times 10^{-5}$, $K_{a2} = 5 \times 10^{-9}$ and $K_{a3} = 4 \times 10^{-12}$. The value of $\frac{[\text{A}^{3-}]}{[\text{H}_3\text{A}]}$ at equilibrium in an aqueous solution originally having $0.2 \text{ M} – \text{H}_3\text{A}$ is
💡 Solution & Explanation
The overall equilibrium constant $K_{eq} = K_{a1} K_{a2} K_{a3} = (2 \times 10^{-5})(5 \times 10^{-9})(4 \times 10^{-12}) = 40 \times 10^{-26} = 4 \times 10^{-25}$. The expression is $K_{eq} = \frac{[\text{H}^+]^3[\text{A}^{3-}]}{[\text{H}_3\text{A}]}$. $[\text{H}^+]$ is primarily determined by the first dissociation: $[\text{H}^+] \approx \sqrt{K_{a1} C} = \sqrt{2 \times 10^{-5} \times 0.2} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \text{ M}$. Rearranging for the required ratio: $\frac{[\text{A}^{3-}]}{[\text{H}_3\text{A}]} = \frac{K_{eq}}{[\text{H}^+]^3} = \frac{4 \times 10^{-25}}{(2 \times 10^{-3})^3} = \frac{4 \times 10^{-25}}{8 \times 10^{-9}} = 0.5 \times 10^{-16} = 5 \times 10^{-17}$. Therefore, correct answer is A.