What is the concentration of unionized acetic acid in the solution? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The concentration of unionized acid $[\text{HA}]$ can be found from $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$. We established $[\text{H}^+] \approx 1.41 \times 10^{-7}\text{ M}$ and $[\text{A}^-] \approx C = 7 \times 10^{-8}\text{ M}$ (since it is nearly 100% dissociated). Substituting these into the equilibrium expression: $2.0 \times 10^{-5} = \frac{(1.41 \times 10^{-7})(7 \times 10^{-8})}{[\text{HA}]}$. Solving for $[\text{HA}]$ gives $\frac{9.87 \times 10^{-15}}{2.0 \times 10^{-5}} \approx 4.9 \times 10^{-10}\text{ M}$. Therefore, correct answer is B.