Which of the following result(s) is/are correct for the equilibrium state in a solution originally h — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The strong acid HCl provides $[\text{H}^+] \approx 0.1\text{ M}$ (Option A). For acetic acid, $K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \implies 1.8 \times 10^{-5} = \frac{[\text{CH}_3\text{COO}^-](0.1)}{0.1} \implies [\text{CH}_3\text{COO}^-] = 1.8 \times 10^{-5}\text{ M}$ (Option B). The degree of dissociation $\alpha = \frac{[\text{CH}_3\text{COO}^-]}{C} = \frac{1.8 \times 10^{-5}}{0.1} = 1.8 \times 10^{-4}$ (Option C). The $[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{0.1} = 10^{-13}\text{ M}$. Since water is the only source of hydroxide, the $[\text{H}^+]$ derived from water equals the $[\text{OH}^-]$, which is $10^{-13}\text{ M}$ (Option D). Therefore, correct answer is A,B,C,D.