The molar ratio of dimer to monomer for acetic acid in water (neglecting the dissociation of acetic — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let the monomer concentration be $x$ and dimer be $y$. Mass balance: $x + 2y = 0.1\text{ M}$. In water, $K = \frac{y}{x^2} = 3.6 \times 10^{-2}$. Since $K$ is very small, very little dimerization occurs, so $x \approx 0.1\text{ M}$. The dimer concentration $y = (3.6 \times 10^{-2})(0.1)^2 = 3.6 \times 10^{-4}\text{ M}$. The ratio of dimer to monomer is $y/x = \frac{3.6 \times 10^{-4}}{0.1} = 3.6 \times 10^{-3} = \frac{36}{10000} = \frac{9}{2500}$. Therefore, correct answer is C.