Saccharin () is a weak acid represented by formula, HSac. A mole amount of saccharin is dissolved in — Ionic Equilibrium Chemistry Question
Question
Saccharin ($K_a = 2 \times 10^{-12}$) is a weak acid represented by formula, HSac. A $4 \times 10^{-4}$ mole amount of saccharin is dissolved in 200 ml water of pH, 3.0. Assuming no change in volume, the concentration of $\text{Sac}^-$ ions in the resulting solution at equilibrium is
💡 Solution & Explanation
Initial concentration of saccharin (HSac) = $4 \times 10^{-4} \text{ mol} / 0.2 \text{ L} = 2 \times 10^{-3} \text{ M}$. The solution pH is fixed at 3.0, meaning $[H^+] = 10^{-3} \text{ M}$ (the dissociation of HSac is negligible and doesn't change this). From the equilibrium expression, $K_a = \frac{[H^+][\text{Sac}^-]}{[\text{HSac}]} \implies 2 \times 10^{-12} = \frac{10^{-3} \times [\text{Sac}^-]}{2 \times 10^{-3}}$. Solving for $[\text{Sac}^-]$ yields $4 \times 10^{-12} \text{ M}$. Therefore, correct answer is A.