Saccharin (Ka = 2 × 10^-12) is a weak acid represented by formula, HSac. A 4 × 10^-4 mole amount of saccharin is dissolved in 200 ml water of pH, 3.0. Assuming no change in volume, the concentration o

Question

Saccharin (Ka = 2 × 10^-12) is a weak acid represented by formula, HSac. A 4 × 10^-4 mole amount of saccharin is dissolved in 200 ml water of pH, 3.0. Assuming no change in volume, the concentration of Sac- ions in the resulting solution at equilibrium is

Prashant Kotian · Accepted Answer

Correct answer: A. Moles of HSac = 4 × 10^-4 mol. Volume = 200 mL = 0.2 L. Analytical concentration of HSac, C = 4 × 10^-4 / 0.2 = 2.0 × 10^-3 M. The pH of water is 3.0, which means [H+] is kept constant at 10^-3 M. Ka = [H+][Sac-] / [HSac] => 2 × 10^-12 = 10^-3 × [Sac-] / (2 × 10^-3 - [Sac-]). Since Ka is extremely small, [Sac-] << 2.0 × 10^-3, so [HSac] approx 2.0 × 10^-3 M. 2 × 10^-12 = 10^-3 × [Sac-] / (2.0 × 10^-3) => [Sac-] = 4.0 × 10^-12 M.

💡 Solution & Explanation

Practice 22,000+ questions like this