Lactic acid, HC3H5O3, produced in 1 g sample of muscle tissue was titrated using phenolphthalein as — Electrochemistry Chemistry Question
Question
Lactic acid, HC3H5O3, produced in 1 g sample of muscle tissue was titrated using phenolphthalein as indicator against OH^– ions which were obtained by the electrolysis of water. As soon as OH^– ions are produced, they react with lactic acid and at complete neutralization, immediately a pink colour is noticed. If electrolysis was made for 1158 s using 50.0 mA current to reach the end point, what was the percentage of lactic acid in muscle tissue?
💡 Solution & Explanation
Charge passed Q = 50.0 × 10^–3 A × 1158 s = 57.9 C. Moles of OH^– produced = 57.9 / 96500 = 0.0006 mol. Because 1 mole of OH^– neutralizes 1 mole of lactic acid (monoprotic), moles of lactic acid = 0.0006 mol. Molar mass of lactic acid (HC3H5O3) = 90 g/mol. Mass of lactic acid = 0.0006 × 90 = 0.054 g. Percentage in 1 g sample = (0.054 / 1) × 100 = 5.4%. Therefore, correct answer is A.