Question

15mL of 0.40M hydrochloric acid were added to a solution of an antacid. 4.1mL of 0.50M sodium hydroxide were used to compensate for the excess of HCl. How many equivalents of excessive HCl were titrated?

Answer 1

Reaction of Hel and Noon, Hea can) + Naon Nard(s) + H2O. from POAC Rule. moles of Hel & males of Nach Cofficient cofficient moles of HC2 - Molaruelty x volume moles of Hcl = 401 ml x 0.5 moles of HCl = 2.05 mmol steh I find fotel moles of Hc. Total mees of HCl = Molavity x volume - 0.4 x 15ml = 6 mmol steh I excessive equivalance that were titrated. Remaning moles - Total moles & moles that Reacts Naon = 6 - 2.05 = 3.85 m mot equivalent of HCl = moles = 3.80 = 3.85x10 equivalance Thank you -factor - 3.85mmal