Question

A 0.8746 g sample of an unknown acid, HA, is reacted with NaOH (aq). The reaction that occurs is as follows: HA (aq) + NaOH (aq) =====> NaA (aq) + H2O (l) If the complete reaction of the HA sample requires exactly 28.62 mL of 0.1927 M NaOH (aq), what is the molecular mass of HA ?

Answer 1

Sol) from the balanced reaction: Na A (aq) + H20(1) HA (aq) + NaOH(aq.) I mole of HA Cag:) reacts with 1 mole of NaOH(aq) to produce 1 mole of Na Alaq.) and 1 mol of H₂O(1) moles of NaOH(aq) in 28.62 mL of 0.1927 M we know 1 m = 1 mol * Moles = 28.62 ml x 0.1927 mol al moles (28.62x10-3 L) X (0-1927 mel) & moles = 5.51x10-3 moles of NaOH(aq) Since NaOH(aq.) and HA (aqil have equimolar quantities (1:1,given from balanced reaction's statement) so moles of NaOH(aq) = moles of HA (aq) a moles of HA Cag. ? = 5.51 x 10-3 moles of HA Caqi) We know moles= mass in ga molar mass in g/mol + 5.51x10 -3 moles = 0.8746 g molar mass ing/mal & molar mass in g/mol = 0.87469 5.51x10-3 mol +/molar mass in glmal = 158 17 g/mol