Question

METHOD 1: TITRATION WITH BURET AND INDICATOR. Section:08 Score: /20 Trial 1 Trial 2 Initial buret reading 10.21 ml 1.95 ml 95mL Trial 3 2.75mL Final buret reading 33-35 ml 25.65mL 26.55 mL NaOH (aq) Volume 23:14 mL 23.7 ml 23.8mL 0.09978 M NaOH (from the container) Calculated moles of NaOH(aq) base used to titrate the acid in 5.00 mL aliquot of acid: 0.002301 mot (show your calculations) NaoH 0.02314 Naut 0.09978 mol NaOH NaOH = 0.002309 mol NaOH Calculated M (mols / L) of the unknown HCl(aq) Trial 1 Trial 2 Trial 3 (Show a sample calculation for one trial) 0.002309 mol NaOH ullmol Ha moim 0.002309 mol HCl 0.00500L -1.066XIOM HOT Average concentration of unknown acid concentration (M)

Answer 1

Acid-Base Titration : (HCl vs NaOH)

Molarity of NaOH = 0.09978 M

# Trial 1 : - Volume of NaOH consumed (titarted with 5 ml of Acid ): 23.14 ml  

Moles of NaOH used = (23.14 ml / 1000 ml/L)* 0.09978 mol / L =0.002309 mol

Since, 1 mol of NaOH = 1 mol of HCl

So, mol of HCl reacted = 0.002309 mol

Molarity of Unkown HCl solution :   0.002309 mol / (5 ml / 1000 ml /L) = 0.4618 M

# Trial 2 : - Volume of NaOH consumed (titarted with 5 ml of Acid ): 23.7 ml  

Moles of NaOH used = (23.7 ml / 1000 ml/L)* 0.09978 mol / L =0.002365 mol

Since, 1 mol of NaOH = 1 mol of HCl

So, mol of HCl reacted = 0.002365 mol

Molarity of Unkown HCl solution :   0.002365 mol / (5 ml / 1000 ml /L) = 0.4729 M

# Trial 3 : - Volume of NaOH consumed (titarted with 5 ml of Acid ): 23.8 ml  

Moles of NaOH used = (23.8 ml / 1000 ml/L)* 0.09978 mol / L =0.002375 mol

Since, 1 mol of NaOH = 1 mol of HCl

So, mol of HCl reacted = 0.002375 mol

Molarity of Unkown HCl solution :   0.002375 mol / (5 ml / 1000 ml /L) = 0.475 M

Average Molarity of Unkown HCl solution : 0.467 M