Question

need to calculate tritation values

EXPERIMENT 11 LABORATORY REPORT Brand of Antacid Rugbu Cost of antacid 4.5 ¢ per tablet Average mass of tablet_1-30129_per tablet Concentration of HCl solution (M) 0.0979 Concentration of NaOH solution (M) .1006 Data Trial 1 Trial 2 Mass of sample (g) 0-104 Initial volume of HCl (mL 10.116 49.1 12101 4601 24.1 Final volume of HCl (mL) Initial volume of NaOH (mL 49.6 48.2 Final volume of NaOH (mL) 34.4 24.2 140

correction trial 2 final volume of NaOH is 30.2ml

Answer 1

Conc. of HCl = 0.0979 M

Conc. of NaOH = 0.1006

#1 Mass of sample - 0.116 g

Vol. of HCl added = 28.0 ml

moles of HCl added initially : (28.0 ml / 1000 ml/L)* 0.0979 mol / L =0.00274 mol

vol. of NaOH added for titration = 15.2 ml

Moles of NaOH used = (15.2 ml / 1000 ml/L)* 0.1006 mol / L =0.001529 mol

Since, 1 mol of NaOH = 1 mol of HCl

So, mol of HCl reacted with NaOH = 0.001529 mol

Moles of HCl neutralized by Antacid : 0.00274 mol-0.001529 mol = 0.0012 moles

Mass of tablet : 0.116 g

Moles of HCl neutralized by per g of Antacid tab. = 0.0104 moles

#2 Mass of sample - 0.104 g

Vol. of HCl added = 22.0 ml

moles of HCl added initially : (22.0 ml / 1000 ml/L)* 0.0979 mol / L =0.002154 mol

vol. of NaOH added for titration = 18 ml

Moles of NaOH used = (18 ml / 1000 ml/L)* 0.1006 mol / L =0.001811 mol

Since, 1 mol of NaOH = 1 mol of HCl

So, mol of HCl reacted with NaOH = 0.001811 mol

Moles of HCl neutralized by Antacid : 0.002154 mol-0.001811 mol = 0.000343 moles

Mass of tablet : 0.104 g

Moles of HCl neutralized by per g of Antacid tab. = 0.0033 moles

Average Moles of HCl neutralized by per g of Antacid tab. = 0.0069 moles

Average Moles of HCl neutralized by per Antacid tab. = 0.0089 moles

Cost of antacid per tablet : 4.5 $

Cost antacid per tablet per Moles of HCl neutralized = 1.00 * (4.5 $/per 0.0089 mol) = 505.6 $/mol