lab .docx - Post Lab 7 Calculations 1 Calculate the average...

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Post Lab # 7 Calculations: 1. Calculate the average molarity of the NaOH solution from your three standardization trials. M NaOH = moles KHPh/ V NaOH in liters Flask One: (0.381 g KHPh/204.23 g/mol KHPh)/ (0.01699 liters NaOH) = 0.1098 mol/L NaOH Flask Two: (0.382 g KHPh/204.23 g/mol KHPh)/ (0.01699 liters NaOH) = 0.1100 mol/L NaOH Flask Three: (0.384 g KHPh/204.23 g/mol KHPh)/ (0..01723 liters NaOH) = 0.1.091 mol/L NaOH Average: [(0.1098 + 0.1100 + 0.1091) mol/L NaOH]/3 = 0.1096 mol/L NaOH 2. Calculations to determine the effectiveness of the antacid. a. Calculate the average number of moles of HCl neutralized per antacid tablet. Moles acid neutralized = (moles HCl added) - (moles NaOH for back titration) = (M HCl x V HCl ) – ( M NaOH x V NaOH ) Tums: Tablet One : [(1.0001 mol HCl/L) (0.01228 L HCl)] – [(0.1096 mol NaOH/L) (0.01742 L NaOH) ]= 0.0104 moles HCl Tablet Two : [(1.0001 mol HCl/L) (0.01199 L HCl)] – [(0.1096 mol NaOH/L) (0.01740 L NaOH)]= 0.0101 moles HCl Average number of moles of HCl for both trials: [(0.0104 + 0.0101) moles HCl] /2 = 0.0103 moles HCl Gelusil (Partner):
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Tablet One : [(1.0001 mol HCl/L) (0.00799 L HCl)] – [(0.1096 mol NaOH/L) (0.01690 L NaOH) ]= 0.0061 moles HCl Tablet Two : [(1.0001 mol HCl/L) (0.00799 L HCl)] – [(0.1096 mol NaOH/L) (0.01719 L NaOH)]= 0.0061 moles HCl Average number of moles of HCl for both trials: [(0.0061 + 0.0061)moles HCl]/2 = 0.0061 moles HCl b. Calculate the number of moles of HCl neutralized per gram of antacid (weight effectiveness). Tums: Tablet One: (0.0104 moles HCl/ 1.305 g) = 0.00797 moles HCl/g Tablet Two: (0.0101 moles HCl/ 1.330 g) = 0.00759 moles HCl/g Average: [(0.00797 + 0.00759) HCl/g]/ 2 = 0.00778 moles HCl/g Gelusil (Partner): Tablet One: (0.0061moles HCl/ 1.197 g) = 0.00510 moles HCl/g
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