1.ISG Antacid # 1.52 Antacid #2 Antacid #3 PART DATA tro henoth Antacid brand na

Question

1.ISG Antacid # 1.52 Antacid #2 Antacid #3 PART DATA tro henoth Antacid brand name Regular or Extra Strength eaula Antacid active ingredients Mass of antacid Molarity of HCI solutionOO Volume of HCI solution5m_ Molarity of NaOH solution .IO0m .tom hnital bue readingDomcm Initial buret Final buret reading CHEMICAL REACTION Write a balanced chemical equation for the titration reaction: CALCULATIONS Initial moles of HCI Volume of Na0H used 0.20ml3%.70mL 3.20 ml Volume of NaOH used 0.20m Moles of NaOH used Moles of HCI neutralized by the NaOH Moles of HCI neutralized by the antacid Effectiveness of Antacid millimol HCl neutralized grams of antacid Which antacid is most effective? 128 ATTACH CALCULATIONS, BAR GRAPH, RUBRIC, AND ANSWERS TO POST-LAB QUESTIONS

Explanation / Answer

The coloumns can be completed as:

For all antacids:                                    

Initial moles of HCl = Molarity X volume in litres = 2 X 0.005 = 0.01 moles

Moles of NaOH used = Molarity X volume in litres

For antiacid 1: Moles of NaOH = 0.1 X 0.2 / 1000 = 0.00002 moles

For antacid 2: moles of NaOH = 0.1 X 36.7 / 1000 = 0.00367 moles

For antacid 3: moles of NaOH = 0.1 X 3.2 / 1000 = 0.00032 moles

Moles of HCl neutrlized b naOH

For antiacid 1: Moles neutralized = 0.00002 moles

For antacid 2: Moles neutralized= 0.00367 moles

For antacid 3: Moles neutralized = 0.00032 moles

moles of HCl neutralized by antacid will be

For antiacid 1: Moles neutralized = 0.01 - 0.00002 moles = 0.00998 = 9.98 millimoles

For antacid 2: Moles neutralized= 0.01 - 0.00367 moles = 0.00633 = 6.33 millimoles

For antacid 3: Moles neutralized =0.01 - 0.00032 moles = 0.00968 = 9.68 millimoles

Effectiveness of antacid

= Millimoles of acid neutralized / grams of antacid

For antiacid 1: 9.98 / 1.56 = 6.397

For antacid 2: 6.33 / 1.53 = 4.137

For antacid 3: 9.68 / 1.262 = 7.670

so most effective is antacid # 3

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