dnuation of the Cost Effectiveness of Antacids Make sure to show all calculation

Question

dnuation of the Cost Effectiveness of Antacids Make sure to show all calculations. This pre-lab is due Objectives For this lab, you will be evaluating two commercially available antacids by 1) Determining the number of moles of H' neutralized per gram of each antacid 2) Calculating the cost effectiveness of each antacid Background Acid indigestion is a common ailment caused by the overproduction of stomach ac provide some relief from the symptoms of acid indigestion. They are generally made up of bases such as Mg(OH)s AI(OH)s, and CaCO, that can react with HCl as shown id, HCL Over-the-counter antacids some mixture of weak below Mg(OH)2 (aq) + 2 HCI(aq) MgCl2 (aq) + 2 H2O() Al (OH)3 (aq) + 3 Hci(aq)-AlCl3(aq) + 3 H20() Caco, (aq) + 2 Hci(aq) CaCl2 (aq) + CO2(g) + H20(!) In this lab, we will perform a special type of titration, a back titration, in order to determine the number of moles of H neutralized per gram of antacid. We first mix the antacid tablet with an excess of HCL. The H' that does not react with the antacid (the excess HCl) is then titrated with a standardized NaOH solution in the presence of an indicator The indicator will change color when the endpoint is reached. The end point is defined as the volume of OH-needed to see a color change. Because only the tinlest excess of OH over H can cause the color change of an indicator, the end point is a close approximation of the equivalence (note that the difference between the end point and th e equivalence point is known as the titration error). At the equivalence point, the number of moles of OH added is equal to the number of moles of excess H' that had not been neutralized by the antacid. By knowing the total moles of HCI added, the number of moles of H+ neutralized by the antacid can thus be calculated. total moles of H+ = moles of H+ neutralized by antacid + moles of H+ neutralized by NaOH Because the antacid may include both OH and Co,, it is not possible to calculate the number of moles of each of these ion species independently. Instead, the number of moles of H+ neutralized by the antacid is determined. The amount of antacid required to neutralize one mole of H+ neutralized is said to be one "equivalent. total equivalents of antacid total moles of Ht neutralized he more cost-effective antacid is the one that costs fewer dollars per equivalent.

Explanation / Answer

1) moles of acid HCl added = molarity x volume = 0.1034 M x 0.03236 L = 3.35 x 10^-3 mols

2) moles of NaOH added = 0.1506 M x 0.01172 L = 1.76 x 10^-3 mols

3) difference between added HCl and added NaOH = 1.58 x 10^-3 mols

4) moles of HCl reacted with antacid = 1.58 x 10^-3 mols

Equivalents of antacid = 1.58 x 10^-3

5) number of equivalents of antacid per g of antacid = 1.58 x 10^-3/0.326 g = 4.85 x 10^-3 Eq/g

6) 100 tablet = $5.99

1 tablet = 650 mg antacid

4.85 x 10^-3 Eq/g antacid, so 650 mg would have = 3.15 x 10^-3 Eq. anatcid

Cost per equivalent of antacid = 5.99 x 3.15 x 10^-3/100 = 1.9 x 10^-4 $/eq

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