Which of the antacids tested was the best buy and Why? A student carried out the

Question

Which of the antacids tested was the best buy and Why? A student carried out the same experiment using a bottle of "Nick's Famous Antacid Pills". Each bottle cost $1.00 and contained 48 pills. The pill the student tested weighed 1.462g. After adding 25.0 ml of 0.800-MHCI to the tablet, the student titrated the excess HCI with 3.52 ml of 1.02-M NaOH. Find the number of moles of HCI added to the tablet. Find the number of moles of NaOH reacted in the titration. Find the number of moles of excess HCI neutralized by NaOH. Find the number of moles of HCI neutralized by the antacid tablet. Find the number of moles of HCI neutralized per penny of antacid tablet. A student added 8.74g of Mg(OH)_2 to 306-ml of 1.0M HCI. Which reactant was the limiting reactant? What was the pH after the reaction was over?

Explanation / Answer

= 0.800M x (25/1000)L

= 0.02 moles

Excess HCl was titrated with 3.52ml of 1.02M NaOH

Volume of excess HCl-

M1 V1 = M2 V2

0.800M x V1 = 1.02M x 3. 52 ml

V1 = (1.02M x 3. 52 ml) / 0.800M

V1 = 4.49ml

Moles of excess HCl = Molarity of HCl x volume of HCl in L

         = 0.800M x (4.49/1000)L

          = 0.00359 moles

              Number of moles of NaOH reacted in titration = 0.00359 moles

   c. Number of moles of Excess HCl neutralized by NaOH = Excess HCl was titrated with 3.52ml of 1.02M     NaOH

M1 V1 = M2 V2

0.800M x V1 = 1.02M x 3. 52 ml

V1 = (1.02M x 3. 52 ml) / 0.800M

V1 = 4.49ml

Moles of excess HCl = Molarity of HCl x volume of HCl in L

         = 0.800M x (4.49/1000)L

          = 0.00359 moles

d.Number of moles of HCl neutralized by antacid tablet

= Number of moles of Hcl added to tablet - Number of moles of Excess HCl neutralized by NaOH

= 0.02 moles - 0.00359 moles

= 0.01641moles

100 pennies equals one dollar (US).

Cost of 48 antacid tablet = $1.00 = 100 pennies

Cost of 1 tablet = 100/48 pennies = 2.08 pennies

2.08 pennies neutralized 0.01641moles

So 1 penny neutralized 0.01641moles /2.08 = 0.00788 moles

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