Question

4. A student analyzed an antacid tablet from a bottle of 48 tablets purchased at a discount store for $1.00. The mass of the tablet was 1.462g. After adding 25.00mL of 0.8000/M HCl solution to the tablet, the student back-titrated the excess HCI with 3.52 mL of 1.019M NaOH solution. (a) Calculate the number of moles of HCI added to the tablet (b) Calculate the number of moles of NaOH required for the back-titration (c) Calculate the number of moles of HCI neutralized by the antacid tablet (d) Calculate the mass effectiveness of the tablet (e) Calculate the cost of one antacid tablet (f) Calculate the cost effectiveness of the tablet

Answer 1

We know antacid contains base. Now to neutralize the base present in an antacid, we have to perform duoble titration.

In the 1st step, we have to add acid in excess to nuetralize all the base present in the antacid tablet.

In the 2nd step, exact amount of another base is added to neutralize the extra acid added in the 1st step.

Now given-

a-

Volume of HCl added = 25 mL

Concentration of HCl solution = 0.8000 M

So mols of HCl added = concentration * volume

= 0.8000 M * 25 mL

= 0.8000 mol/1000 mL * 25 mL

= 0.02 mols

b-Similarly

Volume of NaOH added = 3.52 mL

Concentration of NaOH solution = 1.019 M

So mols of NaOH added = concentration * volume

= 1.019 M * 3.52 mL

= 1.019 mol/1000 mL * 3.52 mL

= 0.00358 mols

C-

That means we have 0.00358 mols of HCl left as extra after neutralization of base in antacid

So mols of HCl used to neutralize antacid = 0.02 mols - 0.00358 mols

= 0.01642 mols

d-

Now an antacid has both OH- and CO32- presnet as base in it. So untill we dont know which is present in what amount, we can't exactly measure the effectiveness.

But we know to neutralize one OH- , we need 1 mole HCl

And to neutralize one CO32- , we need 2 mole HCl

So the antacid which has more OH- will be more mass effective.