A commercial antacid tablet weighing 1.9564g is dissolved in excess HCl(aq) solu

Question

A commercial antacid tablet weighing 1.9564g is dissolved in excess HCl(aq) solution. The concentration and volume of the HCl solution are .7222M and 25.00mL respectively. The solution was subsequently titrated, and it took 13.87 M NaOH solution to just reach the blue endpoint of thymol blue indicator. Calcutlate the number of millimoles of HCL per tablet the antacid tablet could neutralize. If the antacid tablet cost $0.04, calculate the millimoles aced neutralized per penny. If the acid neutralizer in the tablet is calcium carbonate, CaCO3,, how many milligrams of CaCO3 is in the tablet?

Explanation / Answer

In order to solve this problem you need to realize following two things

a) HCl is added in excess to neutralize all of the antacid tablet and

b) the remaining HCl is being titrated to determine how much excess remains in solution. Therefore, you first need to determine the initial number of moles of acid you used:

Ci = (0.7222 M)x(25.00 mL) = 18.05 mmoles

Now determine the amount of HCl titrated taking into account the following balanced equation:

HCl + NaOH ----> NaCl + H2O

Cxs = (0.2212 M)x(13.87 mL NaOH) x [(1 mmol HCl)/(1 mmol NaOH)] = 3.068 mmoles HCl

Now determine the number of mmoles of HCl used to neutralize all of the antacid tablet:

Canti = Ci - Cxs = 18.05 - 3.068 = 14.98 mmoles

Canti is the answer to the problem. Each tablet can neutralize 14.98 mmoles of HCl; in other words, the neutralization of HCl is given by 14.98 mmoles HCl/tablet, which can be changed via conversions to:

(14.98 mmoles HCl/tablet) x [(1 tablet)/(0.04 $ )] = 374.5 mmoles HCl/$

Finally, the amount of CaCO3 is determined taking into account the following balanced equation:

CaCO3 + 2 HCl ---> CaCl2 + CO2 + H2O

Thus the amount of CaCO3 is:

(0.01498 moles HCl) x [(1 mole CaCO3)/(2 moles HCl)] x [(100.09 g)/(1 mole CaCO3)] = 0.7497 g or 749.7 mg CaCO3

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