Question

50.00 mL of 0.125 M HCl solution was added to a 0.500 g sample of an antacid. The back titration with 0.106 M standard NaOH solution consumes 14.75 mL to reach the end-point. Calculate the effectiveness the antacid?

Answer 1

The tablets of antacid contain CaCO₃ that reacts with HCl according to the following equation:

CaCO₃ + 2HCl $\rightarrow$ CaCl₂ + H₂O + CO₂$\uparrow$ .

When an excess of HCl is added, back titration is needed to get the number of the moles of HCl reacted with antacid. This back titration is performed using NaOH:

NaOH + HCl $\rightarrow$ NaCl + H₂O.

As one can see, the excess moles of HCl equals the number of the moles of NaOH used to reach the end-point:

$n(HCl, excess) = n(NaOH)$

$n(HCl, excess) =cV = 0.106\cdot0.01475$

$n(HCl, excess) = 0.0015635$ mol.

Then, the number of the moles that reacted with antacid is the number of the moles of HCl added to antacid minus the excess:

$n(HCl) = n(tot)-n(HCl,excess)$

$n(HCl) = cV - 0.0015635$

$n(HCl) = 0.125\cdot0.050 - 0.0015635 = 0.0046865$ mol.

Using the stoichiometry of the reaction of HCl with CaCO₃, we deduce that the number of the moles of CaCO₃ reacted is two times less than that of HCl:

$n(CaCO_3) = \frac{n(HCl)}{2} = \frac{0.0046865}{2} = 0.002343$ mol, or 4.69 mmol of active ingredient per gram of the sample.

Therefore, the mass of CaCO₃ (its molar mass is 100.09 g/mol) contained in the tablet is:

$m = nM = 0.002343\cdot100.09 = 0.2345$ g, or 235 mg of CaCO₃ per the sample.

Therefore, the mass percent of CaCO3 in antacide sample is:

$\omega = \frac{0.2345}{0.500}\cdot100\% = 46.9\%$ , or 469 mg of active ingredient per gram of the sample.