an antacid tablet weighs 1.200 g. The active ingredient is CaCO3. This tablet wa

Question

an antacid tablet weighs 1.200 g. The active ingredient is CaCO3. This tablet was dissolved in 25.0 mL of 1.0 M HCl and back-titrated with 6.8 mL of 1.0 M NaOH. How much active ingredient is in this tablet? an antacid tablet weighs 1.200 g. The active ingredient is CaCO3. This tablet was dissolved in 25.0 mL of 1.0 M HCl and back-titrated with 6.8 mL of 1.0 M NaOH. How much active ingredient is in this tablet? an antacid tablet weighs 1.200 g. The active ingredient is CaCO3. This tablet was dissolved in 25.0 mL of 1.0 M HCl and back-titrated with 6.8 mL of 1.0 M NaOH. How much active ingredient is in this tablet?

Explanation / Answer

When CaCO3 dissolved in HCl, some of the HCl reacts with CaCO3 and excess unreacted HCl is back titrated with NaOH. So, substract the amount of HCl back titrated from the total amount of HCl. It gives the amount of HCl reacts with CaCO3.

HCl + NaOH   -----> H2O + NaCl

Mole ratio of HCl and NaOH is 1:1.

Millimoles of NaOH added = volume NaOH x Molarity of NaOH = 6.8 mL x 1.0 M = 6.8 mmol

Millimoles of HCl back titrated = 6.8 mmol NaOH x ( 1 mol HCl/1 mol NaOH) = 6.8 mmol HCl

Millimoles of total HCl = 25.0 mL x 1.0 M = 25.0 mmol

Moles of HCl reacted with CaCO3 = 25.0 mmol - 6.8 mmol = 18.2 mmol = 18.2 x 10-3 mol = 0.0182 mol

CaCO3 + 2 HCl ------> H2CO3 +CaCl2

Mole ratio of CaCO3 and HCl is 1:2.

Moles of CaCO3 = 0.0182 mol HCl x ( 1 mol CaCO3 / 2 mol HCl) = 0.00910 mol CaCO3

Mass of CaCO3 = moles of CaCO3 x molar mass of CaCO3 = 0.00910 mol x (100 g/mol) = 0.910 g

So, the mass of active ingredient is 0.910 g.

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