Question

1.3188g of antacid is weighed and mixed with 75.00 mL of excess 0.1746 M HCl. The excess wcid required 27.20 mL of 0.09767 M NaOH for back titration. Calculate the amount of CaCO3 in the tablet.​

Answer 1

The amount of CaCO₃ in the tablet : 0.5219 g

Further explanation

Reaction

1. CaCO₃(in antacid) + 2HCl⇒ CaCl₂+H₂CO₃

2. Titration ⇒ HCl+NaOH⇒NaCl+H₂O

• Reaction 2

For titration :

M₁V₁.n₁=M₂V₂n₂(n=acid/base valence⇒NaOH/HCl=1)

mol HCl=mol NaOH = excess HCl

$\tt 27.2\times 0.09767=2.657~mlmol$

• Reaction 1

mol HCl

$\tt 75\times 0.1746=13.095~mlmol$

Moles of HCl used (reacted with CaCO₃) :

initial HCl - excess HCl =

$\tt 13.095-2.657=10.438~mlmol=0.010438~mol$

From reaction 1 :

mol HCl : mol CaCO₃ = 2 : 1

$\tt mol~CaCO_3=\dfrac{1}{2}\times 0.010438=0.005219$

mass of CaCO₃ :

$\tt 0.005219\times 100~g/mol=0.5219~g$