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* 0.09767=2.657~mlmol`

• Reaction 1

mol HCl

`\tt 75* 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=(1)/(2)* 0.010438=0.005219`

mass of CaCO₃ :

`\tt 0.005219* 100~g/mol=0.5219~g`