2g of a limestone sample containing CaCO3 as a basic substance was dissolved in 50 cm of 1.0 HCl.the resulting solution was made up of to 250 cm with water 25cm of this solution required 11.6cm of 1.0M Noah for neutralisation.calculate the CaCO3 present in the limestone sample the ratio of HCl to CaCO3 from the balanced reaction is 2:1

This question is an example of a back titration. CaCO₃ is dissolved in excess HCl. Any HCl that does not react with CaCO₃ reacts with NaOH when during the titration.

Note

There are 2 reactions taking place:

=> When CaCO₃ reacts with HCl and when HCl reacts with NaOH.

1. CaCO_{3 (s)} + 2 HCl_(aq)==> CaCl_{2 (aq)} + CO_{2 (g)} + H₂O _(l)
2. HCl _(aq)+ NaOH _(aq) ==> NaCl_(aq) + H₂O_(l)

Step 1: calculate the original number of moles of HCl

Volume of HCl = 50 ml = 0.05 L

Concentration of HCl = 1.0 M = 1.0 mol/L

Moles = Concentration x Volume = 0.05 L x 1.0 mol/L = 0.05 mol

Step 2: Calculate moles of HCl that reacted with NaOH

Given:

Concentration of NaOH = 1.0 M = 1.0 mol/L

Volume of NaOH = 11.6 ml = 0.0116 L

Moles of NaOH = 0.0116 L x 1.0 mol/L = 0.0116 mol

The equation for the reaction between NaOH and HCl is:

HCl _(aq)+ NaOH _(aq) ==> NaCl_(aq) + H₂O_(l)

From the equation, mol ratio of HCl to NaOH is 1:1.

Thus:

Moles of HCl = Moles of NaOH = 0.0116 mol

Step 3: Calculate moles of HCl that reacted with CaCO₃

Moles of HCl that reacted with CaCO₃ = Initial moles of HCl - Moles of HCl that reacted with NaOH

= 0.05 mol - 0.0116 mol = 0.0384 mol

Step 4: Calculate moles of CaCO₃ present in reaction when CaCO₃ is dissolved in HCl

From the equation:

CaCO_{3 (s)} + 2 HCl_(aq)==> CaCl_{2 (aq)} + CO_{2 (g)} + H₂O _(l)

Mol ratio of HCl to CaCO₃ is 2:1

Thus:

Moles of CaCO₃ = 1/2 x Moles of HCl = 1/2 x 0.0384 mol = 0.0192 mol

Step 5: Calculate mass of CaCO₃

_?Moles of CaCO₃ that reacted with HCl in the initial reaction is 0.0192 mol.

Mass of CaCO₃ = Moles of CaCO₃ x Molar mass of CaCO₃ = 0.0192 mol x 100 g/mol = 1.92 g

Conclusion:

Mass of CaCO₃ present in the sample of limestone is 1.92 g

% mass of CaCO₃ = Mass of CaCO₃/Mass of limestone x 100 = 1.92 g/2 g x 100 = 96%