Question

1)

a) Prepare an EDTA solution by dissolving 0.1 g of magnesium chloride hexahydrate and 3.2g of EDTA in 1 L of water. Also prepare a standard calcium carbonate solution by weighing out 0.4 g of pure calcium carbonate and adding 500 mL of water and a few drops of 1:1 HCl:Water solution. Take 50mL aliquots of the standard calcium carbonate solution, add 10 mL of ammonia/ ammonia chloride buffer, and 5 drops of calmagite indicator, and titrate this with the EDTA solution. A color change occured when 47.8 mL, 48.2, and 49.0 mL of EDTA was added/ titrated to each aliquot. Determine the exact concentration of EDTA.

b) Weigh 0.18g of acid-soluble limestone and add 20 mL of 1:1 HCl:Water solution. Heat this solution to near dryness, add 10mL of 1:10 HCl: Water solution, and dilute to 100mL with water. Take a 50mL aliquot, add 5 mL of 10% hydroxylamine hydrochloride solution, 10 mL of ammonia/ ammonium chloride buffer and 5 mL of 5% potassium cyanide. Then add a few drops of calmagite indicator, and titrate with the same EDTA solution in part a. A color change occured when 50.0 mL, 48.7 mL, 49.5 mL, 49.2 mL of EDTA was added to each of these aliquots. Find the percentage of calcium carbonate in the unknown limestone sample.

Answer 1

1a.) 1^st titration::

CaCo₃   EDTA

molarity: m₁ = 0.002 m₂ = ?

volume: v₁ = 50 ml v₂ = 47.8 ml

no:of moles: n₁ = 1 n₂ = 1

   m₂ = (v₁*m₁*n₂) / (n₁*v₂)

molarity of EDTA = 0.002 M.

2^nd titration:

CaCo₃   EDTA

  molarity: m₁ = 0.002 m₂ = ?

volume: v₁ = 50 ml v₂ = 48.2 ml

no:of moles: n₁ = 1 n₂ = 1

   m₂ = (v₁*m₁*n₂) / (n₁*v₂)

molarity of EDTA = 0.002 M.

3^rd titration:

CaCo₃   EDTA

  molarity: m₁ = 0.002 m₂ = ?

volume: v₁ = 50 ml v₂ = 49.0 ml

no:of moles: n₁ = 1 n₂ = 1

   m₂ = (v₁*m₁*n₂) / (n₁*v₂)

molarity of EDTA = 0.002 M.

The exact concentration of EDTA = 0.002M.

1b.) 1^st titration:

CaCo3 EDTA

molarity: m₁ = ? M m₂ = 0.002M

volume: v₁ = 50 ml v₂ = 48.7 ml

no:of moles: n₁ = 1 n₂ = 1

m₁ = (v₂*m₂*n₁) / (n₂*v₁)

   m₁ = 0.0019 M .

2^nd titration::

CaCo₃ EDTA

  molarity: m₁ = ?M m₂ = 0.002M

volume: v₁ = 50 ml v₂ = 49.5 ml

no:of moles: n₁ = 1 n₂ = 1

   m₁ = (v₂*m₂*n₁) / (n₂*v₁)

molarity of EDTA = 0.0019M.

3^rd titration::

CaCo₃   EDTA

  molarity: m₁ = ?M m₂ = 0.002M

volume: v₁ = 50 ml v₂ = 49.2 ml

no:of moles: n₁ = 1 n₂ = 1

     m₁ = (v₂*m₂*n₁) / (n₂*v₁)

molarity of EDTA = 0.0019M

percentage of limestone in unknown sample is:: 0.0019 gms present in 1000 ml

1000 ml of limestone contains weight of pure limestone = 0.0019 gms

500 ml of limestone contains weight of pure limestone = (0.0019*500) / 1000 = 0.00095 gms

In the problem, it is given that 0.4gms of limestone sample are dissolved in 500 ml of the solution i.,e 0.4gms of the given sample contains the weight of pure limestone = 0.00095gms

100 gms of the sample contains the weight of pure limestone = (0.00095*100) / 0.4  

= 0.2375% of pure limestone.

Therefore 0.4 gms of the given sample contains the percentage of pure limestone = 0.2375%