Chemistry?

2016-10-18 8:52 am

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The amount of carbonate (CO32-) present in a sample of limestone may be determined by firstly, reacting the carbonate with an excess of strong acid. In this reaction CO2 and H2O are the products. The excess acid is then determined by titration with a strong base.

In this example a sample of limestone is dissolved by adding 60 mL of 1.025 M HCl and the solution is transferred to a 250 mL volumetric flask and made up to the mark using distilled water. Aliquots of 10mL of this solution are then titrated against a standard solution of 0.0998 M NaOH and the average titre is 19.65 cm3.
Calculate the molarity of CaCO3 in the 250 mL volumetric flask.

Answer : 2.5 x 10-2 M

回答 (1)

micatkie

2016-10-18 9:29 am

✔ 最佳答案

In the preparation of the solution of CaCO₃ solution, HCl is in excess.

Consider of titration of the 10 mL aliquot against standard NaOH :
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
OR: Mole ratio HCl : NaOH = 1 : 1

No. of moles of NaOH = (0.0998 mol/L) × (19.65/1000 L) = 1.96 × 10⁻³ mol
No. of moles of HCl in the 10 mL aliquot = 1.96 × 10⁻³ mol
No. of moles of unreacted HCl in 250 mL of the solution = (1.96 × 10⁻³ mol) × (250/10) = 4.90 × 10⁻² mol

Consider the preparation of the 250 mL of CaCO₃ solution :
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
OR: CaCO₃(s) : HCl = 1 : 2

Total no. of moles of HCl added = (1.025 mol/L) × (60/1000) = 6.15 ×10⁻² mol
No. of moles of HCl reacted with CaCO₃ = [(6.15 ×10⁻²) - (4.90 × 10⁻²)] mol = 1.25 × 10⁻² mol
No. oof moles of CaCO₃ = (1.25 × 10⁻² mol) × (1/2) = 6.25 × 10⁻³ mol

Molarity of CaCO₃ in the solution = (6.25 × 10⁻³ mol) / (250/1000 L) = 2.5 × 10⁻² M

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