Question

A 50.0 mL sample containing Cd2 and Mn2 was treated with 46.0 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 13.1 mL of 0.0230 M Ca2 . The Cd2 was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 19.3 mL of 0.0230 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?

Answer 1

Step 1:

Know data:

50 mL sample.

EDTA: 46.0 mL of 0.0700M EDTA

First Titration: Ca²⁼ 13.1 mL of 0.023 M Ca²

Second titration: 19.3 mL of 0.0230 M Ca²

Step 2:

Calaculate mmol:

mmol EDTA: mL x M= 46.0 mL x 0.0700 M EDTA =3.22 mmol EDTA

mmol Ca² First titration: ml x M Ca²= 13.1 mL x 0.0230 M Ca²⁼ 0.3013 mmol Ca² = 0.3013 mmol Ca²

mmol Ca² second titration: ml x M Ca²= 19.3 mL x 0.0230 M Ca²⁼ 0.3013 mmol Ca^{2 =} 0.4439 mmol Ca²

Step 3:

Calculate NET mmol EDTA:

NET mmol EDTA: mmol EDTA-mmol Ca² first titration:

NET mmol EDTA First titration: 3.22mmol EDTA- 0.3013 mmol Ca² =2.9187 NET mmol EDTA mmol Cd²

NET mmol EDTA second titration: 3.22mmol EDTA- 0.4439 mmol Ca² =2.7761 NET mmol EDTA mmol Mn²

M Cd²= mmol Cd²/ mL sample= 2.9187 mmol Cd²/50 mL= 0.05837 M Ca²

M Mn²= mmol Mn²/ mL sample=2.7761 mmol Mn²/50 mL=0.055 M Mn²