Question

A 50.0 mL sample containing Cd 2 + and Mn 2 + was treated with 50.0 mL of 0.0400 M EDTA . Titration of the excess unreacted EDTA required 19.5 mL of 0.0270 M Ca 2 + . The Cd 2 + was displaced from EDTA by the addition of an excess of CN − . Titration of the newly freed EDTA required 17.1 mL of 0.0270 M Ca 2 + . What are the concentrations of Cd 2 + and Mn 2 + in the original solution?

Answer 1

Check the explanation

Step-by-step explanation:

VOLUME OF newly freed EDTA

=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA

=(24.9*0.0220)/ 0.0700

=7.825mL

STRENGTH OF Cd2

=( VOLUME OF newly freed EDTA * STRENGTH OF EDTA) / VOLUME OF SAMPLE

=(7.825*0.0700)/50

=0.0109 M

VOLUME OF excess unreacted EDTA

=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA

=(19.5*0.0270)/ 0.0400

=13.16mL

VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2= (17.1-13.16) mL

=3.94 mL

VOLUME OF EDTA REQUIRED FOR Mn2

= (VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2 - VOLUME OF newly freed EDTA )

=40.02-7.825 mL

=32.20 mL

STRENGTH OF Mn2

=( VOLUME OF EDTA REQUIRED FOR Mn2* STRENGTH OF EDTA) / VOLUME OF SAMPLE

=(32.20*0.0700)/50

=0.045 M