Answer:
the concentration of [tex]Cd^{2+}[/tex] in the original solution= 0.0088 M
the concentration of [tex]Mn^{2+}[/tex] in the original solution = 0.058 M
Explanation:
Given that:
The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &
was treated with 64.0 mL of 0.0600 M EDTA
Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+
i.e the strength of the Ca2+ = 0.0310 M
Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+
To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :
Volume of newly freed EDTA = [tex]\frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}[/tex]
= [tex]\frac{14.2*0.0310}{0.0600}[/tex]
= 7.3367 mL
concentration of [tex]Cd^{2+}[/tex] = [tex]\frac{volume \ of \ newly \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}[/tex]
= [tex]\frac{7.3367*0.0600}{50}[/tex]
= 0.0088 M
Thus the concentration of [tex]Cd^{2+}[/tex] in the original solution = 0.0088 M
Volume of excess unreacted EDTA = [tex]\frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}[/tex]
= [tex]\frac{16.1*0.0310}{0.0600}[/tex]
= 8.318 mL
Volume of EDTA required for sample containing [tex]Cd^{2+}[/tex] and [tex]Mn^{2+}[/tex] = (64.0 - 8.318) mL
= 55.682 mL
Volume of EDTA required for [tex]Mn^{2+}[/tex] = Volume of EDTA required for
sample containing [tex]Cd^{2+}[/tex] and
[tex]Mn^{2+}[/tex] -- Volume of newly freed EDTA
Volume of EDTA required for [tex]Mn^{2+}[/tex] = 55.682 - 7.3367
= 48.3453 mL
Concentration of [tex]Mn^{2+}[/tex] = [tex]\frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}[/tex]
Concentration of [tex]Mn^{2+}[/tex] = [tex]\frac{48.3453*0.0600}{50}[/tex]
Concentration of [tex]Mn^{2+}[/tex] in the original solution= 0.058 M
Thus the concentration of [tex]Mn^{2+}[/tex] = 0.058 M
