Answer:
[Cd²âş] = 5.82x10âťÂł M
[Mn²âş] = 0.037M
Explanation:
The titration of EDTA with Cd²âş, Mn²⺠and Ca²⺠are based in the formation of complexes with a relation 1:1, EDTA: metal, so from the titration of the excess unreacted EDTA with Ca²⺠we can calculate the EDTA moles by:
[tex] \eta_{EDTA}_{e} = \eta_{Ca^{2+}} [/tex]
[tex] \eta_{EDTA}_{e} = 15.9 \cdot 10^{-3} L \cdot 0.0130M = 2.07 \cdot 10^{-4} moles [/tex] Â
The moles of EDTA are:
[tex] \eta_{T} = \eta_{e} + \eta_{r} [/tex] Â Â Â (1)
where [tex]\eta_{T}[/tex]: is the total moles of EDTA, [tex]\eta_{e}[/tex]: is the EDTA excess moles and [tex]\eta_{r}[/tex]: is the EDTA moles that react with Cd²⺠and Mn²âş
Hence the EDTA moles that react with Cd²⺠and Mn²⺠are equal to the moles of both metals:
[tex] \eta_{r} = \eta_{Cd^{2+}} + \eta_{Mn^{2+}} [/tex] Â (2)
From equation (1) we can find the [tex] \eta_{r}[/tex]:
[tex] \eta_{r} = \eta_{T} - \eta_{e} [/tex] Â
[tex] \eta_{r} = 58.4 \cdot 10^{-3}L \cdot 0.0400M - 2.07 \cdot 10^{-4} moles = 2.13 \cdot 10^{-3} moles [/tex] Â
[tex] \eta_{Cd^{2+}} + \eta_{Mn^{2+}} = 2.13 \cdot 10^{-3} moles [/tex] Â Â (3)
Now, from the titration of the EDTA free of Cd²⺠with Ca²âş, we can calculate the moles of EDTA that react with the Cd²⺠and hence the Cd²⺠moles:
[tex] \eta_{EDTA} = \eta_{Ca^{2+}} = 22.4 \cdot 10^{-3}L \cdot 0.0130M = 2.91 \cdot 10^{-4} moles [/tex]
[tex] \eta_{Cd^{2+}} = 2.91 \cdot 10^{-4} moles [/tex]
Having the Cd²⺠moles we can now calculate the Mn²⺠moles using equation (3):
[tex] \eta_{Mn^{2+}} = 2.13 \cdot 10^{-3} moles -\eta_{Cd^{2+}} [/tex]
[tex] \eta_{Mn^{2+}} = 2.13 \cdot 10^{-3} moles - 2.91 \cdot 10^{-4} moles = 1.84 \cdot 10^{-3} moles [/tex]
Now, with the Cd²⺠moles and Mn²⺠moles we can find their concentrations in the original solution:
[tex] [Cd^{2+}] = \frac{2.91 \cdot 10^{-4} moles}{50.0\cdot 10^{-3}L} = 5.82 \cdot 10^{-3} M [/tex]
[tex] [Mn^{2+}] = \frac{1.84 \cdot 10^{-3} moles}{50.0\cdot 10^{-3}L} = 0.037M [/tex] Â
I hope it helps you!
