Question

A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden reductor, reducing Cu2+ to Cu+ . The resulting Cu+ solution required 40.15 mL of each of the titrants to reach the endpoint. Calculate the concentration of each titrant.

Answer 1

Concentration of
`Cr_2O_7^(2-)` = 0.03101 M

Concentration of
`MnO_4^-` = 0.03721 M

Step-by-step explanation:

A)

The reduction for
`Cr_2O_7^(2-)` is;

`Cr_2O_7^(2-) + 14 H ^+ _((aq)) + 6 e^- -----> 2 Cr^(3+) _((aq))+7H_2O _((l))`

`Cu^+_((aq)) -----> Cu^(2+) _((aq)) + 1 e^-`

6 moles of
`Cu ^+` = 1 mole of
`Cr_2O_7^(2-)`

number of moles of Cu reacted =
`(mass \ of \ Cu \ wire )/( molecular weigh tof \ Cu wire )`

number of moles of Cu reacted =
`(0.4749)/(63.55)`

number of moles of Cu reacted = 0.00747 mole

number of moles of
`Cr_2O_7^(2-)`reacted =
`(0.00747)/(6)`

number of moles of
`Cr_2O_7^(2-)`reacted = 0.001245 mole

Concentration of
`Cr_2O_7^(2-)` =
`(number \ of moles )/(Volume)`

Given that the volume = 40.15 mL =
`40.15 *10^(-3)`; we have:

Concentration of
`Cr_2O_7^(2-)` =
`(0.001245)/(40.15*10^(-3))`

Concentration of
`Cr_2O_7^(2-)` = 0.03101 M

B)

The reduction for
`MnO_4^-` is;

`MnO_4^- + 8H^+ + 5 e^- -----> Mn^(2+) + 4H_2O`

`Cu^+_((aq)) -----> Cu^(2+) _((aq)) + 1 e^-`

5 moles of
`Cu ^+` = 1 mole of
`Cr_2O_7^(2-)`

number of moles of Cu reacted =
`(mass \ of \ Cu \ wire )/( molecular weigh tof \ Cu wire )`

number of moles of Cu reacted =
`(0.4749)/(63.55)`

number of moles of Cu reacted = 0.00747 mole

number of moles of
`MnO_4^-` reacted =
`(0.00747)/(5)`

number of moles of
`MnO_4^-` reacted = 0.001494 mole

Concentration of
`MnO_4^-` =
`(number \ of moles )/(Volume)`

Given that the volume = 40.15 mL =
`40.15 *10^(-3)`; we have:

Concentration of
`MnO_4^-` =
`(0.001494 )/(40.15*10^(-3))`

Concentration of
`MnO_4^-` = 0.03721 M