One common method for measuring the amount of $\ce{Fe^2+}$ in $\ce{FeSO4}$ is by dissolving it with $\ce{H2SO4}$ and then titrating with $\ce{KMnO4}$.

However, is it possible to measure the amount of $\ce{Mg^2+}$ ion in magnesium sulfate by dissolving it in $\ce{H2SO4}$ and then titrating it with $\ce{KMnO4}$? Why or why not?

  • 1
    $\begingroup$ Do you know about redox reactions? It looks like you'd like to oxidate sth that can't be oxidated... $\endgroup$
    – Mithoron
    Sep 15, 2015 at 0:38

1 Answer 1


The $\ce{Fe^2+}$ concentration can be measured in this fashion since the iron(II) cation can be oxidized by a suitable oxidant, like the permanganate ion $\ce{MnO4-}$.

$$\ce{MnO4- +4H+ +3Fe^2+ -> MnO2 + 2H2O + 3Fe^3+}$$

Since permanganate is purple, and manganese dioxide is an insoluble brown solid, we have a built in indicator for redox titrations like these.

However, magnesium ion content cannot be determined this way since $\ce{Mg^2+}$ is the highest usual oxidation state for magnesium. The electron configuration for magnesium is $1s^2 2s^2 2p^6 3s^2$ or $[\ce{Ne}]3s^2$. The $\ce{Mg^2+}$ ion has the electron configuration of $\ce{[Ne]}$, which is relatively stable. Remove an additional electron from $\ce{Mg^2+}$ to get $\ce{Mg^3+}$, you get an ion with the electron configuration of $1s^2 2s^2 2p^5$, which is the same electron configuration as electron-hungry fluorine atoms, but with a nucleus containing three more protons. $\ce{Mg^2+}$ cannot be chemically oxidized further.

Now, one could use this method to determine the concentration of many other metal ions that can be oxidized, including even $\ce{Mn^2+}$.

$$\ce{3Mn^2+ + 2MnO4- + 2H2O -> 5MnO2 + 4H+}$$

  • $\begingroup$ Thank you so much! It was actually more simple than I thought it would be. Thank you for your very helpful explanation. $\endgroup$
    – Tesla
    Sep 15, 2015 at 13:34

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