In the titration of $\ce{KMnO4}$ with Mohr's salt, I have learned that sulphuric acid prevents oxidation of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ in the solution before the actual titration. How does the acid hampers oxidation?


This is largely due to the extreme insolubility of iron (III) hydroxide, i.e. $\ce{Fe(OH)3}$.

One can write the oxidation reaction, somewhat simplified, as:

$$\ce{Fe^2+(aq) + 6 H2O(l) <=> Fe(OH)3(s) + 3 H3O+(aq)} + e^-$$

The $K_{sp}$ (solubility product) value for $\ce{Fe(OH)3}$ is very small: $2.79 \times 10^{-39}$ (Wikipedia).

From the equilibrium reaction it's easy to understand that acidic conditions (high $[\ce{H3O+}]$) push the equilibrium to the left, thus preventing the oxidation.

It also explains why Mohr's Salt resists oxidation much better than simple ferrous sulphate, as the ammonium ion provides some acidity.

In alkaline conditions the oxidation of ferrous ions to ferric ions is very fast.

  • $\begingroup$ Yes, I understood you used the Le Chatelier's Principle on the reaction. That answers my questions. But why does extreme insolubility has a role to play in this. $\endgroup$ – Utsav Munendra Feb 4 '18 at 17:42
  • $\begingroup$ Well, precisely because of Le Chatelier. As precipitation removes $\ce{Fe^3+}$, the equilibrium shifts toward more $\ce{Fe(OH)3}$. Lower solubility obviously promotes precipitation. $\endgroup$ – Gert Feb 4 '18 at 19:09
  • $\begingroup$ I'll put a bit of 'mathematical meat' on this answer tomorrow. $\endgroup$ – Gert Feb 4 '18 at 22:33
  • 2
    $\begingroup$ Tomorrow never came... $\endgroup$ – Utsav Munendra Feb 9 '18 at 13:03
  • $\begingroup$ Would you still like me to add that part? $\endgroup$ – Gert Feb 9 '18 at 14:52

The H from the acid reacts with oxygen and prevents the oxygen from reacting with the iron. It's easier for the O to bind with H than with Fe.

OK Gert - I deleted "in layman's terms." Of course this is a chemistry site- don't know why I put that in.

  • $\begingroup$ This a site about chemistry: 'layman's terms' aren't needed here. $\endgroup$ – Gert Feb 4 '18 at 22:24

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