I did a titration today for the first time.

A solution of Potassium Permanganate of unknown molarity was titrated against a solution of ammonium iron II sulphate of known molarity.

I have to say, seeing the purple solution turn clear was a lot of fun.

Dilute sulphuric acid solution was used twice during this experiment. First to prepare the ammonium iron II sulphate and sulphuric acid solution, and then 20 ml was added to each sample before titration.

Why did these have to be added ?

Also, could I have used pure sulphuric acid ? Why did it have to be diluted ?

I think the first usage provided the H+ for the reaction to occur but I am not sure. Would pure sulphuric acid not be more efficient ?


2 Answers 2


A little bit of sulphuric acid is used when preparing iron(II) solution. This acid prevents the possible hydrolysis of $\ce{Fe^{2+}}$ ion according to : $$\ce{Fe^{2+} + 2 H_2O <=> [Fe(OH)_2(s) + 2 H^+}$$ In case this reaction occurs, the added sulphuric acid will react and destroy the precipitate $\ce{Fe(OH)_2}$ according to the inverse reaction : $$\ce{Fe(OH)_2 + 2 H^+ -> Fe^{2+} + H_2O}$$

Now, during the titration, the ion $\ce{H^+}$ is needed to carry out the reaction of permanganate on the ion $\ce{Fe^{2+}}$ . Hopefully you know this equation.

And concentrated sulphuric acid is not wanted, because it reacts strongly with water just during its dissolution. It produces a lot of heat.

  • $\begingroup$ Why instead of adding the acid twice can I not just add it once ? Add to prepare and then add some more before titrating. Why not just add the whole lot when preparing ? $\endgroup$
    – Kantura
    Aug 31, 2020 at 17:10
  • 3
    $\begingroup$ @ Kantura. The risk of precipitating $\ce{Fe(OH)_2}$ in the first phase is small. The solution must be slightly acidified. No need to add more. And the acidified solution may be kept many days in closed bottles. But a huge amount of $\ce{H^+}$ ions are needed during the titration with permanganate. The equation shows that 8 $\ce{H^+}$ ions are needed for each permanganate ion to react. Indeed the equation is $$\ce{MnO_4^- + 5 Fe^{2+} + 8 H^+ -> Mn^{2+} + 5 Fe^{3+} + 4 H2O}$$ I repeat : $8$ $\ce{ H^+}$ are needed for one permanganate ion. $\endgroup$
    – Maurice
    Aug 31, 2020 at 18:39

Ammonium iron(II) sulfate or Mohr's salt, is the inorganic compound with the formula $\ce{(NH4)2Fe(SO4)2(H2O)6}$. In analytical chemistry, Mohr's salt is the preferred source of ferrous ions because the solid salt has a long shelf life and the solid crystals resist to oxidation by air: The presence of $\ce{NH4+}$ ion prevents the oxidation reaction of $\ce{Fe^2+ -> Fe^3+ + e-}$, which slows down in lower $\mathrm{pH}$.

This stability of Mohr's salt extends somewhat to the solution phase reflecting the effect of aforementioned $\mathrm{pH}$ on the $\ce{Fe^2+/Fe^3+}$ redox couple. The ammonium ions make solutions of Mohr's salt slightly acidic, which slows this oxidation process (the oxidation occurs more readily at high $\mathrm{pH}$). Thus, during titrations, sulfuric acid is commonly added to solutions of Mohr's salt to reduce air oxidation to ferric iron, thus, the oxidation would take place only in the presence of permanganate solution.

The two half reactions are:

$$\ce{MnO4– + 8H+ + 5e– -> Mn^2+ + 4H2O} \tag1$$ $$\ce{Fe^2+ -> Fe^3+ + e-} \tag2$$ Equations $\left((1) + 5 \times (2) \right)$ would give the net equation: $$\ce{MnO4– + 8H+ + 5Fe^2+ -> Mn^2+ + 5Fe^3+ + 4H2O} \tag3$$

As evidence in the equations, $\ce{KMnO4}$ solution need acid to function smoothly as oxident while Mohr's salt need acid to stay un-oxidized until it meets $\ce{KMnO4}$. The usual acid used in these situation is dilute $\ce{H2SO4}$, which did not interfere with the redox reaction. Keep in mind that dilute acid is usually $\pu{6 M}$, which is good enough for the purpose.


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