## THEORY:

Mohr’s salt is ferrous ammonium sulphate [FeSO4 (NH4)₂SO4. 6H₂O, a double salt] which is more stable than ordinary ferrous sulphate. Hence, it is often used to prepare a primary standard solution of a ferrous salt.

The oxidation half reaction of a ferrous salt is

Fe^{++} – e ———-> Fe^{+++}

1 mol 1 mol

Therefore, Eq. wt. of Mohr’s salt = mol. Wt. / 1 = 391.8 / 1 = 391.8.

By the definition of a normal solution,

1000 ml of (N) Mohr’s salt = 391.8 gm

or, 1000 ml of N/10 Mohr’s salt = 39.18 gm

or, 250 ml of N/10 Mohr’s salt = 9.795 gm

**Standardization of KMnO _{4} against Mohr’s salt:**

It is a redox titration between acidified KMnO_{4} and the ferrous salt

**Formula unit equation :**

2KMnO_{4} + 8H_{2}SO_{4} +10FeSO_{4} ————> K_{2}SO_{4} + 2MnSO_{4} + 8H₂O + 5Fe_{2}(SO4_{4})_{3}

**Ionic equation :**

MnO_{4} ^{– }+8H^{+} + 5e ————–> Mn^{++} + 4H₂O

^{++}– e ———>Fe

^{+++}] 5

Adding, MnO_{4} ^{–}+8H^{+} + 5Fe^{++} ——–>Mn^{++ }+ 4H₂O + 5Fe^{+++}

## APPARATUS:

As in expt 10.2

## CHEMICALS:

No external indicator

## PROCEDURE:

i) Accurately weigh out about 9.8 gm Mohr’s salt on a watch glass.

ii) Transfer it quantitatively into a clean 250 ml measuring flask. Add distilled water till the flask is about half full. Then, add half a test tube of conc. H₂SO_{4} a little at a time swirling the flask after each addition.

iii) Finally, make up the volume up to the mark by the addition of more distilled water and shake it well.

iv) Now, pipette out 10 ml of the Mohr’s salt solution into a clean conical flask, add one test tube full of bench H₂SO_{4}, and titrate it against the KMnO_{4} solution from the burette till a light pink colour persists for a few seconds.

**N.B. Do not heat the solution.**

v) Repeat till you get two concurrent readings.

## RESULTS:

Wt. of Mohr’s salt taken = 9.6 gm

Solution taken in the burette = KMnO_{4} solution

Mohr’s salt solution taken for each titration = 10 ml

Indicator = KMnO_{4} self indicator

End point = from colorless to light pink

BURETTE READINGS:

No. of obs. | Initial reading | Final reading | Vol. of KMnO_{4} | mean |

1 | 10.8 ——> | (approx) | ||

2 | 10.6 | |||

3 | 10.5 | 10.5 | ||

4 | 10.5 |

## CALCULATIONS :

9.795 gm Mohr’s salt in 250 ml makes the solution N/10

therefore, 9.6 gm Mohr’s salt in 250 ml makes the solution 9.6 / 9.795 x N/10 =N / 10 (f = 0.980)

**KMnO _{4} solution Mohr’s salt solution**

V₁ = 10.5 ml V₂ = 10 ml

N₁ = ? N₂ = N/10 (f = 0.980)

By applying the normality equation.

Normality (N₁) = 10x Nx 0.980 / 10.5 × 10 = N/10 (f = 0.980)

therefore, Molarity of the KMnO4 solution is given by

Molarity = Normality / 5 (why?)

= 0.0933 / 5 M = 0.0187 mol/litre.