Science, asked by ranveerchhotu76, 4 months ago

of the
the
Determine the
mo Carit
solution
help of
supplied
22
salt solution.​

Answers

Answered by nk989443984564
0

Answer:

Aim:

To determine the strength of a given potassium permanganate solution against standard ferrous ammonium sulfate (Mohr’s salt) solution.

Theory:

Potassium permanganate is a strong oxidant in the presence of sulfuric acid. Mohr salt is a double salt forming a single crystalline structure having the formula (NH4</sub2SO4. FeSO4. 6H2O. The chemical name for Mohr’s salt is ferrous ammonium sulfate

In this titration Mohr salt acts as a reducing agent and potassium permanganate acts as an oxidising agent. So the reaction between Mohr’s salt and potassium permanganate is a redox reaction. In this redox reaction, ferrous ion from Mohr’s salt gets oxidised and pink coloured of manganese present in potassium permanganate which is in the +7 oxidation state gets reduced to colourless Mn2+ state.

The chemical reaction and the molecular chemical equation is given below.

Reduction half reaction –

2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O]

Oxidation half reaction –

2FeSO4(NH4)2SO4.6H2O + H2SO4 + 5[O] → Fe2(SO4)3 + 2(NH4)2SO4 + 13H2O x 5

Overall reaction –

2KMnO4 + 10FeSO4(NH4)2SO4.6H2O+8H2O → K2SO4+ 2MnSO4+ 5Fe2(SO4)3+ 10(NH4)2SO4+ 68H2O

The ionic equation involved in the process is given below.

Oxidation half reaction – [Fe2+ → Fe3+ – e–] x 5

Reduction half reaction – MnO4– + 8H+ + 5e– → Mn2+ + 4H2O

Overall ionic equation – MnO4– + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

This titration is based upon oxidation-reduction titrations. When ferrous ammonium sulfate solution is titrated against potassium permanganate in the presence of acidic medium by sulfuric acid. Acidic medium is necessary in order to prevent precipitation of manganese oxide. Here KMnO4 acts as a self indicator and this titration is called permanganate titration.

Materials Required:

Mohr’s salt (ferrous ammonium sulfate)

Potassium permanganate solution

Dilute sulfuric acid

Chemical balance

Burette

Burette stand

Pipette

Conical flask

Funnel

Measuring flask

Weighing bottle

White tile

Burnet

Wire gauze

Apparatus Setup:

In burette – KMnO4 solution

In Conical flask – 10ml of Ferrous Ammonium Sulfate (Mohr’s salt) + Sulfuric acid

Indicator – Self indicator (KMnO4)

End Point – Colourless to permanent pale pink colour.

Mohr Salt Titration with KMnO4

Procedure:

(a) Preparation of 0.05M standard solution of ferrous ammonium sulfate:

The quantity of Mohr’s salt required for the 250ml of the solution having a normality of 0.05N can be calculated as follows.

The molar mass of mohr’s salt = 392 g/mol

Strength = Normality x Equivalent weight

= (1/20) x 392 = 19.6 g/L

For preparing 250ml of N/20 Mohr’s salt solution, Mohr salt required

= (19.6/1000) x 250 = 4.9 gm

Weigh an empty watch glass using a chemical balance.

Weigh accurately 4.9gm of Mohr’s salt in a chemical balance.

With the help of a funnel transfer the Mohr’s salt into the measuring flask.

Now wash the funnel with distilled water without removing the funnel from the flask.

Make the solution up to the marked point with distilled water and make sure the Mohr’s salt is fully dissolved.

This solution is 0.05N standard solution of Mohr’s salt.

(b) Titration of potassium permanganate solution against standard ferrous ammonium sulfate (Mohr’s salt) solution:

ously swirl the solution in the flask gently.

Initially, the purple colour of KMnO4 is discharged with ferrous ammonium sulfate. The appearance of a permanent pink colour reveals the endpoint.

Repeat the titration until concordant values are obtained.

Note down the upper meniscus on the burette readings.

Record the reading in the observation table given below in order to calculate the molarity of KMnO4 given.

Observations:

S.No Volume of ferrous ammonium sulfate (Mohr’s salt) used Burette Reading Volume(V) of KMnO4 used

V = (y-x)ml

Initial(x) Final(y)  

Calculations:

(a) Normality of KMnO4 solution:

Consider y ml of given KMnO4 solution are equivalent to 20ml of N/10 Mohr’s salt solution.

According to law of equivalents,

N1V1 = N2V2

N1, N2 are normality of Mohr’s salt and KMnO4 solution respectively.

V1, V2 are volume of Mohr’s salt and KMnO4 respectively.

1/10 x 20 = N2 x y

N2 = 2/y

N = Normality of given KMnO4 solution = 2/y

(b) Strength of KMnO4 solution:

Strength = Normality x Equivalent mass

Equivalent mass of KMnO4 =

= 158/5

= 31.6

= 2/y x 31.6 g/liter

Molarity of KMnO4 solution

N = M x Number of electron gained

N = M x 5

M = N/5 moles/ litre

The strength of and molarity of given KMnO4 solution is found out as 2/y x 31.6 g/l and N/5 moles/liter respectively.

Similar questions