# Titration of Mohr’s Salt

AIM –

(a) To prepare 250ml of M/20 solution of Mohr’s salt.
(b) Using this calculate the molarity and strength of the given KMnO4 solution.

APPARATUS AND CHEMICALS REQUIRED

Mohr’s salt, weighing bottle, weight box, volumetric flask, funnel, distilled water, chemical balance, dilute H2SO4, beakers, conical flask, funnel, burette, pipette, clamp stand, tile, KMnO4 solution.

(a) THEORY

Mohr’s salt having the formula FeSO4.(NH4)2SO4.6H2O has molar mass 392gmol-1. It is a primary standard.

It’s equivalent mass is 392/1 = 392 as its n factor is 1 as per the following reaction:

Fe2+ →  Fe3+ + e

PROCEDURE:

1. Weigh a clean dry bottle using a chemical balance.
2. Add 4.9g more weights to the pan containing the weights for the weighing bottle.
3. Add Mohr’s salt in small amounts to the weighing bottle, so that the pans are balanced.
4. Remove the weighing bottle from the pan.
5. Using a funnel, transfer the Mohr’s salt to the volumetric flask.
6. Add about 5ml. of dilute H2SO4 to the flask followed by distilled water and dissolve the Mohr’s salt.
7. Make up the volume to the required level using distilled water.
8. The standard solution is prepared.

(b) THEORY

1. The reaction between KMnO4 and Mohr’s salt is a redox reaction and the titration is therefore called a redox titration.
2. Mohr’s salt is the reducing agent and KMnO4 is the oxidizing agent.
3. KMnO4 acts as an oxidizing agent in all the mediums; i.e. acidic, basic and neutral medium.
4. KMnO4 acts as the strongest oxidizing agent in the acidic medium and therefore diluted H2SO4 is added to the conical flask before starting the titration.

IONIC EQUATIONS INVOLVED:

Reduction Half:    MnO4 + 8H+ + 5e  →  Mn2+ + 4H2O

Oxidation Half:     5Fe2+  →  5Fe3+ + 5e

Overall Equation:   MnO4 + 8H+ + 5Fe2+  → Mn2+ + 5Fe3+ + 4H2O

INDICATOR– KMnO4 acts as a self-indicator.

END POINT– Colourless to light pink (KMnO4 in the burette)

PROCEDURE

1.       Fill the burette with KMnO4 solution.

2.        Pipette out 10ml. of Mohr’s salt solution into the conical flask.

3.       Add half a test tube of diluted H2SO4.

4.       Keep a glazed tile under the burette and place the conical flask on it.

5.       Note down the initial reading of the burette.

6.       Run down the KMnO4 solution into the conical flask drop wise with shaking.

7.       Stop the titration when a permanent pink color is obtained in the solution.

8.       This is the end point. Note down the final burette reading.

9.       Repeat the experiment until three concordant values are obtained.

OBSERVATION TABLE: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL)

Volume of Mohr’s salt solution taken =

 S. No. Burette Readings Volume of KMNnO4 Initial Final Used (mL) 1 10 18.8 8.8 2 18.8 27.7 8.9 3 27.7 36.5 8.8

Concordant Value = 8.8mL

CALCULATIONS: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL)

Calculation of amount of Mohr’s Salt to be weighed to prepare 100ml M/20 solution:

Molecular Mass of Mohr’s Salt = 392g/mole

1000 cm3 of 1M KMnOrequire 392g Mohr’s Salt.

250 cmof M/40 KMnO4 require =392/40g = 4.9g

Using formula:

N1M1V1 = N2M2V2

Where N1=5 (for KMnO4), V1=8.8mL, M1 =?

N=1 (for Mohr’s salt), V= 10ml, M2 = 1/20M

M1 = [1*(1/20)*10]/[5*8.8] = 1/88M = 0.01M

Strength = M X Molar Mass = 158 *( 1/88) = 1.79g/L

RESULT- (ON RULED SIDE)

The Molarity of KMnO4 =  0.01M

And the strength of KMnO4 =  1.79g/L

You can also get Class XII Practicals on BiologyPhysicsChemistry, and Physical Education.

## 61 Replies to “Titration of Mohr’s Salt”

1. Nishant david charles says:

Truly helps the last movementers …..Like me..!!
and is very handy and cool

thank u it was very helpful in exam time

2. rajida begum says:

thank you very much for helping to me

3. Dt says:

“moment-ers”

2. Arunima Hui says:

Thanks!!

1. Kuldeep Dudi says:

It’s ok

2. rajida begum says:

i also believe you for it

3. Anjali says:

Ya it’s good!

1. rajida begum says:

i think you that is correct

4. Gaurav Singh Oberoi says:

Thank you for fill the observation table that makes me easy to complete my work

5. Gaurav says:

Thanks it’s very useful

6. GARIMA says:

7. Roushan says:

Good for last minute preparation

8. Roushan says:

Really helpful and handy for last minute preparations☺☺☺

9. Devanand Manu says:

Thanks very much!!!

10. faizan says:

its really a helping hand..

12. Ushnika kar says:

Great.

1. naman says:

it is veru good miss

Should explain calculations

14. Anonymous says:

Thanx for starting things so clearly ..very helpful ! 🙂

15. Harshit suneja says:

THANKS OR THE WRITTEN PART BUT THE STRENGTH RANGE MUST IN BETWEEN 1.55 TO 1.6

16. Rahul says:

Can i wright this on my practical copy.
Its true or not

17. Rahul says:

No use any diagram on this topic

1. rajida begum says:

why

18. Pulkit says:

Ye M/40 KA HAI YA M/20 KA

19. Archit sharma says:

20. Archit sharma says:

21. nishanth says:

22. AMAN KUMAR CHOUDHARY says:

Very nice app and much useful for the 12th standard

23. AMAN KUMAR CHOUDHARY says:

Class12 every practical is given and good notes

☺☺☺

24. Khushbu says:

Really hepful thanks for guidance .

25. Anubhav Singh says:

Next time at least give precautions and sources of errors please

26. Rachit patel says:

Helped me a lot….

27. awemaze says:

Thank you sooo much….. You r an angel that helped me with my practicals…. Lots of love from my side …. And thank you sooooooo much ….

28. Prateek Vishwakarma says:

29. Priyansh says:

it really good and it is very helpful for me.

30. Avni Gupta says:

It helps a lot…Thank u so much

31. Ayush says:

Very very useful… Great work 😇👍

32. Divyansh Barnwal says:

I think the observation given is wrong…..so kindly check it once again

33. Rashi srivastava says:

Why we use volume of solution 100ml

34. Ayush nagle says:

Very much thnx NYC

35. Rithesh says:

How the initial value is 10
The reading starts at 0 only know😯

36. heena says:

Thanks for helping!! it is really really very useful to me
Thanks a lot😊😊

37. Ani says:

Thank u but I have in trouble for finding the concentration and molarity of unknown sol of KMnO4 then what will we do for 15 students is there changing the concentration of KMnO4 by adding water. For forming of three groups.

38. Veda prakash says:

It is appreciable that all the readings r correct 🐤🐤🐼🐼🐼🐼🐼🐼

39. Astik chaturvedi says:

I want m/30
Not 20

40. Abhishek jha says:

Bindass.,……..😀

41. Mitali gautam says:

I like it thanks for this information it realy helps alot

42. Debasis maharana says:

Thankyou, it was very helpful for my practical examination

43. Divyansh rungta says: