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Titration of Mohr’s Salt

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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 KMnO4 require 392g Mohr’s Salt.

250 cm3 of M/40 KMnO4 require =392/40g = 4.9g

 Using formula:

N1M1V1 = N2M2V2

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

N2 =1 (for Mohr’s salt), V2 = 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 Biology, Physics, Chemistry, and Physical Education.


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75 Replies to “Titration of Mohr’s Salt”

  32. 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 ….

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  37. I think the observation given is wrong…..so kindly check it once again
    Btw thank u sir for providing it for free …..

    Reply

  42. 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.

    Reply

  53. It helped me a lot. Thanku so much the owner of this page who uploaded these practicals. Thanku very much.

    Reply

  60. THANKS A LOT FROM BOTTOM OF MY HEART FOR GUIDING US♥️♥️♥️♥️♥️♥️♥️♥️♥️😘😘😘😘😘😘

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