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:
-
- Weigh a clean dry bottle using a chemical balance.
- Add 4.9g more weights to the pan containing the weights for the weighing bottle.
- Add Mohr’s salt in small amounts to the weighing bottle, so that the pans are balanced.
- Remove the weighing bottle from the pan.
- Using a funnel, transfer the Mohr’s salt to the volumetric flask.
- Add about 5ml. of dilute H2SO4 to the flask followed by distilled water and dissolve the Mohr’s salt.
- Make up the volume to the required level using distilled water.
- The standard solution is prepared.
(b) THEORY–
- The reaction between KMnO4 and Mohr’s salt is a redox reaction and the titration is therefore called a redox titration.
- Mohr’s salt is the reducing agent and KMnO4 is the oxidizing agent.
- KMnO4 acts as an oxidizing agent in all the mediums; i.e. acidic, basic and neutral medium.
- 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.
Truly helps the last movementers …..Like me..!!
and is very handy and cool
thank u it was very helpful in exam time
thank you very much for helping to me
“moment-ers”
really helps at the last moment :]
really, it does help the procrastinators like us . thanks a lot :]
Thanks!!
It is really helpful!!
Hi
(Redacted by Admin)
It’s ok
i also believe you for it
Too good
Ya it’s good!
j
Kya good
i think you that is correct
Thank you for fill the observation table that makes me easy to complete my work
Thanks it’s very useful
Thanks
Thanks 😊😊😊It is really helpful
Good for last minute preparation
Really helpful and handy for last minute preparations☺☺☺
Thnxx , it was very helpful 😎😃
Thanks very much!!!
Thanks a lot very helpful
its really a helping hand..
Great.
Really helpful
Not good
it is veru good miss
Should explain calculations
Thanx for starting things so clearly ..very helpful ! 🙂
yuj
THANKS OR THE WRITTEN PART BUT THE STRENGTH RANGE MUST IN BETWEEN 1.55 TO 1.6
Helpful that.
Can i wright this on my practical copy.
Its true or not
No use any diagram on this topic
why
Ye M/40 KA HAI YA M/20 KA
Proved very helpful to me..
Accha ji
Really helpful..thnx a lot..
REALLY HELPFUL.
Very nice app and much useful for the 12th standard
Class12 every practical is given and good notes
☺☺☺
Really hepful thanks for guidance .
Confusing
Next time at least give precautions and sources of errors please
Helped me a lot….
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 ….
Its nice.very helpful.
it really good and it is very helpful for me.
It helps a lot…Thank u so much
Very very useful… Great work 😇👍
I think the observation given is wrong…..so kindly check it once again
Btw thank u sir for providing it for free …..
Why we use volume of solution 100ml
Very much thnx NYC
How the initial value is 10
The reading starts at 0 only know😯
Thanks for helping!! it is really really very useful to me
Thanks a lot😊😊
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.
It is appreciable that all the readings r correct 🐤🐤🐼🐼🐼🐼🐼🐼
I want m/30
Not 20
Bindass.,……..😀
I like it thanks for this information it realy helps alot
Thankyou, it was very helpful for my practical examination
It really haepl me a lot please read it
In double titration known &unknown dono k sath titration hota h kya ..?
I was very helpful to complete my practical before submission
It really helps me a lot.
Ya it’s helpful nd gud tooo
It helped me a lot. Thanku so much the owner of this page who uploaded these practicals. Thanku very much.
Most helpful🥰
Thank u so much for this page it is very helpful for us
ye bahut hi lavdayak….👍✌️🥰
Thankyou 😀
Thanku you very much 👍 👍
Lots of THANKS.
IT’S VERY HELPFUL AND EASY IN LANGUAGE.
THANKS A LOT FROM BOTTOM OF MY HEART FOR GUIDING US♥️♥️♥️♥️♥️♥️♥️♥️♥️😘😘😘😘😘😘