Hey @ChemEasy_Eko nice work but you forget to note that if anyone is using Feso4. 7H2O (water of crystallization) then the molar mass to be used is 278gmol-1. Then the end product will 9.9 or 10 cm3. But yours gave 18.10cm3 simple because you used only the molar mass of FeSO4 or maybe your salt is not having water of crystallization. 👍🏾👍🏾👍🏾
Thank you very much ChemEasy, God bless you. Please I'd like to ask, is there a difference between chemistry practical alternative A and chemistry practical alternative B? Thank you once again, very grateful for your page
1. Chemistry Alternative A is a practical examination where students perform actual laboratory experiments. This practical component typically involves using laboratory equipment and chemicals to carry out specific tasks outlined in the exam. While: 2. Chemistry Alternative B is designed for schools or candidates that may not have access to fully equipped laboratory facilities. Instead of a hands-on practical exam, Alternative B features a "practical paper" that is theoretically based. This paper often includes alternative-to-practical questions where students are tested on their understanding of practical concepts without having to perform actual experiments.
Yes. But the uncertainty in the measurements will be high, that's why I recommended using a volumetric flask of 1000cm³ capacity or 1000cm³ measuring cylinder
Wow .. This is so educating Thank you very much sir for teaching I had a tough time preparing this but with this lecture am enlightened God bless you sir
It could be your reagents. If it gives that value after multiple trials then try preparing the reagents fresh or get another one. Or at most go with it
You always give me joy and confident , you are a blessing to your generation. I wish other science educators are like you. I was just waiting for your analysis on this and finally here it is. Pls analyse that of qualitative
The salt you are using is hydrated with 7mole of water . The reduces the quantity of KMnO4 needed. With the hydrated salt the calculated end point is 9.90 . But since the yellow paper didn't talk about the water of crystallisation hence the need for using anhydrous salt
You can use a measuring cylinder. The uncertainty in the measurements is lower when you use a 1000cm³ volumetric flask (about ±0.30) that's why I ll recommend you use a volumetric flask. But if you don't have it you can use the ones you have.
i appreciate your effort and value you bring to the table year in year out. I just have one question does water of crystallization affect redox titration?
Hydrated compounds can behave differently from their anhydrous counterparts in terms of stability and reactivity. Hydrated compounds may dehydrate over time, especially if stored improperly, changing their characteristics. This change can affect their reducing power, altering the effectiveness in redox titrations. Try and get the anhydrous one
1. HCl (0.1M HCl in 500cm3) C1V1 = C2V2 C1 = 12M V1= ? C2 =0.1M V2 = 500cm³ 12M x V1 = O.1M x 500cm³ V1 = 4.2cm³ (Measure 4.2cm³ of HCl and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water) 2. H2SO4 (0.1M H2SO4 in 500cm3) C1V1 = C2V2 C1 = 18.4M V1= ? C2 =0.1M V2 = 500cm³ 18.4M x V1 = O.1M x 500cm³ V1 = 2.7cm³ (Measure 2.7cm³ of H2SO4 and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water) 3. HNO3 (0.1M HNO3 in 500cm3) C1V1 = C2V2 C1 = 15.9M V1= ? C2 =0.1M V2 = 500cm³ 15.9M x V1 = O.1M x 500cm³ V1 = 3.1cm³ (Measure 3.1cm³ of HNO3 and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water) 4. Ammonia NH3 (0.1M of 35% NH3 in 500cm3) C1V1 = C2V2 C1 = 18.1M V1= ? C2 =0.1M V2 = 500cm³ 18.1M x V1 = O.1M x 500cm³ V1 = 2.8cm³ (Measure 2.8cm³ of NH3 and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water) 5. NaOH (0.1M NaOH in 500cm³) Weigh 2g of NaOH into about 400cm³ of water. Stir properly until it dissolves and then dilute to 500cm³ with distilled water 6. BaCl2 (0.1M BaCl2 in 500cm³) Weigh 10.4g of BaCl2 into about 400cm³ of water. Stir properly until it dissolves and then dilute to 500cm³ with distilled water
Thank you for this video. I have a question, I couldn't get the Anhydrous FeSO4 but I have the hepta hydrate form. I'm thinking of using 10g of that bese on some calculations to get the 5.5g required by waec. Hope am still correct?
Hellow. Is there a possibility of heating the mixture of the salts before addition of water . Because sodium carbonate is stable to heat so will not decompose. Why lead nitrate will decompose to give lead oxide, nitrogen(iv) oxide and oxygen.
I think as far a 2024 waec is concerned, 10.10g of FeSO4.7h2O should be use for 2L(2dm3) and not 1L(1dm3) as that's what the requirement is. Also, he recommended the anhydrous sulphate to be used instead of hydrated one.
I will suggest you to acidify the iron(II) sulfate solution when preparing it, not just at the point of titration by the student. Acidify it immediately after preparing it. 10mL or above of 1M H2SO4 ll be enough to create an acidic environment per dm³ of FeSO4
I've run the titration three times already and till arrived at 9cm3. Am new at this and I don't want to be the reason the students would have a problem
Sir I ran the process again and got 8.5cm3. I wasn't the one that bought the reagents am just suspecting if what was bought is hydrated FeSO4. Please help 🙏
Yes...but note this please❓❓❓❓ Anhydrous ferrous sulfate (FeSO₄) is typically pale green in its solid form. However, when dissolved in water, it can appear brownish instead of green due to oxidation. Anhydrous ferrous sulfate contains ferrous ions (Fe²⁺), which are pale green in colour, but when exposed to air or oxygen in water, ferrous ions can undergo oxidation to form ferric ions (Fe³⁺) which impart a yellow-brown color to the solution, resulting in a brownish appearance.
Thanks! Using same equation C1V1 = C2V2 C1 = 18.4M (initial concentration of 98% H2SO4 stock solution) V1 = ? (the volume you need to make a 2M acid soln) C2 = 2M ( final concentration) V2 = 200cm³ (assuming you want to prepare it to 200cm³, you can use any volume of your choice like 500cm³ or even 1000cm³) Substituting into the eqn above: 18.4M x V1 = 2M x 200cm³ V1 = 2M x 200cm³ / 18M V1 = 21.74cm³ ***Remember to add acid to water. Steps 1. Add about 100cm³ of distilled water to a measuring cylinder or beaker 2. Add the acid (21.74cm³) to the water 3. Make up the volume to 200cm³ by adding more distilled water to it.
In a titration involving potassium permanganate (KMnO4) and iron(II) sulfate (FeSO4), it is the iron(II) sulfate solution that needs to be acidified. The reason for acidifying FeSO4 is to ensure that the redox reaction between KMnO4 and FeSO4 proceeds correctly and to prevent side reactions. Acidification ensures that KMnO4 does not form undesirable manganese oxides, such as MnO2, which can occur in less acidic or neutral conditions.
Thanks a lot for this masterpiece. I have a question please. Does the volume and concentration of the sulfuric acid affect the titration in any way? Thanks for your response in anticipation.
Yes! The concentration and volume of the sulphuric acid does affect the outcome of the titration. The acid is added to create an acidic environment. The acidic environment helps to prevent the premature oxidation of iron(II) (Fe²⁺) to iron(III) (Fe³⁺) by dissolved oxygen in the solution or by other oxidation sources. If the acid is too dilute, it may not adequately maintain the acidic environment throughout the titration. This can lead to oxidation of iron(II) to iron(III) by atmospheric oxygen, skewing the results of the titration. Insufficient acid volume can result in inadequate acidity, leading to similar issues as above. Ensure the concentration is 1M as shown in the video and the volume 5-10ml
One question Sir. After dissolving the knmo4 and feSo4 in 1L. Do we now share into 150cm³ bottles per student? Please I need response. Thanks. You video is amazing
I really love your teaching the lord will continue to help you. Please i need an advice from you, our don't have measuring scale for now. Please can you tell me how many spatula of kmno4 will give me 1.58g of it nd how many spatula of feso4 will give me 5.5g of it i can dissolve it by myself using your method you taught here. Please i will appreciate if i can get a response from you regarding that. I will also tell all my students to subscribe to this UA-cam channel.thanks
Oh really! There's no way you can get the accurate mass this way. The best you can do is advice then to buy immediately or go to a nearby school and request to use their measuring scale. Thanks !!!
Hey @ChemEasy_Eko nice work but you forget to note that if anyone is using Feso4. 7H2O (water of crystallization) then the molar mass to be used is 278gmol-1. Then the end product will 9.9 or 10 cm3. But yours gave 18.10cm3 simple because you used only the molar mass of FeSO4 or maybe your salt is not having water of crystallization. 👍🏾👍🏾👍🏾
For those who don't understand
Thanks a lot for this. You will earn a unique badge in this channel soon 😄
It is....I working towards meeting the criteria
This is my first time of watching your video. You are indeed amazing, well detailed. God bless you sir. Thanks for sharing your knowledge with all .
You are welcome. Thanks a lot
I love your teaching so much, so straight forward
Thanks a lot, I appreciate you
U dey give correct analysis.. that's y as a teacher I always take my time watch your videos to improve on mine... Thank u Soo much..!
I'm grateful for this wonderful comment. Thanks a lot
God will bless u and increase ur knowledge
Amen. Thanks 👍🙏
Thank you so much .I'm really impressed.
You are welcome. Thanks for watching
Thank you sir ,this is awesome 😮
Thank you very much ChemEasy, God bless you.
Please I'd like to ask, is there a difference between chemistry practical alternative A and chemistry practical alternative B?
Thank you once again, very grateful for your page
1. Chemistry Alternative A is a practical examination where students perform actual laboratory experiments. This practical component typically involves using laboratory equipment and chemicals to carry out specific tasks outlined in the exam.
While:
2. Chemistry Alternative B is designed for schools or candidates that may not have access to fully equipped laboratory facilities.
Instead of a hands-on practical exam, Alternative B features a "practical paper" that is theoretically based. This paper often includes alternative-to-practical questions where students are tested on their understanding of practical concepts without having to perform actual experiments.
Great Job
Well detailed
Thanks a lot
You can use 250cm³ beaker, if you measure 4 of the beaker will give you 1000cm³
Yes. But the uncertainty in the measurements will be high, that's why I recommended using a volumetric flask of 1000cm³ capacity or 1000cm³ measuring cylinder
❤@@ChemEasy_Eko
Wow ..
This is so educating
Thank you very much sir for teaching
I had a tough time preparing this but with this lecture am enlightened
God bless you sir
You are welcome. Thanks for watching
Please sir, i followed this preparation but my titre value was 13.40cm because my values were around 13.
It could be your reagents. If it gives that value after multiple trials then try preparing the reagents fresh or get another one. Or at most go with it
You always give me joy and confident , you are a blessing to your generation. I wish other science educators are like you. I was just waiting for your analysis on this and finally here it is. Pls analyse that of qualitative
Thanks a lot for this mind blowing comment. Practical Demonstration of the qualitative analysis and likely questions will be uploaded today.
Nice job
Thanks 🙏👍
This is so very well detailed.
Thanks a lot
Wonderful
Thanks a lot
Thank you so much sir ❤️
You are welcome
Im getting 10.0 as end point o 🙄. Where do you think my problem is from?
The salt you are using is hydrated with 7mole of water . The reduces the quantity of KMnO4 needed. With the hydrated salt the calculated end point is 9.90 . But since the yellow paper didn't talk about the water of crystallisation hence the need for using anhydrous salt
Thanks @dorisonwenna5957
Thank you what if you don’t have this volumetric flask of 1000ml can we use a measuring cylinder to dissolve or even a beaker
You can use a measuring cylinder. The uncertainty in the measurements is lower when you use a 1000cm³ volumetric flask (about ±0.30) that's why I ll recommend you use a volumetric flask. But if you don't have it you can use the ones you have.
@@ChemEasy_Eko thank you very much .
You are welcome Sir!!!
Is there particular volume of H2SO4 that must be measured to get it diluted
Can’t wait for physics
Me too😊
@@kasarachikalu5079 you will be a teacher…or🤔..student ba..
What of one is using hydrated iron sulphate.Is it wrong?.
I think you should use the anhydrous. The instructional material says FeSO4 and not FeSO4.7H2O
THANKS 😊
Good work
Thanks Sir
Does the hydrated form of Iron sulfate need to be acidified by H2SO4? And is there a fixed volume of sulfuric acid to be added?
Yes it does need to be acidified. No fixed volume of H2SO4 but Try using 2 - 5cm³ of 1M H2SO4
i appreciate your effort and value you bring to the table year in year out.
I just have one question does water of crystallization affect redox titration?
Hydrated compounds can behave differently from their anhydrous counterparts in terms of stability and reactivity. Hydrated compounds may dehydrate over time, especially if stored improperly, changing their characteristics.
This change can affect their reducing power, altering the effectiveness in redox titrations.
Try and get the anhydrous one
God bless sir
Amen
Thanks
Well done sir I really commend your effort!
Are we to provide equations or waec will provide?
Thanks 🙏👍
Equation will be provided for us, exactly the one you saw in the video
@@ChemEasy_Eko okay thanks
Thanks sir
You are welcome
I have been refreshing your page 😂God bless you for this ❤
Hahahaha thanks. The sample questions and qualitative analysis will be uploaded soon
Thank you so much for this video.
Please i have a question, in the theoretical equations, how did you get the Vb volume of fe to be 25cm3?
It is the volume of the pipette.
Thanks
The 25cm³ is the size of the pipette
This so detailed and simplified, thank you sir.
Permission to copy ,learn and teach my students ,sir??
Thanks! You have my full permission
You're the best thing to happen to titration. Stay elevated.
Hahahaha
Wow.... thanks so much
Thank you so much sir. When dilutung the Kmno4, are we to use deionized water?
Yes. Use deionized water or distilled water
Thumbs up sir 👍👍. But is this legit sha ?
Thanks
It is...
Please sir how do i prepare suphuric acid for this 2024 to acidified the iron ii sulphate
1. HCl (0.1M HCl in 500cm3)
C1V1 = C2V2
C1 = 12M
V1= ?
C2 =0.1M
V2 = 500cm³
12M x V1 = O.1M x 500cm³
V1 = 4.2cm³ (Measure 4.2cm³ of HCl and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water)
2. H2SO4 (0.1M H2SO4 in 500cm3)
C1V1 = C2V2
C1 = 18.4M
V1= ?
C2 =0.1M
V2 = 500cm³
18.4M x V1 = O.1M x 500cm³
V1 = 2.7cm³ (Measure 2.7cm³ of H2SO4 and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water)
3. HNO3 (0.1M HNO3 in 500cm3)
C1V1 = C2V2
C1 = 15.9M
V1= ?
C2 =0.1M
V2 = 500cm³
15.9M x V1 = O.1M x 500cm³
V1 = 3.1cm³ (Measure 3.1cm³ of HNO3 and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water)
4. Ammonia NH3 (0.1M of 35% NH3 in 500cm3)
C1V1 = C2V2
C1 = 18.1M
V1= ?
C2 =0.1M
V2 = 500cm³
18.1M x V1 = O.1M x 500cm³
V1 = 2.8cm³ (Measure 2.8cm³ of NH3 and add to about 400cm³ of distilled water and then dilute it to 500cm³ with distilled water)
5. NaOH (0.1M NaOH in 500cm³)
Weigh 2g of NaOH into about 400cm³ of water. Stir properly until it dissolves and then dilute to 500cm³ with distilled water
6. BaCl2 (0.1M BaCl2 in 500cm³)
Weigh 10.4g of BaCl2 into about 400cm³ of water. Stir properly until it dissolves and then dilute to 500cm³ with distilled water
Thank you for this video.
I have a question, I couldn't get the Anhydrous FeSO4 but I have the hepta hydrate form. I'm thinking of using 10g of that bese on some calculations to get the 5.5g required by waec. Hope am still correct?
How did you compute your calculations Sir?
@@ChemEasy_Eko percentage composition of FeSO4.7H2O that's is FeSO4
151/278 * 100 = 55%
Hellow. Is there a possibility of heating the mixture of the salts before addition of water . Because sodium carbonate is stable to heat so will not decompose. Why lead nitrate will decompose to give lead oxide, nitrogen(iv) oxide and oxygen.
Possibly, that's a good one...but I'm not very sure they will ask them to do that. Let's see how it goes. Thanks 👍
@@ChemEasy_Eko ok.
Hello sir, please is it wise to replace 5.5g of feso4 with 10.10g of feso4.7h20?
I think as far a 2024 waec is concerned, 10.10g of FeSO4.7h2O should be use for 2L(2dm3) and not 1L(1dm3) as that's what the requirement is. Also, he recommended the anhydrous sulphate to be used instead of hydrated one.
No. I will not recommend that
Please i want to ask is it at the point of each titration that the 5ml of the acid will be added to the iron in the conical flask.
I will suggest you to acidify the iron(II) sulfate solution when preparing it, not just at the point of titration by the student. Acidify it immediately after preparing it. 10mL or above of 1M H2SO4 ll be enough to create an acidic environment per dm³ of FeSO4
@@ChemEasy_EkoOkay, thank you .
Yes
Good day sir, thank you for everything. I just finish running a sample using 1M of H2SO4. And i got 9.00cm3 as my titre value is it ok?
Thanks. Try running it thrice to see if you can get concordant values or reprepare ur reagents
I've run the titration three times already and till arrived at 9cm3. Am new at this and I don't want to be the reason the students would have a problem
Prepare the reagents again and try it, this time be very meticulous. If you get same result then get back to me and I ll guide you on what to do.
Sir I ran the process again and got 8.5cm3. I wasn't the one that bought the reagents am just suspecting if what was bought is hydrated FeSO4. Please help 🙏
Ok pls drop ur Whatsapp number, I ll send you a dm
Is your ferrous sulphate hydrated? Because waec didn't specify anhydrous or hydrated. And the market is full of hydrated ones
It's anhydrous. Buy the anhydrous ferrous sulphate
@@ChemEasy_Ekothe hydrated is yellow, and the anhydrous is green
Yes...but note this please❓❓❓❓
Anhydrous ferrous sulfate (FeSO₄) is typically pale green in its solid form. However, when dissolved in water, it can appear brownish instead of green due to oxidation.
Anhydrous ferrous sulfate contains ferrous ions (Fe²⁺), which are pale green in colour, but when exposed to air or oxygen in water, ferrous ions can undergo oxidation to form ferric ions (Fe³⁺) which impart a yellow-brown color to the solution, resulting in a brownish appearance.
I found your video very helpful. How do I prepare two moles of sulphuric acid
Thanks!
Using same equation C1V1 = C2V2
C1 = 18.4M (initial concentration of 98% H2SO4 stock solution)
V1 = ? (the volume you need to make a 2M acid soln)
C2 = 2M ( final concentration)
V2 = 200cm³ (assuming you want to prepare it to 200cm³, you can use any volume of your choice like 500cm³ or even 1000cm³)
Substituting into the eqn above:
18.4M x V1 = 2M x 200cm³
V1 = 2M x 200cm³ / 18M
V1 = 21.74cm³
***Remember to add acid to water.
Steps
1. Add about 100cm³ of distilled water to a measuring cylinder or beaker
2. Add the acid (21.74cm³) to the water
3. Make up the volume to 200cm³ by adding more distilled water to it.
Thanks. Also how can I acidified the iron Ii sulfate immediately after preparing it
Add about 10 - 20cm³ of the sulphuric acid to it
10-20cm of conc sulphuric acid? Won't it be more than one dm3
Lastly why is the permanganate not acidified?
Well done sir. More power to your elbow sir. I thought Kmno4 needs to be protonate(acidfied)?
In a titration involving potassium permanganate (KMnO4) and iron(II) sulfate (FeSO4), it is the iron(II) sulfate solution that needs to be acidified. The reason for acidifying FeSO4 is to ensure that the redox reaction between KMnO4 and FeSO4 proceeds correctly and to prevent side reactions.
Acidification ensures that KMnO4 does not form undesirable manganese oxides, such as MnO2, which can occur in less acidic or neutral conditions.
@@ChemEasy_Eko
Thanks and God bless you sir
Pls sir why ate u finding the Volume of permanganate twice cause i don't understand 🙏🙏🙏🙏
Pls sir help
Sorry....please I'm yet to understand what you mean. Make it clearer . Thanks a lot
Thanks a lot for this masterpiece. I have a question please. Does the volume and concentration of the sulfuric acid affect the titration in any way? Thanks for your response in anticipation.
Yes! The concentration and volume of the sulphuric acid does affect the outcome of the titration. The acid is added to create an acidic environment. The acidic environment helps to prevent the premature oxidation of iron(II) (Fe²⁺) to iron(III) (Fe³⁺) by dissolved oxygen in the solution or by other oxidation sources.
If the acid is too dilute, it may not adequately maintain the acidic environment throughout the titration. This can lead to oxidation of iron(II) to iron(III) by atmospheric oxygen, skewing the results of the titration.
Insufficient acid volume can result in inadequate acidity, leading to similar issues as above.
Ensure the concentration is 1M as shown in the video and the volume 5-10ml
Also
Remember to freshly prepare your FeSO4 that same time they want to use it.
Thank you so much
One question Sir. After dissolving the knmo4 and feSo4 in 1L. Do we now share into 150cm³ bottles per student? Please I need response. Thanks. You video is amazing
Yes according to WAEC standard. But most school will group the students because of insufficient equipment
Pls sir is this what we will see in our Waec also ?
What you may likely see...I have no idea of what they will set. These are my unique thoughts of what you are likely to see.
I really love your teaching the lord will continue to help you.
Please i need an advice from you, our don't have measuring scale for now. Please can you tell me how many spatula of kmno4 will give me 1.58g of it nd how many spatula of feso4 will give me 5.5g of it i can dissolve it by myself using your method you taught here. Please i will appreciate if i can get a response from you regarding that. I will also tell all my students to subscribe to this UA-cam channel.thanks
I mean our school don't have measuring scale for now
Oh really!
There's no way you can get the accurate mass this way. The best you can do is advice then to buy immediately or go to a nearby school and request to use their measuring scale. Thanks !!!
Thanks a lot. I commend your effort
@@ChemEasy_Eko ok sir I think I will have to do that
Yes
That's the best you can do for now. Thanks