Got this recommended by a comment on another video going over titration calculations and now being a year 2 chemistry student finally understand, thank you!
WOW!!! such great explaination!! okay, for the exam question at the end i think you calculated moles of NaOH wrong 0.1 * (26.5/1000) = 2.65*10^3 not 2.625*10^3 I saw this when i did the question, other than the difference of that value my calculations and technique were same as yours so my % purity of MgCO3 came out be 52.828%
wow i kept seeing these on past papers and couldn't do it, and was never taught it either. I finally fully understand now, you made is so easy to understand!! Thank you so much and God bless you!!
Its Formulae, Equations and Amount of Substance for Edexcel. Its a few core ideas combined- which is what makes them tricky. They are just a certain way of using the titration calculations and combining it with a reacting Mass calculation and often also dilutions. So they often aren't a set part of the specification by themselves. They're a tricky combination of three parts of the course
Sir I was just wondering if this is on the AQA spec since I've never heard/gotten taught this before so its quite new to me if anything? is this as likely to be examined as a normal titration question?
@alizahmohammed7751 Yes, this is AQA... it's the Amount of Substance topic, along with redox titrations from year 2. Its a few core ideas combined- which is what makes them tricky. They are just a certain way of using the titration calculations and combining it with a reacting Mass calculation and often also dilutions. So they often aren't a set part of the specification by themselves. They aren't for AQA for instance. They're a tricky combination of three parts of Amount of Substance
@@chemistrytutor that makes so much sense thank you! i ended up watching the video and it being much easier to comprehend! thank you so much for the effort you put in your videos they're literally saving my grades whilst also making chemistry really enjoyable :)
This video was amazing and explained really well. I appreciate you making this video and teaching others about chemistry with such fluency and clarity. Keep it up, and never stop, people need you.
Good question. The vast majority of acid-base titrations can be simplified down to a reaction between H+ and OH- ions. In the example you use, the Na+ and Cl- start and finish as ions dissolved in solution. This makes them spectator ions, and so they havent changed and so aren't relevant to the actual reaction itself
Its Amount of Substance. Its a few core ideas combined- which is what makes them tricky. They are just a certain way of using the titration calculations and combining it with a reacting Mass calculation and often also dilutions. So they often aren't a set part of the specification by themselves. They're a tricky combination of three parts of Amount of Substance
Which course is that? AQA? it could come up in any a level exam. They often don't call it a back titration. It's effectively a couple of different skills smashed together. So you mostly don't see the words back titration on a specification, but there's nothing in them that isn't in a normal titration and then with a bit of %yield, reacting Mass tagged on at the end
I picked a number just to illustrate my point so I had something to work with that wasn't 'x' In an exam, you'd be given some numbers and then find moles using: moles = conc x vol
Difficult for me to say without a specific example. Depends what other information you've been given. Might be you do it from a series of titre volumes, or from a concentration and volume of an acid. I think I'd need more details to say for sure
@@chemistrytutor the question says , A 0,6g sample of K2CO3 is dissolved in enough water to make a 200ml solution A. A 20ml aliquot of solution A is taken and put into a conical flask .To the flask is added 20 ml of 0,17M of HCl The resulting solution is then titrated with 0,1048 M NaOH. How many ml of NaOH are used? The use of molarity confuses me ,Id prefer if they used no of moles , please help
@@amahlentlangulela8793 work out moles of HCl using 20/1000 x 0.17. Then moles NaOH is the same as they react in a 1:1 ratio. Then calculate volume using this new moles and the concentration of 0.1048 Molarity is a bit of an old fashioned term, but it basically can be treated exactly as concentration
this is the best explanation on back titrations , tysm for all your hard work!
Thank you for your kind feedback 😀
I'm really pleased the video helped
Got this recommended by a comment on another video going over titration calculations and now being a year 2 chemistry student finally understand, thank you!
Excellent news! I'm really pleased it's helped 😊
Damn, actually, I'm gonna stick around 😭🤝🏾
Great news! I'm glad it was helpful 😊
WOW!!! such great explaination!!
okay, for the exam question at the end i think you calculated moles of NaOH wrong
0.1 * (26.5/1000) = 2.65*10^3
not 2.625*10^3
I saw this when i did the question, other than the difference of that value my calculations and technique were same as yours so my % purity of MgCO3 came out be 52.828%
@eeshalkhan9340 thanks for the feedback. The burette volume was 26.25cm3 though, so the moles of NaOH is 2.625 x10^-3
Tryna understand back titration 5 hours before exam..
Why didn’t I search for this video 1 month ago🥲
Well, you found it before it was too late! 😀
I'm really pleased it's useful!
@@chemistrytutor I think I did well in this test anyway,you saved my 12-mark questions,tysm ❤️
@@LaLala-jc1zp excellent! I'm not going to take the credit- well done to you! 😀
Hard carrying my chem
😀👍
wow i kept seeing these on past papers and couldn't do it, and was never taught it either. I finally fully understand now, you made is so easy to understand!! Thank you so much and God bless you!!
That's awesome! I'm really pleased it's made a difference for you! 😀
Is this for AQA AS or for A2?
@@toxins5803 it's AS! But it always comes up in A-level papers too
@@puddleduck1405 Oh okay, thank you for the prompt reply
This is by far the best explanation on back-titration I've found so far which i can clearly understand, thank you so much!!
You're very welcome. Glad it's useful 😊
thank you for this I feel confident in back titrations... finally!!
That's brilliant news! Well done 👏
I'm pleased it was helpful
Thank you for everything you do for us.
You're very welcome 😀
thank you
😃
Do we need to know this for edexcel a level chemistry because it doesn’t explicitly say it in the spec
Its Formulae, Equations and Amount of Substance for Edexcel. Its a few core ideas combined- which is what makes them tricky.
They are just a certain way of using the titration calculations and combining it with a reacting Mass calculation and often also dilutions.
So they often aren't a set part of the specification by themselves. They're a tricky combination of three parts of the course
.. i love u
Glad it's useful 😊
I can't thankyou enough❤❤.
😀
Sir I was just wondering if this is on the AQA spec since I've never heard/gotten taught this before so its quite new to me if anything? is this as likely to be examined as a normal titration question?
@alizahmohammed7751 Yes, this is AQA... it's the Amount of Substance topic, along with redox titrations from year 2. Its a few core ideas combined- which is what makes them tricky.
They are just a certain way of using the titration calculations and combining it with a reacting Mass calculation and often also dilutions.
So they often aren't a set part of the specification by themselves. They aren't for AQA for instance. They're a tricky combination of three parts of Amount of Substance
@@chemistrytutor that makes so much sense thank you! i ended up watching the video and it being much easier to comprehend! thank you so much for the effort you put in your videos they're literally saving my grades whilst also making chemistry really enjoyable :)
@@alizahmohammed7751 that's great to know. Best of luck 👍
can u do a lecture about EDTA titration calculation? you're a very food tutor
I have plans to do this 😀
food tutor 😂
*good tutor AHHAHA sorry@@JessieAllen-r6m
You are the best teacher I’ ever met😢waiting for your video about electrolysis!!! Want to use my mother tongue to thank u 謝謝你！！！你教的太好啦❤️
Thank you 😊
I'm really pleased the teaching is helping! 😀
i finally understand how back titration works, thank you so much for the incredibly clear explanation !
That's great to hear!
Thanks for the feedback 😀
This was so helpful, thank you!
Excellent! Thanks for the feedback 😀
This video was amazing and explained really well. I appreciate you making this video and teaching others about chemistry with such fluency and clarity. Keep it up, and never stop, people need you.
You're very welcome!
Thank you for your kind feedback 😀
Thank you for your help! Made me confident in back titration!!!!! 🥰🥰🥰
I'm so glad! Thanks for the feedback 😀
Thanks
Very welcome 🙏
Thanks! But why at 7:43 why did u use the ions from the OH and H why not the Na and Cl?
Good question. The vast majority of acid-base titrations can be simplified down to a reaction between H+ and OH- ions.
In the example you use, the Na+ and Cl- start and finish as ions dissolved in solution. This makes them spectator ions, and so they havent changed and so aren't relevant to the actual reaction itself
Well explained, God bless you and give you more wisdom
Thank you for your kind words 😊
What topic in aqa is back titrations in because in amount of substance there is only the normal one?
Its Amount of Substance. Its a few core ideas combined- which is what makes them tricky.
They are just a certain way of using the titration calculations and combining it with a reacting Mass calculation and often also dilutions.
So they often aren't a set part of the specification by themselves. They're a tricky combination of three parts of Amount of Substance
Thank you
😃 very welcome
are these still relevant to our course ??
Which course is that? AQA? it could come up in any a level exam.
They often don't call it a back titration. It's effectively a couple of different skills smashed together. So you mostly don't see the words back titration on a specification, but there's nothing in them that isn't in a normal titration and then with a bit of %yield, reacting Mass tagged on at the end
Thank you , it was helpful
😀 very welcome
7:34 how do you know we used 2 moles NaOH ?
I picked a number just to illustrate my point so I had something to work with that wasn't 'x'
In an exam, you'd be given some numbers and then find moles using:
moles = conc x vol
@@chemistrytutor thank you!
are these used to calculate % by mass too?
That's definitely another thing they can be for, yes. Once you're in to the swing of them they can be useful for quite a few things
@@chemistrytutor Thanks for letting me know! They are definitely very useful
@rachaelkenyon4712 thank you 😊
thank u ssooooo muchhhhh
You're very welcome 😀
How do you combat back titration when they ask you to calculate the volume of the base that you titrated with ?
Difficult for me to say without a specific example. Depends what other information you've been given. Might be you do it from a series of titre volumes, or from a concentration and volume of an acid. I think I'd need more details to say for sure
@@chemistrytutor the question says , A 0,6g sample of K2CO3 is dissolved in enough water to make a 200ml solution A.
A 20ml aliquot of solution A is taken and put into a conical flask .To the flask is added 20 ml of 0,17M of HCl
The resulting solution is then titrated with 0,1048 M NaOH.
How many ml of NaOH are used?
The use of molarity confuses me ,Id prefer if they used no of moles , please help
@@amahlentlangulela8793 work out moles of HCl using 20/1000 x 0.17.
Then moles NaOH is the same as they react in a 1:1 ratio.
Then calculate volume using this new moles and the concentration of 0.1048
Molarity is a bit of an old fashioned term, but it basically can be treated exactly as concentration
@@chemistrytutor thank you very much, I appreciate it 🙂
@@amahlentlangulela8793 😃
Mr you ate that up!!!
😀
Last year I was doing Foundation year - your videos helped me a lot. Thanks 💙
Thanks for the feedback, I'm really pleased it was useful for you 👍
Thank you very much for this discrete lesson. I am using it to prepare for my classes - Pharmaceutical Analysis
That's for the feedback, and good luck!
Amazing video ❤
Thank you 😊
thank you so much, this was such a clear explaination i was struggling so much on the past paper but now it finally makes sense!!☺
Excellent! It's great to know that it's helped 😀
I appreciate it, also appreciate the past paper question
You're very welcome 🙏
Thanks for the feedback 😀
this was so useful thank you!
Excellent! Thank you for the feedback 😀
i love you
😁
Thanks you so much it was really useful .
Brilliant! It's a tough topic so I hope it was clear!
@@chemistrytutor it was really clear.
@@meerabfatima2709 excellent! Thanks 😊
THANK YOU SO MUCH SAVING ME ONCE AGAIN 🫶 you got me from an E to a C i cant thank you enough
Thanks for the feedback 😀
I'm really glad it's helping ☺️