You are a legend. Guys i reccomend practicing with past questions because the video info might go out of your head. More practice --> more likely to remember steps
Also, wise words: "Persistence is the key. You've just gotta keep on keepin' on. If you bang your head against the wall, eventually that wall will definitely fall down." -Harold Walden
Haha, thanks for comment Ryan, it is good to hear that the videos (and some of my weird philosophy) is having a positive effect. I usually only get comments where people don't understand how I have explained something so it is really nice to get a strictly positive comment. But yeah, just keep on keeping on and if you would like me to try and explain something in the future, hit me up. I am always looking for new video ideas. :)
Ni2+ can be analysed by a back titration using standard Zn2+ at pH 5.5 with xylenol orange indicator. A solution containing 25.00 mL of Ni2+ in HCl is treated with 25.00 mL of 0.05388 M Na2EDTA. The solution is neutralised with NaOH, and the pH adjusted to pH 5.5 with acetate buffer. The solution turns yellow when a few drops of indicator are added. Titration with 0.02129 M Zn2+ requires 15.02 mL to reach the end point. What is the molarity of the Ni2+ solution? How would you approach this question?
Idk if it was a mistake or not, but when you said the hcl that was consumed was 3.0967*10 to the -2, in the next step, it became 3.0967*10 to the -3. I am a little bit confused
Thank you so much harold walden! I have a question , why do you multiply 9.033 ×10(-4) mol by 10 ? Just because of your solution volume i mean 10 ml ? I cant really understand the logic of that !
Hi, I need some help on a question: It goes the same as a normal back titration question but tells me that a resulting solution was made up to 250 cm3 with water. what would i do at this point? Would I work out the question normally?
Tanisha you are right, technically it should be g/mol, I am just charging through it and it is only the name of the metal that I care about so I have just written grams because I know that it is on a ‘per mole’ basis. Cheers for alerting me to it, I’ll be more careful next time.
+Matin Hussain just gotta keep on putting my videos together. I get so hyped when people like them. Just make sure you subscribe, like my videos and share them with people who you think could use them. I actually helps a lot. Thanks for the love :)
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You are a legend.
Guys i reccomend practicing with past questions because the video info might go out of your head.
More practice --> more likely to remember steps
I think your exactly right. Doing practice questions that mimic exam problems are great for ensuring that you can apply your knowledge of the theory.
thank you for your explanation!! it's made back titration so much more easier to understand now 😊!!
You're amazing! Thank you so much for this demonstration! Very informative.
Also, wise words:
"Persistence is the key. You've just gotta keep on keepin' on. If you bang your head against the wall, eventually that wall will definitely fall down." -Harold Walden
Haha, thanks for comment Ryan, it is good to hear that the videos (and some of my weird philosophy) is having a positive effect. I usually only get comments where people don't understand how I have explained something so it is really nice to get a strictly positive comment.
But yeah, just keep on keeping on and if you would like me to try and explain something in the future, hit me up. I am always looking for new video ideas.
:)
Haha, I have worn that saying out. I use it all the time !!!
Thank you very much for finding a lost person like myself, after being lost on finding...... in the back titration process.👋👋👋👋👋👏👏👏👏👏👏👏
Beautifully explained... thanks a lot
Best video out there man, thanks heaps
Thanks mate for the positive comment. I really appreciate it 🤓
first time understanding back titration thanks sir
No problem, glad I could help 🤓
great video! Perfect way of explaining it.
Glad it helped 🤓
Ni2+ can be analysed by a back titration using standard Zn2+ at pH 5.5 with xylenol orange
indicator. A solution containing 25.00 mL of Ni2+ in HCl is treated with 25.00 mL of 0.05388 M
Na2EDTA. The solution is neutralised with NaOH, and the pH adjusted to pH 5.5 with acetate
buffer. The solution turns yellow when a few drops of indicator are added. Titration with
0.02129 M Zn2+ requires 15.02 mL to reach the end point. What is the molarity of the Ni2+
solution? How would you approach this question?
Amazing video sir!!! Wish you were my teacher thank you!!!
Thanks man, it is really nice to get positive feedback on my videos.
This is an excellent video
Thanks for the positive feedback 🤓
Thanks for the calculation sir, the qestion was part in my exams you helped alot GOD BLESS YOU keep the good work going lit
Thanks I really appreciate it 🙏🏽
Idk if it was a mistake or not, but when you said the hcl that was consumed was 3.0967*10 to the -2, in the next step, it became 3.0967*10 to the -3. I am a little bit confused
good spot, so at 9:21 when I wrote 3.0967*10^(-3) that should be 3.0967*10^(-2). Thanks for letting me know :)
same problem, maybe he made a mistake
@@HaroldWalden Doesn't that affect your final answer? Am getting different figures.
MUPYA KANKINZA your right. I made a mistake, thanks for letting me know :)
@@HaroldWalden what then should be the final element? Please help.
well done you hit the nail
Glad I could help... thanks for the positive feedback :)
Wow you explainend that really well , thankyou 🤗
Thank you so much harold walden! I have a question , why do you multiply 9.033 ×10(-4) mol by 10 ? Just because of your solution volume i mean 10 ml ? I cant really understand the logic of that !
Because the 9.033x10^(-4) is for a 10ml aliquot and we need the number of moles in a 100ml sample. So we multiply it by 10.
Hope that helps 🤓
Yup ! Understood :) thank you 😊
H Mnejad sweet thanks for watching one of my old videos. I hope they have improved since.
Thanks a lot ! Really helped this one
Not a problem, glad I could help :)
Hi, I need some help on a question:
It goes the same as a normal back titration question but tells me that a resulting solution was made up to 250 cm3 with water. what would i do at this point? Would I work out the question normally?
Yep, exactly the same using 250mL of water
THANK U! Grt vid. There is a mistake, on 11:22 Mr shouldn't be in g(grams) it should rather be in g mol^-1 (grams per mol)
Tanisha you are right, technically it should be g/mol, I am just charging through it and it is only the name of the metal that I care about so I have just written grams because I know that it is on a ‘per mole’ basis.
Cheers for alerting me to it, I’ll be more careful next time.
Oh Alright > < Still Great video!
Pleasure to do so :D
Thank you sir!
where can i find more questions like this
wow! That was great!
I didn’t think that questions involving titrations could ever be great but I am really glad that the video helped 🤓
u need more views
+Matin Hussain just gotta keep on putting my videos together. I get so hyped when people like them.
Just make sure you subscribe, like my videos and share them with people who you think could use them. I actually helps a lot.
Thanks for the love :)