thank you for this! I ended up getting x=1 by assuming the molar ratio between hcl and na2co3 was 2:1 because i wasn't sure how to construct an equation where one of the reactants has .H20. Is it fine to ignore the .H20 when writing out equations like this or does it only apply to titration calcs?
No worries, well done! I always just leave the .xH2O out of the chemical equation since the water molecules are just going to pop up on the product side of the equation also so I view them similar to spectator ions. This does not only apply to titrations, it just makes writing out the equation easier :)
Because we used the mean titre volume of NaOH to calculate the moles of NaOH, it is exactly equal to the moles of HCl in the conical flask since they react in a 1:1 ratio - it is the endpoint volume where we will get the colour change. If that doesn't make sense, brush up on the fundamentals of titrations and mole equations :) Best of luck!
There is excess HCl remaining after it has reacted with the Na2CO3. In the acids and bases topic you need to keep in mind the concept of conjugate acid-base pairs, and that they are Bronsted Lowry acids/bases that either need to donate or accept a H+ ion. If the NaOH reacts with the salt, there are no H+ ions to be donated, so there is no Bronsted Lowry acid present, this will never be the case in a titration! In a titration question, there is always going to be a neutralisation reaction occuring (acid + base reaction). Another way of looking at it is that NaCl reacting with NaOH is not energetically favourable, and so they will never react. The most that could happen is that if you heated the solution, they would separate into their constituent ions, but once cooled would re-form NaOH and NaCl.
You should watch the back titration video by ‘The Chemistry Teacher’ first because he kinda explains the basics of how these calculations works and then rewatch this video, it will make a lot more sense.
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All of that for x=1 is crazy 😭😭😭 thanks for the video
hahaha ikr! no worries :)
omg thank you so much! Amazing and easy to understand explanation throughout the entire vid!! :D
Happy to help, I'm glad it helped u understand :)
Watched the video twice and it make sense....thank you so much
It's a difficult one! I'm glad u understand it, nice!
Amazing video!!!!! Thank you very much for taking your time out to make this video :)
No worries! Be sure to like and share with anyone studying chem that would find it helpful 😊
Can you do more of these.
Im still get confused about what goes in what and how we use this to get moles and stuff 😅
I’m about to cry
honestly same, we have around 30ish hours its doable, Crying and all
thank you for this! I ended up getting x=1 by assuming the molar ratio between hcl and na2co3 was 2:1 because i wasn't sure how to construct an equation where one of the reactants has .H20. Is it fine to ignore the .H20 when writing out equations like this or does it only apply to titration calcs?
No worries, well done! I always just leave the .xH2O out of the chemical equation since the water molecules are just going to pop up on the product side of the equation also so I view them similar to spectator ions. This does not only apply to titrations, it just makes writing out the equation easier :)
Thanks for the video, it helps alot!
if the waters cancel out at 9:59, why include it on the RHS?
I don’t understand why Hcl is = to 3.99x10^-3
Because we used the mean titre volume of NaOH to calculate the moles of NaOH, it is exactly equal to the moles of HCl in the conical flask since they react in a 1:1 ratio - it is the endpoint volume where we will get the colour change. If that doesn't make sense, brush up on the fundamentals of titrations and mole equations :) Best of luck!
How can we demoralise these students😂😂😂
😂
Ive commented a lot now, but WOWW, i looked everywheree for a video on this question, your acc a lifesaver🥹🥹🥹🥹‼️‼️
God damn back titrations
agreed 😂
Bro I dont understand. Why is naoh reacting with the hhcl. Wouldnt it react with the salt. Becuz the resulting solution of hcl and na2co3 had reacted
There is excess HCl remaining after it has reacted with the Na2CO3. In the acids and bases topic you need to keep in mind the concept of conjugate acid-base pairs, and that they are Bronsted Lowry acids/bases that either need to donate or accept a H+ ion. If the NaOH reacts with the salt, there are no H+ ions to be donated, so there is no Bronsted Lowry acid present, this will never be the case in a titration! In a titration question, there is always going to be a neutralisation reaction occuring (acid + base reaction). Another way of looking at it is that NaCl reacting with NaOH is not energetically favourable, and so they will never react. The most that could happen is that if you heated the solution, they would separate into their constituent ions, but once cooled would re-form NaOH and NaCl.
Alsoo could you do question 11.1 on the 2017 paper 1 pleaseee
Sure I can add it to the list of question requests haha, but I will not be able to do the video until after paper 1, crazy busy atm!
i wasn't able to get these marks.
no problem, it was a really tough question!
@@easymodeexams33 cheers
I’m so lost
You should watch the back titration video by ‘The Chemistry Teacher’ first because he kinda explains the basics of how these calculations works and then rewatch this video, it will make a lot more sense.
@@kellynan THANKSSS
Im soo failing 😭