IB`NOT`ez Posted August 7, 2016 Report Share Posted August 7, 2016 Hi, For months now, I still can't deduce the answer to this question and with a reexamination coming soon, I'd really appreciate any help anyone can provide. The question, which is from the 2016 Chemistry Specimen Paper, and is attached as an image to this post. The answer is 22.10cm^3, but I can't derive the answer from anywhere. With the question asking to "calculate", I've done a manner of different things from averaging the titre values, doing M1V1 = M2V2, just subtracting the second titre from the first titre -- but I still don't understand how the answer is derived. The markscheme says something like (hate the new mark scheme format so much as it's crazy unclear, for me at least) ‹(22.05+22.15)(0.5) =›22.10‹cm3› and I'm really confused as to where those numbers came from. Also, does anyone have any recommendations for good IB Chemistry videos that teaches students on how to solve titration problems, perhaps with an example? I've looked through MSJChem, Richard Thornley and couldn't find any videos on how to solve titration problems. Apart from them, Google only showed a video by someone called "Jeffrey Yankel", and another video that's for IGCSE. There's also Khan Academy but since they're not explicitly for IB Chemistry, I'm a little hesitant to use them (worse comes to worse, I'll look into the Khan Academy one, but hoping if anyone has better suggestions). Thanks for your time, appreciate any help! Reply Link to post Share on other sites More sharing options...
Msj Chem Posted August 7, 2016 Report Share Posted August 7, 2016 To calculate the average titre, the second and third titres have been added together and divided by the number of trials, which is two. The first trial is not used because it is a rough trial and is not used to calculate the average. When calculating an average volume from a titration, you should use concordant values which are within 0.10 cm3 of each other. Reply Link to post Share on other sites More sharing options...
IB`NOT`ez Posted August 7, 2016 Author Report Share Posted August 7, 2016 (edited) 1 hour ago, Msj Chem said: To calculate the average titre, the second and third titres have been added together and divided by the number of trials, which is two. The first trial is not used because it is a rough trial and is not used to calculate the average. When calculating an average volume from a titration, you should use concordant values which are within 0.10 cm3 of each other. Ahh, I got the numbers by doing that now. Thank you -- a bit of a depressing thought that after one year of IB Chemistry, I still didn't know this x_x Edited August 7, 2016 by IB`ez Reply Link to post Share on other sites More sharing options...
mrwengchemistry Posted August 7, 2016 Report Share Posted August 7, 2016 The complete answer to your question is here: I have a lot of example titration calculation example specifically for IB on http://www.mrwengibchemistry.com Please also look for Vit C titration in the search. It is very IB. I cover a lot more than msjchem and Richard at the moment including IA resources. Reply Link to post Share on other sites More sharing options...
IB`NOT`ez Posted August 9, 2016 Author Report Share Posted August 9, 2016 On 8/8/2016 at 6:32 AM, mrwengchemistry said: The complete answer to your question is here: I have a lot of example titration calculation example specifically for IB on http://www.mrwengibchemistry.com Please also look for Vit C titration in the search. It is very IB. I cover a lot more than msjchem and Richard at the moment including IA resources. Thank you, I'll look at those videos Reply Link to post Share on other sites More sharing options...
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