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Chemistry Predictions/Thoughts (1 Viewer)

ManifestationIsKey

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what was the iodic acid and sulfamic acid question, were we supposed to know the formulas for it?
No you just had to calculate their pH by plugging in the numbers. But since it wan't equal to that 1.151 or whatever it was, it can't be a strong acid and doesn't fully ionise. Btw which one was stronger?
 

O_Starkindler!

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No you just had to calculate their pH by plugging in the numbers. But since it wan't equal to that 1.151 or whatever it was, it can't be a strong acid and doesn't fully ionise. Btw which one was stronger?
light work. yeah iodic acid was the stronger one (the 0.1 molL-1 one) since you need less of it to obtain the same pH as the sulfamic acid
 

help2024

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what if i calculated degree of ionisation and did the ph do I get full marks?
 

radioheadfan42

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What do you reckon you guys got raw?
In the negatives YEAHHHH!!

kidding, but in all seriousness - as i tend to do - didnt get to finish it, so i think a maximum of 30-40 for me. If that. But hey! I finished more than in trials, and i got my depth study that i did pretty okay in so- win?

Not an academic win, but a personal win! XD
 

radioheadfan42

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Me when anything that isnt module 8 or 7: :mad:

module 6 can explode I HATE ACIDS AND BASES!!! I have a grudge.
 

potpal

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why am i always like 3 raw marks away from making a band 6/e4 😭 happened in 4u too and now chem fml
 

hughjanus_

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The reason that a Keq value of 0.53 is incorrect is due to the fact that you can’t do 0.1 - [BrCH2COOH](aq) to calculate [BrCH2COOH](octanol). This is because of the keyword “an AQUEOUS solution of 0.1M bromoacetic acid is shaken with an equal volume of octan-1-ol. This means that the concentration of bromoacetic acid in the entire solution is 0.05 M. Thus, instead of calculating the concentration of bromoacetic acid in water this way, you must instead replace [BrCH2COOH](aq) in the Ka expression with (0.1 - 9.18*10^3 - x) where x is [BrCH2COOH](octanol).

Because the concentration of bromoacetic acid in octanol and in water are not contributing to a singular molarity, it isn’t possible to do a simple 0.1 - [BrCH2COOH](aq) to solve the question.
 

appleali

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guys i only answered mc but i think i got full marks can i state rank?
 

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