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CHEM 1AA3 Study Guide - Final Guide: Sodium Benzoate, Benzoic Acid, Pi Bond


Department
Chemistry
Course Code
CHEM 1AA3
Professor
Pippa Lock
Study Guide
Final

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CHEM 1AA3 EXAM REVIEW - GOOD LUCK ON THE EXAM!!
Note: A lot of the questions on these practice exams are recycled in practice tests
for test 1 and 2.. somewhere you can find full solutions for a good number of them.
IS THE EXAM IN THE SAME ROOM AS WHERE WE WRITE
OUR MIDTERMS?? (it doesn’t say on Avenue)
Yes, its in BSB 147 @ 6 PM today! :) Thanks GOOD LUCK TO EVERYONE!
2011 Exam:
1) B - A buffer must have limiting strong (acid/base) with excess weak (acid/base)
[one must be an acid and one a base] or both weak that are conjugates of each
other.
2) D - I got E? How do you incorporate sodium benzoate into the calculation?
- Hmm i got A - I got A too!
- How did you guys work out this question?
Well, since they give you the Ka for benzoate, i assumed it was a buffer system
(so buffer equation) with benzoate. So go through each of the reactants, they’re
either going to increase or decrease the moles conjugate base or acid. pka
+log[(moles benzoate - moles HCL +moles NaOH + moles NaBr)/(moles benzoate
+ moles HCl - moles NaOH - moles NaBr)]. Thats how i got A as an answer though,
other people got different answers so i could be wrong.
- Oh jeeze, that’s pretty complicated.. how did you know which moles increased
and decreased?
I got D: React the strong acid and base together first: see what moles of acid are
left over: react that with the weak base and find your buffer pH (i got exactly 4.40
so i don’t know if he has multiple answers)
- But when you react NaOH and HCl, you get NaCl, which is a salt, right? So how
do you know what to react that with?
- so: moles of NaOH= 0.02 mol and moles of HCl= 0.045 mol: when you do a
before/after chart, you will have 0.025 mol remaining (the other moles will be
neutralized they have no affect on your pH)
- Yeah, I got that :) But then what?
so then i just reacted the 0.025 mol of HCL with ur sodium benzoate (which will
produce a buffer)
- OHH, I see now! Thank you :D But how did you know the mol. formula for sodium
benzoate?
- not necessary: you have concentration and volume
- what is used as the volume here to calculate the final concentrations of the HCl
and sodium benzoate? the total v?
- I think so..
if you use the total volume it doesn’t give you the answer...i’m confused on how to

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go from moles of the HCL and sodium benzoate to get the concentrations for the
H-H equation :S
- Don’t stress over it, there’s only 3 acid/base questions on the exam
it’s frustrating! it seems so simple but i can’t get it haha
- The approach you guys took to arrive to the answer D = 4.40 seems correct.
But how did you guys solve it? The concentration of HCl that I carried over
was (0.025 mol)/(0.145L) = 0.1724 M. However when I went about to solve it
using HH, I didn’t get the right answer..I would greatly appreciate it if
someone can explain it. Thank You!
-how is there 0.025 moles of HCL left over? there’s more moles of NaOH
than HCl isnt there? (0.200M)(0.100L) = 0.02 moles of NaOH and
(0.045)(0.100L) = 4.5e-3 moles of HCl?
Nevermind I’m an idiot, its 1.00L of HCl x 0.045M :S
3) E - Could someone explain their process for this question?
From the titration, you know moles of NaOH = moles HCl at the equivalence point,
right? So solve for the moles, divide that by the volume of initial HCl which is 100,
that gives you the conc of HCl, multiply that by the volume used when the
precipitation happened (3.34 mL) to find the moles, then divide that by the total
volume in the protein beaker which is 3.34 + 225, that gives you the [H3O] which
you can take the log of to get the pH. - Thank you! :)
4) C - Why? I think it’s E? Help?
Yeah, pretty sure it’s C... iv is true
I think it’s E, can’t be C (v) is true, CCl4 is non-polar, CHCl3 has a dipole moment
making it polar
i also think the answer is E, but when i google it, i found that hydrogen bonds are
broken when water boils, which makes me confused.
Notice how it says polarizable, not polar? What does this mean?
- No one can figure that out :(
Just my thought process, suggestions?:
- (i) At a given temperature, HF has a lower vapour pressure than HCl
True: HF has a lower vapour pressure than HCl because HCl wants to dissociate
because it is a strong acid, leading it to have low IMF, a low boiling point, and a
high vapour pressure
- (ii) CH3OCH3 has a higher boiling point than CH3CH2OH
False: CH3CH2OH has hydrogen bonding, and therefore has a higher boiling
point
- (iii) CH3CH2CH2CH2CH2OH has a higher viscosity than CH3CH2OCH3
True: The more IMF, the higher the viscosity

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- (iv) Hydrogen bonds are broken when water boils
True: Googled it
- (v) CCl4 is less polarizable than CHCl3
***True/False?***
CCl4 is nonpolar, and CHCl3 is polar- therefore...?
This is true as well. So is iv. This question must be a mistake.
^ I agree, since polar molecules have dipole-dipole interaction while non-polar
have just dispersion. - making the statement for (v) true.
5) C- Why? C makes sense theoretically, but when you calculate the pKa using
the Henderson- Has. or by doing the -log Ka you get a pH of approx. 3.55.
- you have excess KOH, pH can’t be less than 7
- But how did you get to the answer? (Educated guess, if there is a buffer solution
the pH range is usually 4-10, but someone should share the real way)
why can’t you just use the 3.55 pH? i don’t understand why this is incorrect
- Me neither!
- assume a 1L sample/ work in moles instead of [ ]? get a pH in the range of
answer option C pH= 8.35
- But even if you work in moles, the ratio of [A-]/[HA] will be the same!
there was a similar question to this one in test 1, #16 in version 2. i got it wrong
though and the answer key doesn’t really show a full solution. could someone
please post how to do these?! :) thanks
using an indicator: therefore want to find pH at equivalence
BEFORE: (assume a 1L sample total)
acetic acid: 0.150M*1L= 0.150mol
KOH: 0.150M*1L= 0.150mol < The concentration of KOH in the question is 0.20M.
Why did you put 0.150M? (finding pH at equivalence weak acid and weak base
have to be equal?- this process isn’t really right LOL, but it made the most sense -
Lmao okay!
AFTER:
acetate: 0.150mol
Find pH of 0.150mol/1L of acetate (given ka value therefore find kb)
will calculate the pOH when using these values
pH= 14-pOH
pH= 8.35
Not too sure if this process is right, but logically at equivalence point its going to
be weakly basic- so the other basic option is too basic?
- It makes sense, but there’s a lot of odd assumptions and guessing haha :(
i feel like we’re overlooking a very simple way to answer this question...i’m going
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