Longshanks

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Not necessarily. The only fear I have with TBR is their practice passages are very rewarding if you use their "tricks", so I'll see what happens with practice exams very soon. I got passages 1 & 3 perfect but 2 I absolutely bombed, made silly detail errors, last night.

Quick rules:
when pH< pKa : compound is protonated
pH > pKa: deprotanted
pH = pkA at half equivalence(strong base with weak acid / strong acid with weak base)
(pH titrant + pka whats being titrated)/2 = pH end point for weak acid/base
10:1 - 1:10 rule
of course the mother the rules all this: pH= pKa + log [A-/HA] & pH= 1/2 pKa - 1/2 log [HA]

Those should get you through any acid-base or buffer passage. After you play around with them for a bit and do the examples and practice problems, you'll be able to start eye balling answers, there's always one or two choices that are completely far from being correct usually with acid-base questions so you can get rid of those immediately (ex- its an acid solution and the pH is greater than 7, some fool is titrating an acid with acid, etc)

If you have difficulty with the buffer chapter make sure you've mastered the acid/base chapter, they got hand in hand naturally.
 
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LostInStudy

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I agree with you Meli. I felt like I really mastered the material as I read the chapter but they ended up being the worst scores I had up to that point. I had been getting above 80% on everything until I hit acids and got my ass kicked (I got a 74%) and then I got to Buffers and Titrations which just added insult to injury because I got a 68% on that section. Usually, I'll move on to something later while I study at night but that night I looked up and studied everything I could find on those 2 subjects. It helped for my confidence but that section has more tricky questions/questions you don't HAVE to know in it. I felt that going through the chapter I had everything in the goals section mastered and I got the questions regarding those fundamental concepts right. I just missed a lot of tricky questions and for some reason I couldn't add that day.

Basically, if you have the stuff from the beginning for the chapter mastered then you are good and just review and try to learn from it and move on.

Hope this helped,

-LIS
 

unDRdog

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Did anyone else find this section of BR really difficult?!?
no...it really launched out of the acid/base chapter so i would try and solidify acid/base then move back into buffer/titrations. Keep trying you;ll own it with practice
 

Longshanks

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:confused:
I got pH endpoint (for titrating weak acid with strong base) = 14 +.5 log (kb *[A-] )
I quoted their shortcut approximation, on p.306 of the physics book (at least the version from this spring). In their example they admit that the pH is off by .15 using that equation. But on a multiple choice exam that should work fine. Their approximation equation should make sense too since it is used again a few pages later to estimate what would be a good indicator, and in the best case the pH at equivalence should = pKa of the indicator.

I haven't tried yours yet, but that would prob work too as their other suggested equation is pOH = 1/2 pKb - 1/2 log [A-] (which I think is the same equation as yours just written for pOH which you can then just convert to pH). Yours is probably more exact, but if there answer choices aren't close together (and they usually aren't for acid-base questions) their approximation should work, granted you know the pH of what you were using.

If you don't have that information it would probably be annoying to do extra math to get that so just doing pOH = 1/2 pKb - 1/2 log [A-] and knowing pH + pOH =14, just taking the difference to get the pH. Which, I'm not great at math, but looking at your equation now is the same exact thing I believe, except adding the 14 to the one side of the equation so you get pH as your answer.
 
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wanderer

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I quoted their shortcut approximation, on p.306 of the physics book (at least the version from this spring). In their example they admit that the pH is off by .15 using that equation. But on a multiple choice exam that should work fine.

I haven't tried yours yet, but that would prob work too as their other suggested equation is pOH = 1/2 pKb - 1/2 log [A-] (which I think is the same equation as yours just written for pOH which you can then just convert to pH)
Can you give an example?

The equation I posted I derived on my own, but it has worked for the few examples I've tried.

for example 50 mL of .5 M acetic acid + 50 mL of .5 M NaOH = .25 M A-
pKa is around 4.7, pKb is around 9.3 so pH = 14 + .5 * log ( 10^-9.3 * 2.5*10^-1)
pH = 14 + .5 * log ( 2.5^-10.3 which is around 1 * 10^-10)
pH=9
I suppose I could rearrange it to: 14 -.5 (pkB+ p[A-]) which is the same as your other equation.....
 
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Longshanks

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Can you give an example?

The equation I posted I derived on my own, but it has worked for the few examples I've tried.

for example 50 mL of .5 M acetic acid + 50 mL of .5 M NaOH = .25 M A-
pKa is around 4.7, pKb is around 9.3 so pH = 14 + .5 * log ( 10^-9.3 * 2.5*10^-1)
pH = 14 + .5 * log ( 2.5^-10.3 which is around 1 * 10^-10)
pH=9
I suppose I could rearrange it to: 14 -.5 (pkB+ p[A-]) which is the same as your other equation.....
This is why I hate logs. I can't think of an example right now, other than just typing one from the book, but i doubt that's legal. If it works it works.
 

LostInStudy

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Both are correct. One gives you pOH (BR version) and the other just directly converts pOH to PH in the same step (Wanderer version). BR states that these equations can only be used if the pKa is between 2 and 12 and the concentration of the reagent has to be more than the Ka.

-LIS