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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 practiceDid anyone else find this section of BR really difficult?!?
(pH titrant + pka whats being titrated)/2 = pH end point for weak acid/base
I got pH endpoint (for titrating weak acid with strong base) = 14 +.5 log (kb *[A-] )
Can you give an example?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.....
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.