Book says Benzoic acid, C6H5CO2H, undergoes dissociation in water at 25*C with the following thermodynamic parameters:
H* = -67 cal mol-1
S* = -19.44 cal deg-1 mol-1
Write the equation for this dissociation process. What kind of reaction is it? Calculate Ka and pKa for benzoic acid. How does Benzoic acid compare in
strength with acetic acid, whose pKa = 4.7
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C6H5CO2H + H2O <-----> C6H5CO2(-) + H3O(+)
This process is a typical Bronsted-Lowry acid-base reaction. In order to calculate the equilibrium constant, we must first determine G* at 25*C, or
298 Kelvins:
So basically my answer is G* = +5.726 kcal mol-1
I've got that far, but I don't know how to find Ka using that.
I know how to change Ka into pKa and compare Benzoic acid to acetic acid whose pKa=4.7, but how do I get Benzoic acid's K or Ka with only what
they've given me. Organic Chemistry Structure and Function fourth edition by K.Peter C. Vollhardt & Neil E. Schore isn't a bad book but it's
mostly vague on any lessons. It doesn't go into comprehensible details like some of the first books I've read to get this far, nore does it include
enough exercises for practice. Thank you for your time!
It just says Using G* = -1.36 log K, we obtain log Ka = -4.2,
How did they get "-1.36 log K" from
G* = +5.726 kcal mol-1
and how did they obtain the Ka from it?sparkgap - 2-12-2012 at 23:44
You've computed ΔG°, so it's a matter of determining the proper units to use for the gas constant R, and then solving for Ka. As for
pKa, the thing to know about p-functions in general is that
pX=-log10 X
(e.g. pH is the negative common logarithm of the hydrogen ion concentration, pKsp is the negative common logarithm of the solubility
product, etc.)
I should hope these hints should be sufficient.
sparky (~_~)
P.S. Some logarithm identities might also be of use; in particular, logb x = (loga x)/(loga b)...
[Edited on 3-12-2012 by sparkgap]yoyoils - 3-12-2012 at 10:12
can you tell me how to answer it please or give me an example with how to use the calculator. Because I learn better that wayyoyoils - 3-12-2012 at 10:26
You said I've computed G*, and so it's a matter of determining the proper units to use for the gas constant R.
With that example, I will do my best to assume
ΔG°=-RT*ln Ka
ΔG°=[(negative gas constant)](multiplied?)[Absolute Temp in Kelvins] *(multiply?) ln Ka
I see ΔG°= negative number multiplied by absolute temp, but i don't know what else. Do I use LN or eX buttons on the calculator?
I learned how to do logarithms because it was simply the log button and negative log button to compute pH,pOH,Ka,pKa,Kb,pKb. It's just difficult for
me without help on how to use the calculator.watson.fawkes - 3-12-2012 at 13:08
