Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: light on December 06, 2004, 07:43:18 AM
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my book gives an example of how to find the concentration of OH- for transforming 99% benzoic acid to benzoate anion. it is difficult to understand how they get the answer, since there are no specified concentrations for the benzoic acid and benzoate anion, besides the way they write it is also difficult to understand. this is how they write:
the ability to separate a strong from weak acids depends on the acidity constants of the acids and the basicity constants of the bases as folllows. in the first equation, consider the ionization of benzoic acid, which has an equilibrium constant Ka of 6.8x10-5. the conversion of benzoic acid to the benzoate anion in the fourth equation is governed by the equilibrium constant K (Eq.5), obtained by the combining the third and fourth equations.
C6H5COOH + H2O -> <- C6H5COO- + H3O+ (Eq. 1)
Ka= [C6H5COO-][H3O+]/[C6H5COOH]=6.8x10-5, pKa=4.17 (Eq. 2)
Kw=[H3O+][OH- ]=10-14 (Eq. 3)
C6H5COOH + OH- -> <- C6H5COO- + H2O (Eq. 4)
K=[C6H5COO-]/[C6H5COOH][OH- ]=Ka/Kw=6.8x10-5/10-14=6.8x109 (Eq. 5)
if 99% of the benzoic acid is converted to C6H5COO- :
[C6H5COO-]/[C6H5COOH]=99/1 (Eq. 6)
then from Eq. 5 the hydroxide ion concentration would need to be 6.8x10-7M.
my question, how do they get [OH- ]= 6.8x10-7M ? ???
hope for any ideas and guidance!
thanks for helping!
...and yes i have tried, but can not figure out how do they think :'(
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[H3O+]=[Kw/[OH-]
Equation 5 should be
Ka=([C6H5COO-]/[C6H5COOH])(Kw/[OH-])
=>
[OH-]=Ka/Kw=([C6H5COO-]/[C6H5COOH])(Kw/Ka)=1.456x10-8 ~ 1.5x10-8
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I think:
from [C6H5COO-]/[C6H5COOH]=99/1 (Eq.6)
and
[C6H5COO-]/[C6H5COOH][OH- ]=6.8x109 (Eq. 5)
then [OH -] ={ [C6H5COO-]/ [C6H5COOH]}/6.8x109
[OH -] = 99/1*6.8x109 = 10-7/6,8 M (approximately)
I think your book has made a mistake : ( answer : 6,8x10-7)
don't worry about the mistake.
But no need specified concentrations for the benzoic acid and benzoate anion, because 99%x(C)/1%(C) = 99/1.
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thanks, guys! :Lighten: