How TO Calculate The pH of Aspirin Solution And Calculate The pH of Stoichiometric Point In Aspirin Titration
October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
A laboratory technician wants to determine the aspirin content of a headache pill by acid-base titration. Aspirin has a Ka of 3.0 x 10-4 M. If the pill is dissolved in water to give a solution about 0.005 M, what is the pH of this solution? (Neglect dilution effects.)
If the solution in the problem above is then titrated against KOH solution, what will be the pH at the stoichiometric point.
Chemistry Tutor Site - Chemistry Homework Answer
With this Ka and concentration, you can’t use the usual formula for weak-acid equilibrium
HA -> H+ + A-
HA dissociated into H+ and A-, if the X mol of the HA is dissociated thus the H+ and A- would be X mol too. Thus
X2 / [Acid conc] = Ka
because you can’t neglect the amount of acid actually reacted to H+ and the Cation. So you have the more general equation
X2 /[0.005-X]= 3×10-4
You can solve this by trial/error or by the general quad. eqtn method. Roughly, X=0.0027
At the stoichiometric point, all acid is dissociated, and the problem is equivalent to dissolving 0.005 M potassium salicylate in water. In that case,
Salicylate- + H2O = Acid + OH-
In this situation, the equilibrium expression is
[Acid][OH-]/[Salicylate] = Kb
Since KbKa=10×10-15, Kb= 3.3×10-11/p>
If X is the amount of [OH-] formed at equil., then
X2/[0.005]= 3.3×10-11
X is roughly 4×10-7. From this you can compute pOH= 7-log 4= 6.4. Then since pH+pOH=14, pH=7.6.
Entry Filed under: Titration
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