Question

Since the patient was brought into the emergency room only two hours

after the overdose, you suspect that her stomach might contain undissolved

aspirin that is continuing to be absorbed. The fact that she is

experiencing severe respiratory ankylosis 10 hours after the ingestion

confirms your suspicion and you decide to use a gastric lavage at pH 8.5

to effectively remove any undissolved aspirin. This treatment solubilizes

the aspirin so that it can easily be removed from the stomach.

a. Calculate the percentage of protonated and unprotonated forms of salicylic acidic the pH of the stomach, which is usually around 2.0?

b) Calculate the percentage of protonated and unprotonated forms of salicylic acid at the pH of gastric lavage. How does the gastric lavage at pH 8.5 facilitate the removal of aspirin from the stomach? Explain. (Note: Assume that the pKa values for the carboxylate group in salicylic acid and acetylsalicylic acid are the same.)

Answer 1

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Answer 2

#2.(a): Given pH = - log[H⁺(aq)] = 2.0

=> [H⁺(aq)] = 10^-2.0 = 1.0*10^-2 M or 0.01 M

For aspirin, Ka = 10^-2.97 = 1.0715*10^-3

Chemical equation for the dissociation of Aspirin (C₉H₈O₄) is:  

--C₉H₈O₄(aq) <-----> C₉H₇O₄^-(aq) + H⁺(aq) ; Ka = 1.0715*10^-3

E: [C₉H₈O₄(aq)] ------ [C₉H₇O₄^-(aq)], 0.01 M

(Protonated) ------- (unprotonated)

Ka = 1.0715*10^-3 = [C₉H₇O₄^-(aq)] * [H⁺(aq)] / [C₉H₈O₄(aq)]

=> 1.0715*10^-3 = ([C₉H₇O₄^-(aq)] * 0.01) / [C₉H₈O₄(aq)]

=>  [C₉H₈O₄(aq)] / [C₉H₇O₄^-(aq)] = 0.01 / 1.0715*10^-3

=>  [C₉H₈O₄(aq)] / [C₉H₇O₄^-(aq)] = 9.3325

So protonated form is 9.3325  times larger than the unprotonated form i.e if protonated form is 9.3325X, then unprotonated form is X.

Hence % of protonated form = [9.3325X / (9.3325X+X)]*100 = (9.3325X / 10.3325 X)*100 = 90.3 % (Answer)

% of unprotonated form = 100 - 90.3 = 9.7% (Answer)

(b): pH of gastric lavage = 8.5

=> - log[H⁺(aq)] = 8.5

=> [H⁺(aq)] = 10^-8.5 = 3.16*10^-9 M

Chemical equation for the dissociation of Aspirin (C₉H₈O₄) is:  

--C₉H₈O₄(aq) <-----> C₉H₇O₄^-(aq) + H⁺(aq) ; Ka = 1.0715*10^-3

E: [C₉H₈O₄(aq)] ------ [C₉H₇O₄^-(aq)], 3.16*10^-9 M

(Protonated) ------- (unprotonated)

Ka = 1.0715*10^-3 = [C₉H₇O₄^-(aq)] * [H⁺(aq)] / [C₉H₈O₄(aq)]

=> 1.0715*10^-3 = ([C₉H₇O₄^-(aq)] * 3.16*10^-9) / [C₉H₈O₄(aq)]

=>  [C₉H₈O₄(aq)] / [C₉H₇O₄^-(aq)] = 3.16*10^-9 / 1.0715*10^-3

=>  [C₉H₈O₄(aq)] / [C₉H₇O₄^-(aq)] = 2.95*10^-6  

or [C₉H₇O₄^-(aq)] /  [C₉H₈O₄(aq)] = 1 / 2.95*10^-6 = 338838

So unprotonated form is 338838 times larger than the protonated form i.e if unprotonated form is 338838X, then unprotonated form is X.

Hence % of unprotonated form(C₉H₇O₄^-(aq)) = [338838X / (338838X+X)]*100 = (338838X / 338839X)*100 = 99.999705 % (Answer)

% of protonated form (C₉H₈O₄(aq)) = 100 - 99.999705 % = 2.95*10^-6 % (Answer)