Question

A laboratory technician wants to determine the aspirin content of a headache pill by acid-base titration. Aspirin has a K_a of 3.0 x 10⁻⁴. If the pill is dissolved in water to give a solution about 0.0050 M, what is the pH of this solution? (Neglect dilution effects.)

2.97

If the solution in the problem above is then titrated against KOH solution, what will be the pH at the stoichiometric point

.

Answer 1

Let asprin, a weak acid is represented by HA

HA Ž Htta 0.005 0.005-2 a n Ka = [HA][AT [HA] 3x10 = x² 0.005-n Solving the quadratic equation, x² + 3x107) n - 0 0 15 x 10 =0 x -3x10Y 113 x 1004)*744160.0158109) - 2 x > 0.001084 Hence, [ht] = 0.001084 M pH = -log [tt = -log[000lo 84] = 2.96

Concentration of H+=0.001084

pH=-log[H+]

pH=2.96

(B)

Ka=3 *10-4

pKa =-log(ka)

pKa=3.52

At stoichiometric point all of asprin will be neutralised by strong base KOH which will result in formation of potassium salicylate.

Then this will be case of hydrolysis of salt of weak acid and strong base

Moreover assuming concentration of KOH same as Asprin as concentration of KOH is not given

Then, concentration of salt will be half of concentration of acid as total volume of solution will be volume of acid + volume of base

As concentration of both acid and base are same so concentration of salt will be halfed of acid Or Asprin

Hence, concentration

=0.005/2

=0.0025M

pH of salt of strong base and weak acid is given by

pH=1/2{pka+pkw +log(concentration of salt) }

Rest solution you can see in the pic

pH=7.45

HAT KOK A + H₂O KAT H₂O HA + oH pH = I lpk wt pka t log C) Xa= 3 x 164 - log Ka = -log(3x104) pka - 3.52 Since concentration of KOH is not it to be same as. H A = 0.005 given , assuming con Concentration of salt = 0.005 -0.0025 pHat (14 + 3:52 + log (0.0025)) = 1 / 14.917) pH = 7.45