The chemical name for aspirin is acetylsalicylic acid It is believed that the an

Question

The chemical name for aspirin is acetylsalicylic acid It is believed that the analgesic and other desirable propene in of aspirin are due not to the aspirin itself, but rather to the simple compound salicylic acid. C_6 H_4(OH)CO_2H, which result from the breakdown of aspirin in the stomach. Assume you have 250 mL of a 0.018 M solution of salicylic acid and titrate it with 0.010 M NaOH. What is the pH at the halfway point of the titration? What the pH at the equivalence point? K_a for the acid equals 1.35 = 10^-3 (at the halfway point) (at the equivalence point)

Explanation / Answer

Solution:- Salisylic acid is a monoprotic acid so it react with NaOH in 1:1 mol ratio.

for salicylic acid, 25.0 ml x (1L/1000ml) x (0.018mol/L) = 0.00045 mol

at equivalence point moles of NaOH will also be 0.00045.

Now we will calculate the volume of NaOH used to reach the end point..

0.00045 mol x (1L/0.010mol) = 0.045 L or 45 ml.

Halfway point is the point at which exactly half of the equivalence point volume of base is added. from calculations 45 ml of NaOH were used to reach the equivalence point. So, volume used at halfway point = 45ml/2 = 22.5ml

so, moles of NaOH used will also be half that is 0.000225.

excess moles of salicylic acid = 0.000225

moles of conjugate base formed = 0.000225

Here we have a buffer solution so pH could be calculated using Handerson equation..

pH = Pka + log(base/acid)

Ka is given as 1.35 x 10-3

Pka = - log 1.35 x 10-3 = 2.87

pH = 2.87 + log(0.000225/0.000225)

pH = 2.87

At equivalence point, there will be conjugate base(salt) in the solution,

moles of conjugate base formed = 0.000225

total volume = 25.0 ml + 45.0 ml = 70.0 ml = 0.070 L

molarity of conjugate base = 0.000225 mol/0.070 L = 0.00321 M

let's present conjugate base as A-. The reaction of it with water could be written as...

A-(aq) + H2O(l) <-------> HA(aq) + OH-(aq)

I 0.00321 0 0

C -X +X +X

E (0.00321 - X) X X

Kb = [HA][OH-]/[A-]

Kb = Kw/Ka

Kb = (1.0 x 10-14)/(1.35 x 10-3) = 7.41 x 10-12

7.41 x 10-12 = [X)2]/(0.00321 - X)

since the Kb is very low so the X on the bottom could be neglected.

7.41 x 10-12  = [X)2]/(0.00321]

on solving this,

X = 1.54 x 10-7

This X is the concentration of OH- so we will calclate pOH with the help of this.

pOH = - log(1.54 x 10-7)

pOH = 6.81

pH = 14 - 6.81

pH = 7.19

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