Question

Aniline hydrochloride, [C6H5NH3]Cl, is a weak acid. Its conjugate base is the weak base aniline, C6H5NH2. Ka = 2.40 X 10-5. Calculate pH at points if 50 mL of .100 aniline hydrochloride is titrated with .185 M NaOH Calculate the pH (a)before titration begins (b)at E.P (c) at the midpoint (d) after 20 mL of NaOH (e) after 30 mL of NaOH

Answer 1

a)Initially, we have aniline hydrochloride in the titration flask. We shall denote it by HA. HA will dissociate as follows.

The concentrations after dissociation are as follows.

Now we calculate the concentration x using Ka.

We calculate the pH using the H+ concentration calculated.

b) The equivalence volume should be calculated first.

At the equivalence point, we have the conjugate base A- of the weak acid. It hydrolyses to form,

This is the hydrolysis of the conjugate base formed.

The concentration of A- at the equivalence point,

Now continuing the calculation for the concentration of OH-.

Calculate the pOH value and then the pH.

c) At the midpoint half of the HA is neutralized. Thus,

Using the Henderson-Hasselbach equation,

d) At 20ml of NaOH, the concentration of HA remaining is,

As the dissociation of HA is negligible, using henderson-hasselbach equation,

e)After 30ml of NaOH, an excess of OH- will be present in the titration flask,