Question

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.47.

a) Determine the concentration of C6H5NH3+ in the solution if the concentration of C6H5NH2 is 0.200 M. The pKb of aniline is 9.13.

b) Calculate the change in pH of the solution, delta pH , if 0.393 g NaOH is added to the buffer for a final volume of 1.55 L. Assume that any contribution of NaOH to the volume is negligible.

Answer 1

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.47.

a) Determine the concentration of C6H5NH3+ in the solution if the concentration of C6H5NH2 is 0.200 M. The pKb of aniline is 9.13.

Solution :-

pH= 5.47

pOH = 14 – pH = 14 – 5.47 = 8.53

Henderson equation

pOH= pkb + log [acid /base]

8.53 = 9.13 + log [acid / 0.200]

8.53 -9.13 = log [acid / 0.200]

-0.6 = log [acid / 0.200]

Antilog [-0.6] =[acid / 0.200]

0.2511 = [acid / 0.200]

[acid] = 0.2511 * 0.200 M

[acid] = 0.0502 M

So the concentration of the C6H5NH3+ = 0.0502 M

b) Calculate the change in pH of the solution, delta pH , if 0.393 g NaOH is added to the buffer for a final volume of 1.55 L. Assume that any contribution of NaOH to the volume is negligible.

Solution :- lets calculate the moles of NaOH

Moles of NaOH = 0.393 g / 40.0 g per mol = 0.009825 mol

Now lets calculate the moles of each species in the buffer

Moles of C6H5NH2 = 0.200 mol per L * 1.55 L = 0.31 mol

Moles of C6H5NH3+ = 0.0502 mol per L * 1.55 L = 0.07781 mol

After adding the NaOH it will react with conjugate acid and forms the base so

New moles of C6H5NH2 = 0.31 mol + 0.009825 mol = 0.319825 mol

New moles of C6H5NH3+ = 0.07781 mol – 0.009825 mol = 0.067985 mol

Now lets calculate the new molarities

[C6H5NH2] = 0.319825 mol / 1.55 L = 0.20634 M

[C6H5NH3+] = 0.067985 mol / 1.55 L = 0.04386 M

Now lets calculate the pOH

pOH = pka + log [acid /base]

        = 9.13 + log [0.04386 / 0.20634]

        = 8.46

pH= 14 – pOH

     = 14 – 8.46

     = 5.54

So the change in the pH = final pH – initial pH

                                           = 5.54-5.47

                                           = 0.07