A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.55

Question

Determine the concentration of C6H5NH+3 in the solution if the concentration of C6H5NH2 is 0.260 M. The p𝐾b of aniline is 9.13.

[C6H5NH+3]= 0.054 (solved)

i cant seem to solve part b which is

B. Calculate the change in pH of the solution, ΔpH, if 0.352 g NaOH is added to the buffer for a final volume of 1.35 L.Assume that any contribution of NaOH to the volume is negligible.

ΔpH=

Answer 1

The NaOH will react with the conjugate acid (C₆H₅NH₃⁺) to reduce it's amount and to increase the amount of free base (C₆H₅NH₂). Moles of NaOH added = 0.352 g x 1 mol NaOH/40 g = 0.0088 moles. Concentration of OH- added = 0.0088 moles / 1.35 L = 0.00652 M.

C₆H₅NH₃⁺ + OH^- ==> C₆H₅NH₂ + H2O

@equilibrium we will have...

0.054 M - 0.00652 M = 0.0475 M C₆H₅NH₃⁺

0.260 M + 0.00652 M = 0.267 M C₆H₅NH₂

Using the Henderson Hasselbalch equation

pOH = pKb + log [conj.acid]/[base]

pOH = 9.13 + log (0.0475/0.267) = 9.13 + (-0.75)

pOH = 8.38

pH = 14 - 8.38

pH = 5.62

∆pH = 5.62 - 5.55 = +0.07 pH units