A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochlo

Question

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.25

sapling earning A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.25 a) Determine the concentration of CeH5NH3 in the solution if the concentration of CeH5NH2 is 0.290 M. The pKb of aniline is 9.13. Number M C H-NH b) Calculate the change in pH of the solution, ApH, if 0.377 g NaOH is added to the buffer for a final volume of 1.25 L. Assume that any contribution of NaOH to the volume is negligible. Number ApH Map

Explanation / Answer

a)

pH = 5.25

=> pOH =14 - 5.25 = 8.75

pKb = 9.13

We know that

pOH of a basic buffer = pKb + log (Salt / Base)

Salt = C6H5NH3Cl

Base = C6H5NH2

=> 8.75 = 9.13 + log (Salt / 0.29)

=> log (Salt / 0.29) =

=> Salt / 0.29 = 0.41687

=> Salt = 0.121 M = [C6H5NH3Cl]

b)

Mass of NaOH added = 0.377 g

=> Moles fo NaOH added = 0.377 / 40 = 0.009425 moles

Volume of solution = 1.25 L

=> [NaOH] = [OH-] added = 0.009425 / 1.25 = 0.00754 M

pOH after addition of NaOH = 9.13 + log [ (0.121 - 0.00754) / (0.29 + 0.00754) ]

=> pOH = 8.71

=> pH = 14 - 8.71 = 5.29

=> delta pH = 5.29 - 5.25 = 0.04

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