A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochlo
ID: 1089036 • Letter: A
Question
A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.59. 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, pH, if 0.397 g NaOH is added to the buffer for a final volume of 1.10 L. Assume that any contribution of NaOH to the volume is negligible.
I'm stuck on the part b, if all the steps to part b can be posted that would help a lot!
Explanation / Answer
a)
pH = 5.59
pOH = 8.41
pKb = 9.13
pOH = pKb + log [salt / base]
8.41 = 9.13 + log [salt / 0.200]
[salt / 0.200] = 0.1905
concentration of salt = 0.0381
concentration of [C6H5NH3] = 0.0381 M
b)
moles of NaOH = 0.397 / 40 = 0.009925 mol
moles of base = 0.2 x 1.10 = 0.22
moles of salt = 0.0381 x 1.10 = 0.0419
pOH = pKb + log [salt - C / base + C]
= 9.13 + log [0.0419 - 0.009925 / 0.22 + 0.009925]
pOH = 8.27
pH = 5.73
pH = 5.73 - 5.59
pH = 0.137
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