A buffer solution contains 0.397 M C6H5NH3Br and 0.327 M C6H5NH2 (aniline). Determine the pH change when 0.090 mol HBr is added to 1.00 L of the buffer.

pH = 4.39

Explanation:

pH of Aniline buffer is determined using the H-H equation for weak bases:

pOH = pKb + log [C6H5NH3Br] / [C6H5NH2]

Where pKb of aniline is 9.30 and [] could be taken as moles of each species

In 1.00L of buffer, initial moles of each species are:

[C6H5NH3Br] = 0.397 moles

[C6H5NH2] = 0.327 moles

The reaction of HBr with C6H5NH2 is:

C6H5NH2 + HBr → C6H5NH3Br

Thus, moles of HBr added are moles subtracted of C6H5NH2 and moles of C6H5NH3Br produced:

[C6H5NH3Br] = 0.397 moles + 0.090mol = 0.487 mol

[C6H5NH2] = 0.327 moles - 0.090mol = 0.237mol

Replacing:

pOH = 9.30 + log [0.487 mol] / [0.237mol]

pOH = 9.61

And as:

pH = 14 - pOH

pH = 4.39

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