Caculate the pH of a solution prepared by mixing 93 mL of a 0.25 M aqueous aniline solution (C6H5NH2, Kb = 4.3 x 10-10) with 100. mL of a 0.11 M aqueous nitric acid solution.

Question

I got a pH of 4.68 when I solved it but my teacher got a pH of 9.30 and I'm confused of how she got 9.30.

Answer 1

C6H5NH2 + HNO3 ==> C6H5NH3+ + NO3-

moles aniline = 0.093 L x 0.25 mol/L = 0.02325 moles

moles HNO3 = 0.1 L x 0.11 mol/L = 0.011 moles

C6H5NH2 + HNO3 ==> C6H5NH3+

0.02325........0.011.............0..........Initial

-0.011.........-0.011..........+0.011.....Change

0.01225..........0................0.011.....Equilibrium

We can use a modification of the standard Henderson Hasselbalch equation

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

pKb = -log Kb = 9.37

pOH = 9.37 + log (0.011/0.01225) = 9.37 + log 0.898

pOH = 9.37 + (-0.0467)

pOH = 9.32

pH = 14 - pOH

pH = 4.68

I have to agree with you, and not with your teacher.