Question

# Find thepH after 0.25 moles of HI is added to an aqueous solution of 0.50 moles...

Find thepH after 0.25 moles of HI is added to an aqueous solution of 0.50 moles of aniline (Kb= 3.9 × 10–10) _______________.

(a)3.41(b)4.59(c)10.59(d)11.39(e)17.41

Aniline, C6H5NH2 is a weak base and reacts with HI as below.

C6H5NH2 (aq) + HI (aq) ----------> C6H5NH3+Cl- (aq)

As per the stoichiometric equation,

1 mole C6H5NH2 = 1 mole C6H5NH3+ (conjugate acid of weak base, C6H5NH2)

= 1 mole HI.

Therefore,

0.25 mole HI = 0.25 mole C6H5NH2 = 0.25 mole C6H5NH3+

Therefore,

mole(s) of unreacted C6H5NH2 = (0.50 – 0.25) mole = 0.25 mole.

Let V L be the volume of the solution.

Use Henderson-Hasslebach equation for base.

pOH = pKb + log [C6H5NH3+]/[C6H5NH2]

=====> pOH = -log (Kb) + log [(0.25 mole)/(V L)]/[(0.25 mole)/(V L)]

=====> pOH = -log (3.90*10-10) + log (1.00)

=====> pOH = 9.409 + 0.0 = 9.409

We know that

pH + pOH = 14

Therefore,

pH = 14 – pOH

= 14 – 9.409

= 4.591 ≈ 4.59

Option (b) is the correct answer.

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