Question

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

Answer 1

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 = pK_b + log [C₆H₅NH₃⁺]/[C₆H₅NH₂]

=====> pOH = -log (K_b) + 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.