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