Aniline, C6H5NH2, reacts with water according to the equation: C6H5NH2(aq) + H2O(l) ⇌ C6H5NH3+(aq) + — Acid-Base Equilibrium Chemistry Question
Problem Context
Aniline, C6H5NH2, reacts with water according to the equation: C6H5NH2(aq) + H2O(l) ⇌ C6H5NH3+(aq) + OH–(aq)
In a 0.180 M aqueous aniline solution the [OH–] = 8.80×10–6.
Write the equilibrium constant expression for this reaction.
Model Answer
Kb = [C6H5NH3+][OH–] / [C6H5NH2]
Determine the value of the base ionization constant, Kb, for C6H5NH2(aq).
Model Answer
Kb = (8.80×10–6)(8.80×10–6) / 0.180 = 4.3×10–10
Calculate the percent ionization of C6H5NH2 in this solution.
Model Answer
% ionization = (8.80×10–6 / 0.180) × 100% = 4.9×10–3 %
Determine the value of the equilibrium constant for the neutralization reaction;
C6H5NH2(aq) + H3O+(aq) ⇌ C6H5NH3+(aq) + H2O(l)
Model Answer
C6H5NH2 + H3O+ ⇌ C6H5NH3+ + H2O
so K = Kb / Kw = (4.3×10–10) / (1.0×10–14) = 4.3×104
Find the [C6H5NH3+(aq)] / [C6H5NH2(aq)] required to produce a pH of 7.75.
Model Answer
For a pH = 7.75, the pOH = 6.25 so
[OH-] = 10^-6.25 = 5.62×10–7 M.
4.3×10–10 = [C6H5NH3+][OH-] / [C6H5NH2] so,
[C6H5NH3+] / [C6H5NH2] = 4.3×10–10 / 5.62×10–7 = 7.65×10–4
Calculate the volume of 0.050M HCl that must be added to 250.0 mL of 0.180 M C6H5NH2(aq) to achieve this ratio.
Model Answer
The HCl is a strong acid that will protonate the aniline, so to get the HCl required, we need the amount of C6H5NH2 required multiplied by the value of the ratio from (i):
7.65×10–4 × 0.250 L × 0.180 M C6H5NH2 = 3.44×10–5 mol HCl
Now determine the volume of reagent:
3.44×10–5 mol HCl × (1 L / 0.050 mol HCl) = 6.88×10–4 L = 0.688 mL