Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C)is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution. A)4.87 B)9.13 C)9.74 D)10.74

Question

Quizplus · Accepted Answer

Step 1: Write the equation for the dissociation of aniline.
C6H5NH2(aq) + H2O(l) ↔ C6H5NH3+(aq) + OH-(aq)

Step 2: Write the expression for the equilibrium constant (Kb) for the above reaction.
Kb = [C6H5NH3+][OH-]/[C6H5NH2]

Step 3: Calculate the concentration of aniline in the solution.
Moles of aniline = 3.90 g / 93.13 g/mol = 4.18 × 10^-2 mol
Volume of solution = 100 mL = 0.1 L
Concentration of aniline = moles/volume = 4.18 × 10^-2 mol / 0.1 L = 0.418 M

Step 4: Calculate the concentration of hydroxide ions (OH-).
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
Assume that initially, [OH-] = 0 (since aniline is a weak base and will not completely dissociate into ions).
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
4.3 × 10^-10 = x^2 / (0.418 - x)
Assuming x << 0.418, then (0.418 - x) ≈ 0.418.
x = [OH-] = 6.416 × 10^-6 M

Step 5: Calculate the pOH and pH of the solution.
pOH = -log[OH-] = -log(6.416 × 10^-6) = 5.19
pH = 14 - pOH = 8.81

Step 6: Apply the correction factor for the ionization of water.
Since aniline is a weak base, we need to correct for the ionization of water. We can assume that [H+] = [OH-] = 6.416 × 10^-6 M (at 25°C). Therefore, pKa = 14 - pKb = 14 - 4.87 = 9.13
pH = pKa + log([C6H5NH3+]/[C6H5NH2]) = 9.13 + log(0.017) = 9.13 + (-1.77) = 7.36
But remember that we need to apply the correction factor for the ionization of water.
pH = 7.36 + log([H+]/[OH-])
Since [H+] = [OH-] = 6.416 × 10^-6 M,
pH = 7.36 + log(1) = 7.36

Therefore, the pH of the aniline solution is 7.36. The closest answer choice is B) 9.13, but this is not the correct answer.

Aniline, (C6H5NH2,Kb = 4

Aniline, (C₆H₅NH₂,K_b = 4