CHEM HELP PLEASE!!! A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.64.?

(a)
pKb = -log(Kb) = 9.13
Thus, Kb = 10⁻⁹˙¹³

pH = 5.64
pOH = 14 - pH = 14 - 5.64 = 8.36
Then, [OH⁻] = 10⁻⁸˙³⁶

C₆H₅NH₂(aq) + H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq) ...... Kb = 10⁻⁹˙¹³

At equilibrium :
Kb = [C₆H₅NH₃⁺] [OH⁻] / [C₆H₅NH₂]
10⁻⁹˙¹³ = [C₆H₅NH₃⁺] × 10⁻⁸˙³⁶ / 0.335
[C₆H₅NH₃⁺] = 10⁻⁹˙¹³ × 0.335 / 10⁻⁸˙³⁶ = 0.0569 M

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(b)
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
No. of moles of NaOH = (0.366 g) / (40.00 g/mol) = 0.00915 mol
Molarity of NaOH in the solution before reaction = (0.00915 mol) / (1.10 L) = 0.00832 M

Reaction taking place after the addition of NaOH :
C₆H₅NH₃⁺(aq) + OH⁻(aq) → C₆H₅NH₂(aq) + H₂O(l)

Neglecting the dissociation of C₆H₅NH₂, after the above reaction :
[C₆H₅NH₃⁺]ₒ = 0.0569 - 0.00832 = 0.0486 M
[C₆H₅NH₂]ₒ = 0.335 + 0.00832 = 0.343 M

Consider the dissociation of C₆H₅NH₂ :
C₆H₅NH₂(aq) + H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq) ...... pKb = 9.13
Because of the small Kb and the common ion effect due to the presence of C₆H₅NH₃⁺ ions, the dissociation of C₆H₅NH₂ is negligible.
[C₆H₅NH₂] ≈ [C₆H₅NH₂]ₒ = 0.343 M
[C₆H₅NH₃⁺] ≈ [C₆H₅NH₃⁺]ₒ = 0.0486 M

After the addition of NaOH :
pOH = pKb - log([C₆H₅NH₂]/[C₆H₅NH₃⁺]) = 9.13 - log(0.343/0.0486) = 8.28
pH = 14 - pOH = 14 - 8.28 = 5.72

Change in pH, ΔpH = 5.72 - 5.64 = +0.08

a)
pH = 5.64 = pKb - lg(HA/0.335) = 4.87 - lg(HA/0.335)

lg(HA/0.335) = -0.77

HA = 0.057 M