When aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=7.4 \times 10^{-10}\right)\), reacts with a strong acid, its conjugate acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\), is formed. Calculate the \(\mathrm{pH}\) of a \(0.100 \mathrm{M}\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) and compare it with the \(\mathrm{pH}\) of acetic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.86 \times 10^{-5}\right)\)

Using the given Kb of aniline, we can find out the Ka of its conjugate acid, C₆H₅NH₃⁺, using the relationship between Ka, Kb, and Kw: Ka × Kb = Kw Kw = 1.0 × 10⁻¹⁴ (self-ionization constant of water) Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (7.4 × 10⁻¹⁰) Ka = 1.35 × 10⁻⁵ Now we have the Ka value for C₆H₅NH₃⁺.

Since 0.100 M C₆H₅NH₃⁺ is a weak acid, we can use an equilibrium expression to determine the H⁺ concentration. The reaction can be represented as follows, with HA representing C₆H₅NH₃⁺ for simplicity: HA <-> H⁺ + A⁻ At the beginning, the concentrations will be [HA]=0.100 M and [H⁺]=[A⁻]=0. Then, HA will dissociate by a small extent, x, to produce x mol of H⁺ and A⁻ ions. So, final concentrations in the solution would be: [HA] = 0.100 - x [H⁺] = x [A⁻] = x Using Ka expression: Ka = ([H⁺][A⁻]) / [HA] 1.35 × 10⁻⁵ = (x × x) / (0.100 - x) We can assume x is quite small, so (0.100 - x) is approximately equal to 0.100. 1.35 × 10⁻⁵ ≈ x² / 0.100 x² ≈ 1.35 × 10⁻⁶ x ≈ 1.16 × 10⁻³ The H⁺ concentration, x, is approximately 1.16 × 10⁻³ M.

Now that we have the H⁺ concentration, we can calculate the pH using the pH formula: pH = -log[H⁺] pH ≈ -log(1.16 × 10⁻³) pH ≈ 2.94 The pH of the 0.100 M C₆H₅NH₃⁺ solution is approximately 2.94.

The Ka value of acetic acid is given as 1.86 × 10⁻⁵. Using this value, we will calculate the pH of a 0.100 M acetic acid (CH₃COOH) solution. The process for acetic acid is similar to that of C₆H₅NH₃⁺: CH₃COOH <-> H⁺ + CH₃COO⁻ Ka = ([H⁺][CH₃COO⁻]) / [CH₃COOH] 1.86 × 10⁻⁵ ≈ x² / 0.100 x² ≈ 1.86 × 10⁻⁶ x ≈ 1.36 × 10⁻³ The H⁺ concentration in the acetic acid solution is approximately 1.36 × 10⁻³ M. pH_acetic = -log(1.36 × 10⁻³) pH_acetic ≈ 2.87 The pH of the 0.100 M acetic acid solution is approximately 2.87. Comparing the pH values: pH(C₆H₅NH₃⁺) = 2.94 pH(acetic acid) = 2.87 The pH of the 0.100 M C₆H₅NH₃⁺ solution is slightly higher than that of the 0.100 M acetic acid solution.