Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

Since the salt contains C₅H₅NH⁺ and F⁻ ions, and we are given that the principal equilibrium is C₅H₅NH⁺(aq) + F⁻(aq) ⇌ C₅H₅N(aq) + HF(aq), we will use this reaction to analyze the species involved.

Given a 0.200 M solution of C₅H₅NHF, the initial concentrations of C₅H₅NH⁺ and F⁻ ions are: \[ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}]_{0} = [\mathrm{F}^{-}]_{0} = 0.200 \, \mathrm{M} \] Since the reaction has not yet taken place, the initial concentrations of C₅H₅N and HF are both zero.

Using the given equilibrium constant K, we can write an equilibrium expression for the reaction: \[ K = \frac{ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}][\mathrm{HF}]}{ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{F}^{-}] } \] During the course of the reaction, the change in concentration of C₅H₅NH⁺ and F⁻ is equal. So, we can denote it as minus-x, which means the decrease in their concentrations. The change in concentration of C₅H₅N and HF is equal as well, and it can be denoted as plus-x, which means the increase in their concentrations.

Now, we substitute the initial concentrations and the change in concentrations into the K expression: \[ K = \frac{ (0.200 + x)(x)}{ (0.200 - x)^2 } \] Given K = 8.2 × 10⁻³, we can solve for x: \[ 8.2 \times 10^{-3} = \frac{x(0.200 + x)}{(0.200 - x)^2} \] This quadratic equation can be solved to find the value of x. After an approximation, we get x ≈ 0.00134.

As we found the change in concentrations, we can calculate the [HF] concentration in the solution: \[ [\mathrm{HF}] = [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}] = x \approx 0.00134 \, \mathrm{M} \] Since HF is a weak acid, we can use its expression for Kₐ to find the H⁺ concentration in the solution: \[ K_\mathrm{a} = 6.6 \times 10^{-4} = \frac{ [\mathrm{H}^{+}][\mathrm{F}^{-}]}{ [\mathrm{HF}] } = \frac{ [\mathrm{H}^{+}]}{ 0.00134 } \] Solving for H⁺ concentration, we get: \[ [\mathrm{H}^{+}] \approx 8.8 \times 10^{-7} \, \mathrm{M} \] Now, we can calculate the pH using the formula: \[ \mathrm{pH} = -\log_{10} [\mathrm{H}^{+}] \] Hence, the pH of the solution is: \[ \mathrm{pH} \approx -\log_{10} (8.8 \times 10^{-7}) \approx 6.06 \] So, the pH of the 0.200 M C₅H₅NHF solution is approximately 6.06.