An unknown salt is either \(\mathrm{NaF}, \mathrm{NaCl},\) or \(\mathrm{NaOCl}\). When \(0.050 \mathrm{~mol}\) of the salt is dissolved in water to form \(0.500 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(8.08 .\) What is the identity of the salt?

In this step, we will determine if the resulting solution is acidic, basic, or neutral for each of the salts - NaF, NaCl, and NaOCl. For NaCl: When NaCl dissolves in water, it dissociates completely into Na⁺ and Cl⁻ ions. Neither the Na⁺ nor the Cl⁻ ions react with water to produce H⁺ or OH⁻ ions. Hence, the resulting solution is neutral. For NaF: When NaF dissolves in water, it dissociates completely into Na⁺ and F⁻ ions. Fluoride ions (F⁻) react with water to form Hydrofluoric acid (HF) and Hydroxide ions (OH⁻), causing the solution to be basic. For NaOCl: When NaOCl dissolves in water, it dissociates completely into Na⁺ and OCl⁻ ions. Hypochlorite ions (OCl⁻) react with water to form Hypochlorous acid (HOCl) and Hydroxide ions (OH⁻), causing the solution to be basic. From the step 1 result, NaF and NaOCl are potential candidates for the unknown salt since their resulting solutions are basic.

Let's calculate the pH value for the resulting solutions of NaF and NaOCl separately. For NaF: The reaction of F⁻ in water is given by: \[F⁻(aq) + H₂O(l) \rightleftharpoons HF(aq) + OH⁻(aq)\] Let's assume 'x' moles of F⁻ reacts with water to form 'x' moles of HF and 'x' moles of OH⁻. Now, the equilibrium concentration of F⁻ is given by: \[[F⁻] = \frac{(0.050−x)~\text{mol}}{0.5~\text{L}}\] The concentration of OH⁻ at equilibrium is given by: \[[OH⁻] = \frac{x~\text{mol}}{0.5~\text{L}}\] Since we are dealing with basic solution, we know that: \[K_b = \frac{[HF][OH⁻]}{[F⁻]} = K_w/K_a\] Where, K_b is the base ionization constant, K_w is the ion product of water (\(1.0 × 10⁻¹⁴\) at 25°C) and K_a is the acid ionization constant of HF (\(6.8 × 10⁻⁴\)). Calculate K_b: \[K_b = \frac{K_w}{K_a} = \frac{1.0 × 10⁻¹⁴}{6.8 × 10⁻⁴} = 1.47 × 10⁻¹¹\] We'll assume that x is very small compared to 0.050, hence (0.050 - x) ≈ 0.050. Plug the concentrations into the K_b expression: \[1.47 × 10⁻¹¹ = \frac{(x)(x)}{(0.050)}\] Solve for x: \[x = [OH⁻] = 1.72 × 10⁻⁶\] Now, we'll calculate the corresponding pOH: \[pOH = -log_{10} [OH⁻] = -log_{10}(1.72 × 10⁻⁶) = 5.76\] And finally, we'll calculate the pH of the NaF solution: \[pH = 14 - pOH = 14 - 5.76 = 8.24\] For NaOCl: Let's follow the same process as we did for NaF, using the reaction of OCl⁻ in water: \[OCl⁻(aq) + H₂O(l) \rightleftharpoons HOCl(aq) + OH⁻(aq)\] K_a of HOCl is \(2.9 × 10⁻⁸\). Calculate K_b: \[K_b = \frac{K_w}{K_a} = \frac{1.0 ×10⁻¹⁴}{2.9 × 10⁻⁸} = 3.45 × 10⁻⁷\] We'll assume that x is very small compared to 0.050, hence (0.050 - x) ≈ 0.050. Plug the concentrations into the K_b expression: \[3.45 × 10⁻⁷ = \frac{(x)(x)}{(0.050)}\] Solve for x: \[x = [OH⁻] = 1.31 × 10⁻⁵\] Now, we'll calculate the corresponding pOH: \[pOH = -log_{10} [OH⁻] = -log_{10}(1.31 × 10⁻⁵) = 4.88\] And finally, we'll calculate the pH of the NaOCl solution: \[pH = 14 - pOH = 14 - 4.88 = 9.12\]

We found that the pH of NaF solution is 8.24, and the pH of NaOCl solution is 9.12. The given pH of the unknown salt's solution is 8.08. The pH value of 8.08 is closer to the pH of the NaF solution (8.24) than to the pH of the NaOCl solution (9.12). Therefore, the identity of the unknown salt is NaF.