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

First, we need to calculate the concentration of the salt in the solution. Given that 0.050 mol of salt has been dissolved in 0.500 L of water, the concentration can be found using the formula: \[Concentration = \frac{moles}{volume}\] \[Concentration = \frac{0.050}{0.500}\]

For NaF, F- will act as a base, reacting with water to form HF and OH-. Using Kb for F- (we use Kb because F- is a base) \[K_b = \frac{[OH^{-}][HF]}{[F^{-}]}\] From the formula, we know that the concentration of OH- is equal to the concentration of HF. Since initial concentration of F- was calculated in step 1, \[K_b = \frac{[OH^{-}]^2}{0.1 - [OH^{-}]}\] We know that F- is a weak base, and its dissociation is small, so we can approximate that: \[K_b = \frac{[OH^{-}]^2}{0.1}\] Now, in order to find Kb, we use the following relationship between Ka and Kb for conjugate acid-base pairs: \[K_a * K_b = K_w\] For HF, Ka = \(7.2 * 10^{-4}\) and Kw = \(1 * 10^{-14}\) \[K_b = \frac{K_w}{K_a} = \frac{1 * 10^{-14}}{7.2 * 10^{-4}} = 1.39 * 10^{-12}\] Now we can solve for the [OH-]: \[1.39 * 10^{-12} = \frac{[OH^{-}]^2}{0.1}\] \[ [OH^{-}]^2 = 1.39 * 10^{-11}\] \[ [OH^{-}] = 3.73 * 10^{-6}\] The pOH of this solution is calculated as: \[pOH = -log[OH^{-}] = -log(3.73 * 10^{-6})\] \[pOH = 5.43\] Now we can find the pH of NaF solution: \[pH = 14 - pOH = 14 - 5.43 = 8.57\]

For NaCl, Cl- is the conjugate base of a strong acid (HCl), and therefore it does not have any noticeable effect on the pH of the solution. Na+ is also the conjugate acid of a strong base (NaOH) and does not have any noticeable effect on the pH. Hence, the solution of NaCl would simply be neutral and have a pH of 7.

For NaOCl, OCl- will act as a base, reacting with water to form HOCl and OH-. Using Kb for OCl-: \[K_b = \frac{[OH^{-}][HOCl]}{[OCl^{-}]}\] From the formula, we know that the concentration of OH- is equal to the concentration of HOCl. Since the initial concentration of OCl- was calculated in step 1: \[K_b = \frac{[OH^{-}]^2}{0.1 - [OH^{-}]}\] We know that OCl- is a weak base, and its dissociation is small, so we can approximate that: \[K_b = \frac{[OH^{-}]^2}{0.1}\] Now, in order to find Kb, we use the following relationship between Ka and Kb for conjugate acid-base pairs: \[K_a * K_b = K_w\] For HOCl, Ka = \(3.2 * 10^{-8}\) and Kw = \(1 * 10^{-14}\) \[K_b = \frac{K_w}{K_a} = \frac{1 * 10^{-14}}{3.2 * 10^{-8}} = 3.125 * 10^{-7}\] Now we can solve for the [OH-]: \[3.125 * 10^{-7} = \frac{[OH^{-}]^2}{0.1}\] \[ [OH^{-}]^2 = 3.125 * 10^{-8}\] \[ [OH^{-}] = 5.59 * 10^{-5}\] The pOH of this solution is calculated as: \[pOH = -log[OH^{-}] = -log(5.59 * 10^{-5})\] \[pOH = 4.25\] Now we can find the pH of NaOCl solution: \[pH = 14 - pOH = 14 - 4.25 = 9.75\]

Now let's compare the calculated pH values for each of the salts to the given pH value (8.08): 1. NaF: pH = 8.57 2. NaCl: pH = 7 (neutral) 3. NaOCl: pH = 9.75 The salt with the pH value closest to the given pH value (8.08) is NaF (pH 8.57). Therefore, the unknown salt is NaF.