An unknown salt is either $\mathrm{NaCN}, \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},$ NaF, NaCl, or NaOCl. When 0.100 mole of the salt is dissolved in 1.00 \(\mathrm{L}\) of solution, the pH of the solution is \(8.07 .\) What is the identity of the salt?

For each salt, we can outline the following dissociation reactions in water: 1. NaCN: \(CN^-(aq) + H_2O(l) \rightleftharpoons HCN(aq) + OH^-(aq)\) 2. NaC2H3O2: \(C_2H_3O_2^-(aq) + H_2O(l) \rightleftharpoons HC_2H_3O_2(aq) + OH^-(aq)\) 3. NaF: \(F^-(aq) + H_2O(l) \rightleftharpoons HF(aq) + OH^-(aq)\) 4. NaCl: (No relevant reaction because it doesn't affect pH) 5. NaOCl: \(OCl^-(aq) + H_2O(l) \rightleftharpoons HOCl(aq) + OH^-(aq)\) The resulting OH- concentration will affect the pH, so we'll focus on that.

For each salt, we can write the base-dissociation constant expression (\(K_b\)) as follows: 1. NaCN: \[K_{b1} = \frac{[HCN][OH^-]}{[CN^-]}\] 2. NaC2H3O2: \[K_{b2} = \frac{[HC_2H_3O_2][OH^-]}{[C_2H_3O_2^-]}\] 3. NaF: \[K_{b3} = \frac{[HF][OH^-]}{[F^-]}\] 5. NaOCl: \[K_{b4} = \frac{[HOCl][OH^-]}{[OCl^-]}\] We'll use \(K_b\) values and stoichiometry to find the \(pOH\) of the solutions.

We have the given pH value: \(pH = 8.07\) The relationship between pH and pOH is given by: \(pH + pOH = 14\) Now we can calculate the pOH: \(pOH = 14 - 8.07 = 5.93\) Then, we can find the concentration of OH-: \[ [OH^-] = 10^{-pOH} = 10^{-5.93} \Rightarrow [OH^-] \approx 1.19 \times 10^{-6} M\]

We will assume that the amount of OH- in all reactions comes only from their corresponding salts. By substituting the values of \([OH^-]\), most of the \(K_b\) expressions can be simplified. For example, in case of NaCN, we have: \[K_{b1} = \frac{[HCN][1.19 \times 10^{-6}]}{[CN^-]} \Rightarrow K_{b1}[CN^-]=[HCN][1.19 \times 10^{-6}]\] Now we will use stoichiometry for each salt and eliminate the corresponding ion concentration. Since we have 0.100 moles of each salt in 1L of water, the initial concentration of each of them is 0.100M. Therefore, we can substitute these values into the expressions derived above. In the end, we should find a salt whose \(K_b\) matches the given OH- concentration. 1. For NaCN: \[K_{b1}(0.100 - x)=x(1.19 \times 10^{-6})\] 2. For NaC2H3O2: \[K_{b2}(0.100 - x)=x(1.19 \times 10^{-6})\] 3. For NaF: \[K_{b3}(0.100 - x)=x(1.19 \times 10^{-6})\] 5. For NaOCl: \[K_{b4}(0.100 - x)=x(1.19 \times 10^{-6})\]

Now we need to check which of these salts give a value of x (1.19 x 10^-6) close to their experimentally determined \(K_b\). After checking a reference table for the \(K_b\) values of the ions and evaluating the expressions, we can conclude that the salt is: NaF, since its \(K_b\) value, F- ions and given concentration match the resulting pH. Identity of the unknown salt: NaF