91影视

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鈦� and Cl鈦� ions. Neither the Na鈦� nor the Cl鈦� ions react with water to produce H鈦� or OH鈦� ions. Hence, the resulting solution is neutral. For NaF: When NaF dissolves in water, it dissociates completely into Na鈦� and F鈦� ions. Fluoride ions (F鈦�) react with water to form Hydrofluoric acid (HF) and Hydroxide ions (OH鈦�), causing the solution to be basic. For NaOCl: When NaOCl dissolves in water, it dissociates completely into Na鈦� and OCl鈦� ions. Hypochlorite ions (OCl鈦�) react with water to form Hypochlorous acid (HOCl) and Hydroxide ions (OH鈦�), 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鈦� in water is given by: \[F鈦�(aq) + H鈧侽(l) \rightleftharpoons HF(aq) + OH鈦�(aq)\] Let's assume 'x' moles of F鈦� reacts with water to form 'x' moles of HF and 'x' moles of OH鈦�. Now, the equilibrium concentration of F鈦� is given by: \[[F鈦籡 = \frac{(0.050鈭抶)~\text{mol}}{0.5~\text{L}}\] The concentration of OH鈦� at equilibrium is given by: \[[OH鈦籡 = \frac{x~\text{mol}}{0.5~\text{L}}\] Since we are dealing with basic solution, we know that: \[K_b = \frac{[HF][OH鈦籡}{[F鈦籡} = K_w/K_a\] Where, K_b is the base ionization constant, K_w is the ion product of water (\(1.0 脳 10鈦宦光伌\) at 25掳C) and K_a is the acid ionization constant of HF (\(6.8 脳 10鈦烩伌\)). Calculate K_b: \[K_b = \frac{K_w}{K_a} = \frac{1.0 脳 10鈦宦光伌}{6.8 脳 10鈦烩伌} = 1.47 脳 10鈦宦孤筡] We'll assume that x is very small compared to 0.050, hence (0.050 - x) 鈮� 0.050. Plug the concentrations into the K_b expression: \[1.47 脳 10鈦宦孤� = \frac{(x)(x)}{(0.050)}\] Solve for x: \[x = [OH鈦籡 = 1.72 脳 10鈦烩伓\] Now, we'll calculate the corresponding pOH: \[pOH = -log_{10} [OH鈦籡 = -log_{10}(1.72 脳 10鈦烩伓) = 5.76\] 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鈦� in water: \[OCl鈦�(aq) + H鈧侽(l) \rightleftharpoons HOCl(aq) + OH鈦�(aq)\] K_a of HOCl is \(2.9 脳 10鈦烩伕\). Calculate K_b: \[K_b = \frac{K_w}{K_a} = \frac{1.0 脳10鈦宦光伌}{2.9 脳 10鈦烩伕} = 3.45 脳 10鈦烩伔\] We'll assume that x is very small compared to 0.050, hence (0.050 - x) 鈮� 0.050. Plug the concentrations into the K_b expression: \[3.45 脳 10鈦烩伔 = \frac{(x)(x)}{(0.050)}\] Solve for x: \[x = [OH鈦籡 = 1.31 脳 10鈦烩伒\] Now, we'll calculate the corresponding pOH: \[pOH = -log_{10} [OH鈦籡 = -log_{10}(1.31 脳 10鈦烩伒) = 4.88\] 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.