Question

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?

Short answer

The unknown salt is NaF. We determined that NaF and NaOCl are basic salts, resulting in basic solutions when dissolved in water. We calculated the pH for NaF and NaOCl solutions as 8.24 and 9.12, respectively. Since the given pH of the unknown salt's solution is 8.08, which is closer to the pH of the NaF solution, we concluded that the identity of the unknown salt is NaF.

Step-by-step solution

Determine if the salts are acidic, basic, or neutral when dissolved in water

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.

Calculate the pH for NaF and NaOCl

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: \[pH = 14 - pOH = 14 - 4.88 = 9.12\]

Identify the unknown salt based on calculated pH values

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.

Key concepts

Acid-Base Chemistry

Acid-base chemistry is all about understanding how acids and bases behave in solution. When a substance dissolves in water, it can affect the pH level depending on whether it is an acid, a base, or neutral. In basic solutions, like those formed when salts like NaF or NaOCl dissolve in water, there are more hydroxide ions (OH⁻) than hydrogen ions (H⁺). This results in a pH greater than 7.
When NaF is dissolved, the fluoride ions (F⁻) interact with water in an equilibrium reaction: \( F⁻ + H_2O \rightleftharpoons HF + OH⁻ \). This produces hydroxide ions, making the solution basic. Similarly, NaOCl forms a basic solution through the reaction \( OCl⁻ + H_2O \rightleftharpoons HOCl + OH⁻ \).
The pH value tells us about the solution's acidity or basicity. A pH of 8.08, as observed in the exercise, is slightly basic, indicating the presence of such ions.

Chemical Equilibrium

Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal, meaning that the concentrations of reactants and products remain constant over time. In this context, the reaction of the ions with water achieves equilibrium, influencing the pH of the solution.
For NaF, equilibrium is reached with the fluoride ions converting to HF and hydroxide ions: \( F⁻ + H_2O \rightleftharpoons HF + OH⁻ \). The extent of this reaction is described by the base ionization constant, \( K_b \).
Similarly, NaOCl establishes an equilibrium: \( OCl⁻ + H_2O \rightleftharpoons HOCl + OH⁻ \). Each salt has a specific \( K_b \) value indicative of its ability to produce hydroxide ions. Calculating these equilibria allows us to understand the pH of the resulting solution, as seen with the pH values around 8.08 and 9.12.

Solution Chemistry

Solution chemistry deals with the study of solutes dissolved in a solvent, often water, to form a solution. Understanding the composition and concentration of a solution is crucial in calculating properties like pH.
The exercise involves dissolving 0.050 mol of a salt in 0.500 L of water. This sets the stage for analyzing how much the dissolved ions will shift the chemical equilibrium and impact the solution's pH. The concentration of ions like OH⁻ at equilibrium is key to this analysis.
In the example, both NaF and NaOCl dissolve to generate OH⁻ ions, affecting the pH. Knowing the molar concentrations and using ionization constants, you can predict the pH by calculating the concentrations of ions in the solution: \( [OH⁻] = \frac{x}{0.5~L} \). This approach allows students to determine the salt behind a given pH.