Question

An unknown salt is either NaF, \(\mathrm{NaCl}\), or \(\mathrm{NaOCl}\). When \(0.050\) 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 identity of the unknown salt is NaOCl, as the [OH-] concentration resulting from NaOCl (5.82 × 10^-6) is the closest to the given [OH-] for the pH of 8.08 (1.20 × 10^-6).

Step-by-step solution

Calculate the molarity of the salt

Since 0.050 mol of the unknown salt is dissolved in 0.500 L of water, we can calculate the molarity of the salt using the equation M = mol/L. In this case, M = (0.050 mol) / (0.500 L) = 0.100 M.

Write the dissolution equations for the unknown salts

To determine the effect of the unknown salt on the pH of the solution, we need to know how each salt dissolves in water. For NaF: \(NaF \rightarrow Na^+ + F^-\) For NaCl: \(NaCl \rightarrow Na^+ + Cl^-\) For NaOCl: \(NaOCl \rightarrow Na^+ + OCl^-\)

Determine the pH effect of each ion

In order to identify which salt would result in a solution of pH 8.08, we need to evaluate the effect of each ion on the pH of the solution. F- can act as a base that will accept a proton from water: \(F^- + H_2O \rightleftharpoons HF + OH^-\) Cl- is the conjugate base of a strong acid, HCl, and therefore, do not significantly affect the pH of the solution. OCl- can act as a weak base, accepting a proton from water: \(OCl^- + H_2O \rightleftharpoons HOCl + OH^-\)

Calculate the [OH-] at pH 8.08

To continue, we need to determine the hydroxide ion concentration ([OH-]) at the given pH. We use the following equation: pOH = 14 - pH So, pOH = 14 - 8.08 = 5.92 Now, we can use the formula [OH-] = 10^-pOH to calculate the [OH-]: [OH-] = 10^-5.92 = 1.20 × 10^-6 M

Use [OH-] to identify the corresponding salt

Now, we need to identify which salt would result in this [OH-] concentration. NaF: Using the Kb expression for the fluoride ion F- and the Kₐ value of HF (Kₐ = 6.6 × 10^-4), we can calculate the Kb value for F-. Kb = Kw / Ka = (1.0 × 10^-14) / (6.6 × 10^-4) = 1.52 × 10^-11 Next, we need to set up an ICE table to solve for the equilibrium concentrations of the ions: Initial: 0.100 0 0 Change: -x +x +x Equilibrium: 0.100-x x x Kb = [HF][OH-] / [F-] 1.52 × 10^-11 = (x)(x) / (0.100-x) Since x is very small compared to 0.100, we can simplify the equation: 1.52 × 10^-11 = (x)(x) / (0.100) Solving for x, which represents the [OH-] of the system, we get x = 3.90 × 10^-6 NaCl: As stated earlier, chloride ion doesn't significantly affect the pH. So, for NaCl, since it's a salt of a strong acid and strong base, the [OH-] would not match the required value of 1.20 × 10^-6 NaOCl: Using the Kb expression for hypochlorite ion, OCl- and the Kₐ value of HOCl (Kₐ = 2.95 × 10^-8), we can calculate the Kb value for OCl-. Kb = Kw / Ka = (1.0 × 10^-14) / (2.95 × 10^-8) = 3.39 × 10^-7 Next, we need to set up an ICE table to solve for the equilibrium concentrations of the ions: Initial: 0.100 0 0 Change: -x +x +x Equilibrium: 0.100-x x x Kb = [HOCl][OH-] / [OCl-] 3.39 × 10^-7 = (x)(x) / (0.100-x) Again, since x is very small compared to 0.100, we can simplify the equation: 3.39 × 10^-7 = (x)(x) / (0.100) Solving for x, which represents the [OH-] of the system, we get x = 5.82 × 10^-6

Compare the [OH-] values and identify the salt

Comparing the [OH-] values, we can see that the concentration of OH- resulting from NaOCl (5.82 × 10^-6) is the closest to the given [OH-] for the pH of 8.08 (1.20 × 10^-6). Therefore, the identity of the unknown salt is NaOCl.

Key concepts

pH Calculations

Understanding how to perform pH calculations is fundamental to mastering acid-base equilibria. The pH of a solution is a measure of its acidity or basicity. It is defined as the negative logarithm of the hydrogen ion concentration: \[ \text{pH} = -\log[H^+] \]In this context, for a basic solution, we often deal with hydroxide ions \([OH^-]\) as well. The relationship between \([H^+]\) and \([OH^-]\) is given by the water dissociation constant:\[ KW = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ at 25°C} \]To find \([OH^-]\) when the pH is known, we use the formula:

• Calculate pOH using \( \text{pOH} = 14 - \text{pH} \).
• Determine the \([OH^-]\) concentration using \([OH^-] = 10^{-\text{pOH}} \).

In the exercise provided, a solution with pH 8.08 corresponds to a pOH of 5.92. From this, \([OH^-]\) can be calculated as \(1.20 \times 10^{-6} \text{ M}\). Understanding these calculations helps in identifying the nature of the aqueous solution.

Salt Hydrolysis

Salt hydrolysis refers to the reaction between water and the ions that constitute a salt. In many instances, this interaction can influence the pH of the solution. When a salt dissolves in water, the anions or cations can act as acids or bases themselves.For example, in the case of sodium fluoride (NaF), when dissolved in water, it dissociates into \(F^-\) and \(Na^+\). The ion \(F^-\) is the conjugate base of a weak acid (HF), and it undergoes the following equilibrium reaction:\[ F^- + H_2O \rightleftharpoons HF + OH^- \]Similarly, sodium hypochlorite (NaOCl) yields the hypochlorite ion \(OCl^-\), which engages in hydrolysis to form:\[ OCl^- + H_2O \rightleftharpoons HOCl + OH^- \]During these reactions:

• The presence of \(OH^-\) raises the pH, making the solution more basic.
• Hydrolysis extent is reflected by the equilibrium constant \(K_b\), related to the tendency of these ions to undergo hydrolysis.

Understanding salt hydrolysis is crucial in explaining why solutions of certain salts exhibit changes in pH when dissolved in water.

Chemical Equilibria

Chemical equilibria form the basis for analyzing the behavior of weak acids and bases in solution. When a salt like NaF or NaOCl dissolves, it reaches a state of equilibrium with its hydrolysis products. To analyze this, we use an ICE (Initial, Change, Equilibrium) table, which helps to determine concentrations of species at equilibrium.Consider the equilibrium for \(OCl^-\):

• Initial concentrations are set before any reaction occurs.
• Changes in concentration are tracked as the system moves toward equilibrium.
• The equilibrium concentrations are determined by solving the equilibrium expression \( K_b = [HOCl][OH^-]/[OCl^-] \).

This step helps to calculate the resulting \([OH^-]\), matching it to observe pH changes. Identifying chemical equilibria involves understanding how different salts affect \(pH\) through their ability to either yield \(H^+\) or \(OH^-\). In the example, comparing these equilibria enabled us to identify NaOCl as the salt responsible for a pH of 8.08.