Question

An unknown salt is either $\mathrm{NaF},\mathrm{NaCl}$, 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 $\mathrm{NaOCl}$, as the given pH of $8.08$ indicates a basic solution and, among the given salts, the equilibrium constant for the dissociation of ${\mathrm{OCl}}^{-}$ in water is significantly higher than that of ${\mathrm{F}}^{-}$, which will produce a higher concentration of OH- ions as observed in the solution.

Step-by-step solution

Write the reactions for each salt when dissolved in water

When a salt is added to water, it dissociates into its ions, and any basic or acidic ions may react with water. For the salts in question, the reactions are as follows: - For $\mathrm{NaF}$: $${\mathrm{F}}^{-}+\mathrm{H}2\mathrm{O}\rightleftharpoons \mathrm{HF}+{\mathrm{OH}}^{-}$$ - For $\mathrm{NaCl}$: $${\mathrm{Cl}}^{-}+\mathrm{H}2\mathrm{O}$$ (No reaction occurs with Cl-) - For $\mathrm{NaOCl}$: $${\mathrm{OCl}}^{-}+\mathrm{H}2\mathrm{O}\rightleftharpoons \mathrm{HOCl}+{\mathrm{OH}}^{-}$$

Use the Ice Table Method and Kb expressions

We will now consider the reactions and determine the equilibrium expressions for their basic ions. Then we will use the given pH value to solve for the concentration of OH- and compare it to what we expect from the respective dissociation reactions. 1. For $\mathrm{NaF}$: The equilibrium expression for the dissociation of ${\mathrm{F}}^{-}$ in water is: $$K_{b1}=\frac{[\mathrm{HF}][{\mathrm{OH}}^{-}]}{[{\mathrm{F}}^{-}]}$$ We will use a table to help determine the equilibrium concentrations for each species: ``` Initial: 0.050 M - - Dissociated: -x +x +x Equilibrium: (0.050-x) M x x ``` 2. For $\mathrm{NaOCl}$: The equilibrium expression for the dissociation of ${\mathrm{OCl}}^{-}$ in water is: $$K_{b2}=\frac{[\mathrm{HOCl}][{\mathrm{OH}}^{-}]}{[{\mathrm{OCl}}^{-}]}$$ The table for this reaction is: ``` Initial: 0.050 M - - Dissociated: -x +x +x Equilibrium: (0.050-x) M x x ```

Calculate the concentration of OH- and compare it to the given pH value

We are given a pH value of 8.08. We can use this to calculate the concentration of ${\mathrm{OH}}^{-}$ ions in the solution: $$\mathrm{pOH}=14-\mathrm{pH}$$ $$\mathrm{pOH}=14-8.08=5.92$$ So, the concentration of OH- ions in the solution is: $$[{\mathrm{OH}}^{-}]={10}^{-\mathrm{pOH}}={10}^{-5.92}$$ Now, we will compare this concentration to the expected equilibrium concentrations from our ice tables in Step 2. 1. For $\mathrm{NaF}$: Using the Kb expression and the equilibrium concentrations from the table, we can write: $$K_{b1}=\frac{x^2}{0.050-x}$$ 2. For $\mathrm{NaOCl}$: Using the Kb expression and the equilibrium concentrations from the table, we can write: $$K_{b2}=\frac{x^2}{0.050-x}$$ Since both of the reactions have the same expression for Kb, we can compare their Kb values directly. From a table of Kb values, we can find the respective values for the ions: - ${\mathrm{F}}^{-}:K_{b1}=1.4\times {10}^{-11}$ - ${\mathrm{OCl}}^{-}:K_{b2}=2.95\times {10}^{-7}$ We know that the solution is basic, and the pH value is greater than 7. Considering the Kb values, it is clear that the reaction involving OCl- will produce a higher concentration of OH- ions than the reaction involving F-. Therefore, the identity of the unknown salt is: $$\boxed{{\displaystyle \mathrm{NaOCl}}}$$

Key concepts

Dissociation of Salts in Water

Understanding how salts dissociate in water is crucial when trying to identify an unknown salt through its chemical behaviour. When ionic compounds (salts) dissolve in water, they break apart into their constituent ions. This process is known as dissociation.

For instance, when NaF dissolves, the fluoride ion (F^{-}) can accept a proton from water (H2O), generating HF (hydrogen fluoride) and hydroxide ions (OH^{-}). On the other hand, sodium chloride (NaCl) is a neutral salt, where its chloride ion (Cl^{-}) does not react with water. However, salt like NaOCl releases hypochlorite ions (OCl^{-}) which can also accept a proton from water and increase the number of hydroxide ions in the solution.

The resultant pH of the solution can thus be influenced by these reactions, as the generation of hydroxide ions makes the solution more basic (increase in pH). By observing the pH change when each salt is dissolved in water, chemists can make predictions about the salt's identity.

pH and pOH Calculations

The pH scale is a measure of the acidity or basicity of an aqueous solution. pH is mathematically defined as the negative logarithm (log) of the hydrogen ion concentration ([H+]), while pOH is related to the hydroxide ion concentration ([OH-]). The two are inversely related and can be connected through the formula: $$\text{pOH}=14-\text{pH}$$

Therefore, by knowing the pH of a solution, one can calculate the pOH and then find the concentration of hydroxide ions present by inverting the logarithm: $$[{\text{OH}}^{-}]={10}^{-\text{pOH}}$$
This is precisely what is carried out in our problem to work backward from the pH value given, 8.08, to figure out the concentration of hydroxide ions in the solution. As the pH is above 7, which is neutral, the solution is basic. It indicates that the solution must contain a significant number of hydroxide ions, a clue that directs us towards a salt that can release or produce OH^- ions when dissolved.

Equilibrium Constant Expressions

In a chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. For the dissociation of salts in water, the equilibrium constant (K) expresses the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their coefficients in the balanced equation.For basic solutions, we use the equilibrium constant for the base, Kb, to measure the strength of a base in terms of its ability to form hydroxide ions in water. A higher Kb value indicates a stronger base, capable of producing more hydroxide ions, which translates to a higher pH of the solution.

Going back to our exercise, the Kb value for the fluoride ion (F^-) and the hypochlorite ion (OCl^-) were key to determining the identity of the unknown salt. By comparing known Kb values with the concentration of hydroxide ions calculated from the pH, we deduce that NaOCl has a considerably larger Kb and, therefore, is the correct identity of the unknown salt as it aligns with the more basic pH of 8.08.