Problem 167 - Chemistry

Question

Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

Step-by-Step Solution

Answer

The pH of the 0.200 M C₅H₅NHF solution is approximately 6.06.

Step 1: Write the Equation for the Principal Equilibrium Reaction

Since the salt contains C₅H₅NH⁺ and F⁻ ions, and we are given that the principal equilibrium is C₅H₅NH⁺(aq) + F⁻(aq) ⇌ C₅H₅N(aq) + HF(aq), we will use this reaction to analyze the species involved.

Step 2: Determine Initial Concentrations of Reactants and Products

Given a 0.200 M solution of C₅H₅NHF, the initial concentrations of C₅H₅NH⁺ and F⁻ ions are: \[ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}]_{0} = [\mathrm{F}^{-}]_{0} = 0.200 \, \mathrm{M} \] Since the reaction has not yet taken place, the initial concentrations of C₅H₅N and HF are both zero.

Step 3: Write the Equilibrium Expression and Define the Change in Concentration

Using the given equilibrium constant K, we can write an equilibrium expression for the reaction: \[ K = \frac{ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}][\mathrm{HF}]}{ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{F}^{-}] } \] During the course of the reaction, the change in concentration of C₅H₅NH⁺ and F⁻ is equal. So, we can denote it as minus-x, which means the decrease in their concentrations. The change in concentration of C₅H₅N and HF is equal as well, and it can be denoted as plus-x, which means the increase in their concentrations.

Step 4: Plug Initial and Change Concentrations into K Expression and Solve for x

Now, we substitute the initial concentrations and the change in concentrations into the K expression: \[ K = \frac{ (0.200 + x)(x)}{ (0.200 - x)^2 } \] Given K = 8.2 × 10⁻³, we can solve for x: \[ 8.2 \times 10^{-3} = \frac{x(0.200 + x)}{(0.200 - x)^2} \] This quadratic equation can be solved to find the value of x. After an approximation, we get x ≈ 0.00134.

Step 5: Calculate the H⁺ Concentration and the pH of the Solution

As we found the change in concentrations, we can calculate the [HF] concentration in the solution: \[ [\mathrm{HF}] = [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}] = x \approx 0.00134 \, \mathrm{M} \] Since HF is a weak acid, we can use its expression for Kₐ to find the H⁺ concentration in the solution: \[ K_\mathrm{a} = 6.6 \times 10^{-4} = \frac{ [\mathrm{H}^{+}][\mathrm{F}^{-}]}{ [\mathrm{HF}] } = \frac{ [\mathrm{H}^{+}]}{ 0.00134 } \] Solving for H⁺ concentration, we get: \[ [\mathrm{H}^{+}] \approx 8.8 \times 10^{-7} \, \mathrm{M} \] Now, we can calculate the pH using the formula: \[ \mathrm{pH} = -\log_{10} [\mathrm{H}^{+}] \] Hence, the pH of the solution is: \[ \mathrm{pH} \approx -\log_{10} (8.8 \times 10^{-7}) \approx 6.06 \] So, the pH of the 0.200 M C₅H₅NHF solution is approximately 6.06.

Key Concepts

Equilibrium Expression

Understanding the equilibrium expression is key in analyzing reactions that don't go to completion. In any chemical reaction, the equilibrium expression helps us understand the balance between the reactants and products at equilibrium. For a general reaction:

• Reactants are on the left side of the arrow.
• Products are on the right side.

The equilibrium expression, often denoted by the equilibrium constant \( K \), is written as a ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation.
In our exercise, the equilibrium expression for the reaction \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q) + \mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) + \mathrm{HF}(a q) \) is given by:\[ K = \frac{ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}][\mathrm{HF}]}{ [\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{F}^{-}] }\]This equation allows us to predict the concentrations of species involved when the reaction reaches equilibrium. The equilibrium constant \( K \) provides a measure of how far the reaction will proceed and is crucial in calculating any changes in concentration during the reaction.

Acid-Base Reaction

An acid-base reaction involves the transfer of protons \((\mathrm{H}^+)\) between a pair of substances. Generally, the substance donating a proton is the acid, while the one accepting it is the base. In our scenario, we deal with a reaction involving the cation \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \) acting as a weak acid and the anion \( \mathrm{F}^{-} \) acting as a base.
The principle reaction for these ions, given as \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} + \mathrm{F}^{-} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} + \mathrm{HF} \), highlights a fundamental acid-base interaction:

• \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \) donates a proton becoming \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} \).
• \( \mathrm{F}^{-} \) accepts a proton becoming \( \mathrm{HF} \).

These reactions help us understand the process of neutralization between acids and bases, where a weak acid and a weak base are involved. The position of equilibrium indicates which side of the reaction is favored. Thus, in calculating the pH, understanding the acid-base reaction allows us to determine the strength and extent each species will contribute to the overall acidity of the solution.

Weak Acid Dissociation Constant

The dissociation constant \( K_a \) is an essential concept when working with weak acids. It quantifies the degree to which a weak acid (like HF in this exercise) dissociates into its ions in a solution. The smaller the \( K_a \) value, the weaker the acid.
For the equilibrium involving HF:\[ \mathrm{HF} \rightleftharpoons \mathrm{H}^{+} + \mathrm{F}^{-}\]The expression for the dissociation constant \( K_a \) is given by:\[ K_a = \frac{ [\mathrm{H}^{+}][\mathrm{F}^{-}] }{ [\mathrm{HF}] }\]The pH of a solution can be calculated using \( K_a \) by determining the concentration of \( \mathrm{H}^+ \) ions in the equilibrium state. Since HF is not a strong acid, only a small fraction of HF molecules dissociate to release \( \mathrm{H}^+ \).When calculating the \( \mathrm{pH} \), knowing the extent of dissociation—a critical factor—is achieved through \( K_a \). This calculation influences how we understand the solution's acidity, shown when solving for \( \mathrm{H}^+ \) concentration, and ultimately helps determine the pH using \( \mathrm{pH} = -\log_{10} [\mathrm{H}^+] \). With this knowledge, we can appreciate how weak acids contribute to system's pH. Such insights show why certain substances behave the way they do in solutions, particularly concerning their acidic or basic nature.