Problem 175 - Chemistry An Atoms First Approach

Question

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) 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 principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$

Step-by-Step Solution

Answer

The pH of the 0.200-M solution of $\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NHF}$ is approximately 1.74.

Step 1: Write the Principal Equilibrium Reaction

Write the principal equilibrium reaction between C5H5NH+ and F-. \( C5H5NH^+(aq) + F^-(aq) \rightleftharpoons C5H5N(aq) + HF(aq) \)

Step 2: Write Down the Equilibrium Constant

Write down the equilibrium constant (K) given for the reaction. We have, \( K = 8.2 \times 10^{-3} \)

Step 3: Set Up an ICE Table

Set up an Initial, Change, and Equilibrium (ICE) table for the reaction. \( \begin{array}{c|ccc} & C5H5NH^+ & +F^- & \rightleftharpoons & C5H5N & +HF \\ \hline I & 0.200 & 0.200 & & 0 & 0 \\ C & -x & -x & & +x & +x \\ E & 0.200-x & 0.200-x & & x & x \\ \end{array} \)

Step 4: Write the Equilibrium Expression

Write the equilibrium expression (using K) and substitute the values from the ICE table. \( K = \frac{[C5H5N][HF]}{[C5H5NH^+][F^-]} \) Plug in the values from the ICE table: \( 8.2 \times 10^{-3} = \frac{x^2}{(0.200-x)^2} \)

Step 5: Solve the Equation for x

Assuming that the reaction proceeds to an equilibrium where x is much less than 0.200, we can solve the equation for x. \( x^2 = (8.2 \times 10^{-3}) (0.200)^2 \) We can now find the x value: \( x = \sqrt{(8.2 \times 10^{-3})(0.200)^2} \) \( x = 0.01807\)

Step 6: Calculate the pH

Now that we have the concentration of HF (x), we can calculate pH using the following equation: \(\phantom{xx} pH = -\log_{10}(\mathrm{H}^+ )\) We know that the concentration of H+ is equal to the concentration of HF because we have a 1:1 relationship between them: \(\phantom{xx} [\mathrm{H}^+] = [HF] = 0.01807 \) Therefore, the pH of the solution will be: \(\phantom{xx} pH = -\log_{10}(0.01807) \) Calculating the pH, \(\phantom{xx} pH \approx 1.74 \) The pH of the 0.200-M solution of C5H5NHF is approximately 1.74.

Key Concepts

Equilibrium Reaction

An equilibrium reaction represents a state in a chemical process where the reactants and products reach a balance. This means the rate at which the reactants turn into products is equal to the rate at which products revert to reactants. In the given exercise, the equilibrium reaction involves * - the acid ion - \( C_5H_5NH^+ \). and the base ion * - \( F^- \). forming products * - \( C_5H_5N \). and * - \( HF \). When equilibrium is reached, the concentrations of these ions in the solution remain constant. The key feature of the equilibrium reaction is its dynamic nature, where ions are still moving and reacting; yet their overall concentrations do not change. Understanding this balance is fundamental in predicting how the solution's pH will respond to various changes or stresses in the system.

ICE Table

The ICE table is a valuable tool used to analyze equilibrium reactions. ICE stands for Initial, Change, and Equilibrium, and it helps break down the process of an equilibrium reaction into manageable parts. You start with the "Initial" concentrations of the compounds. In our problem, both * - * \( C_5H_5NH^+ \). and * * \( F^- \). had the initial concentrations of * * \( 0.200 \). M.Then there is a "Change" happens as the reaction proceeds toward equilibrium, defined by a variable * * \( x \). Finally, "Equilibrium" refers to the stage where concentrations have become constant. At equilibrium, you can express the concentrations as * * \( 0.200-x \). for the reactants and * * \( x \). for the products. This table is not only a structured way to track concentration changes but also serves to simplify solving for unknowns in equilibrium constants and concentration values.

Equilibrium Constant

The equilibrium constant, commonly represented by * - \( K \), is a critical value that quantifies the position of equilibrium in a chemical reaction. It reflects the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation. In our exercise, * - \( K = 8.2 \times 10^{-3} \) indicates that at equilibrium, the rate of conversion of reactants to products is less than their conversion back to reactants, suggesting the reactants are favored at equilibrium.The equilibrium expression is derived from the balanced equation and is given by: \[{K = \frac{[C_5H_5N][HF]}{[C_5H_5NH^+][F^-]}}\].This equation is solved using the concentrations derived from the ICE table. Understanding and calculating * - \( K \) is essential to assess how a reaction behaves and how changes can affect the * - \( pH \) of a solution.

Acid-Base Reaction

An acid-base reaction is where an acid donates a proton (* - \( H^+ \)) to a base. This type of reaction plays a critical role in determining the * - \( pH \) of solutions.In our scenario, * - \( C_5H_5NH^+ \)acts as the proton donor, the acid, while * - \( F^- \) acts as the proton acceptor, the base. Upon reaching equilibrium, products * - \( C_5H_5N \) and * - \( HF \) are formed, influencing the solution’s pH.The resultant * - \( [H^+] \) ion concentration determines the * - \( pH \) of the solution. Calculating this involves using the concentration of * - \( HF \) from the equilibrium expressions.By solving * - \( x = 0.01807 \), the concentration of \* - \( HF \) is directly linked to the * - \( [H^+] \), enabling us to calculate * - \( pH = -\log_{10}(0.01807) \) and find a value of approximately * - \( 1.74 \). This reflects the acidic nature of the solution.