Question

What is the \(\mathrm{pH}\) of a solution obtained by adding \(1.15 \mathrm{mg}\) of aniline hydrochloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\right)\) to \(3.18 \mathrm{L}\) of \(0.105 \mathrm{M}\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right) ?\)

Short answer

The pH of the solution after adding aniline hydrochloride to aniline is calculated using the Henderson-Hasselbalch equation. Apply the details discussed in the steps to get the final pH value.

Step-by-step solution

Calculate Moles of Aniline Hydrochloride and Aniline

Convert mass of the aniline hydrochloride (\(C_{6}H_{5}NH_{3}^{+}Cl^{-}\)) into moles using its molar mass (129.57 g/mol). By the same token, calculate the moles of aniline (\(C_{6}H_{5}NH_{2}\)) present in solution by multiplying its concentration by the volume of the solution.

Use Reaction Stoichiometry to Find Moles of Species Remaining

Both compounds will react to form water and chloride ion. Use stoichiometry to subtract the moles of reactant used in the reaction to find the final moles of each species. This will calculate the amounts of \(C_{6}H_{5}NH_{2}\) and \(C_{6}H_{5}NH_{3}^{+}\) remaining after the reaction.

Calculate Final Concentrations

Divide the final moles of aniline (\(C_{6}H_{5}NH_{2}\)) and aniline hydrochloride (\(C_{6}H_{5}NH_{3}^{+}\)) by the total volume to get their equilibrium concentrations.

Look up pKa

Look up the pKa for aniline. It is approximately 4.60. Remember that pKa = -log(Ka), we will use the pKa value for the Henderson-Hasselbalch equation.

Use Henderson-Hasselbalch Equation

Substitute the concentrations of aniline and aniline hydrochloride, and the pKa into the Henderson-Hasselbalch equation. The formula is \(pH = pKa + log \left(\frac{{[base]}}{{[acid]}} \right)\), where [base] is the concentration of the aniline and [acid] is the concentration of aniline hydrochloride.

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry involving the quantitative relationships between the reactants and products in chemical reactions. By using the coefficients from balanced chemical equations, we can determine the proportions in which chemicals react as well as the quantities in which they are produced or consumed.

In the exercise provided, stoichiometry is crucial to understand how aniline hydrochloride reacts with aniline. Since the reaction proceeds according to a fixed ratio, it helps us deduce how many moles of each compound remain in the solution after the reaction has taken place. This is imperative to calculate the pH, because the pH is influenced by the concentration of the acids and bases present in the solution.

To excel in stoichiometry, one must be proficient in balancing chemical equations, understanding molar relationships, and being able to convert between units such as grams, moles, and molecules.

Molarity and Solution Concentration

Molarity is a measure of the concentration of a solute in a solution, defined as the number of moles of solute divided by the volume of the solution in liters. It is a fundamental concept in chemistry because it provides a direct way to relate the amount of substance to the volume of solution, which is critical in preparing solutions and performing dilutions.

In this exercise, understanding molarity allows us to find the initial concentrations of aniline and aniline hydrochloride in the solution before any reaction takes place. We must also understand this concept to determine the final concentrations after the reaction, once we've found the moles of each species remaining through stoichiometry. These concentrations are then used in the Henderson-Hasselbalch equation to calculate the pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a straightforward way to relate the pH of a solution to the pKa of an acid and the concentrations of an acid and its conjugate base. The pKa is the negative base-10 logarithm of the Ka value, which is the acid dissociation constant - a measure of the strength of the acid.

By using the formula \(pH = pKa + log \left(\frac{[base]}{[acid]} \right)\), we can find the pH if we know the concentrations of the acid and its conjugate base. In our exercise, we use the concentrations of aniline (\(C_{6}H_{5}NH_{2}\)) and aniline hydrochloride (\(C_{6}H_{5}NH_{3}^{+}\)) to find the pH of the solution. It's a useful tool for understanding acid-base balance in a wide range of chemical and biological systems.

Acid-Base Equilibrium

Acid-base equilibrium refers to the state of balance between an acid and its conjugate base in a solution. This concept is central to understanding how solutions buffer changes in pH, a critical aspect in many chemical and biological processes.

In the exercise, the acid-base equilibrium process plays a key role, as we're examining a solution containing both aniline (the base) and aniline hydrochloride (the acid). Their relative concentrations, which change upon mixing and reacting, affect the pH of the solution. Balancing the acid with its conjugate base minimizes changes in the pH, which depends on the acid's dissociation constant (Ka) or pKa.

Understanding acid-base equilibrium is crucial for predicting the behavior of acids and bases when diluted or mixed, and is directly applied in the pH calculation using the Henderson-Hasselbalch equation.