Question

A buffer solution is prepared by adding \(0.125 \mathrm{~mol}\) ammonium chloride to 500. mL of 0.500 -M aqueous ammonia. Calculate the pH of the buffer. If 0.0100 mol HCl gas is bubbled into 500. mL buffer and all of the gas dissolves, calculate the new \(\mathrm{pH}\) of the solution.

Short answer

Initial pH = 9.551; After HCl, pH = 9.810.

Step-by-step solution

Identify Components and Concentrations

Begin by identifying the components of the buffer solution. The buffer consists of aqueous ammonia (NH₃), a weak base, and ammonium chloride (NH₄Cl), its conjugate acid. We have 0.125 mol of NH₄Cl added to 500 mL of 0.500 M NH₃. Convert the volume to liters to make the calculations easier: 500 mL = 0.500 L.

Calculate Initial Concentrations

Calculate the initial concentration of NH₄⁺ ions from NH₄Cl. Since 0.125 mol of NH₄Cl is added to 0.5 L of solution, the concentration of NH₄⁺ is \(\frac{0.125}{0.5} = 0.25\) M. The initial concentration of NH₃ is given as 0.500 M.

Use the Henderson-Hasselbalch Equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: \[pH = pK_a + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right)\]Here, the base is NH₃ and the acid is NH₄⁺. The \(pK_a\) of NH₄⁺ is 9.25. Substitute the concentrations into the equation:\[pH = 9.25 + \log \left(\frac{0.500}{0.250}\right)\]This simplifies to \(pH = 9.25 + 0.301 = 9.551\).

Calculate Moles of HCl Added

When HCl gas is bubbled into the solution, it reacts with the NH₃ in the buffer. Calculate the number of moles of HCl added: 0.0100 mol.

Adjust Concentrations After Reaction with HCl

HCl will react with NH₃ to form NH₄⁺. Thus, the moles of NH₃ decrease by 0.0100 mol, and the moles of NH₄⁺ increase by 0.0100 mol. Adjust the concentrations:- New NH₃ moles: 0.500 mol - 0.010 mol = 0.490 mol- New NH₄⁺ moles: 0.125 mol + 0.010 mol = 0.135 mol- Divide each by the volume (0.5 L) to find new concentrations: - [NH₃] = \(\frac{0.490}{0.5} = 0.980\) M - [NH₄⁺] = \(\frac{0.135}{0.5} = 0.270\) M

Recalculate pH Using Updated Concentrations

Use the Henderson-Hasselbalch equation again with the updated concentrations:\[pH = 9.25 + \log \left(\frac{0.980}{0.270}\right)\]This simplifies to:\(pH = 9.25 + \log(3.63)\)\(pH = 9.25 + 0.560\)\(pH \approx 9.810\).

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental tool in chemistry for calculating the pH of buffer solutions. It provides a clear relationship between the pH, the pKₐ of the acid, and the concentration ratio of the base and its conjugate acid.

This equation is expressed as: \[ pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]This expression allows us to predict how changes in the concentrations of the components of the buffer affect the pH.

• The \( pK_a \) is the negative logarithm of the acid dissociation constant, a measure of acid strength.
• The fraction \( \frac{[\text{base}]}{[\text{acid}]} \) represents the relative concentrations of the base and the acid in the buffer solution.

This equation helps understand buffer capacity, which is how well a buffer can resist changes in pH. The closer the \( pH \) to the \( pK_a \), the greater the capacity of the buffer.

weak base

A weak base is a base that does not fully ionize in water. This means it does not completely convert into its constituent ions. Instead, weak bases exist in a dynamic equilibrium between the undissociated base and the ions.

In the provided exercise, we see ammonia (\( NH_3 \) ) functioning as a weak base.

• A key property of weak bases is their partial ionization in water, forming hydroxide ions (\( OH^- \) ) and their conjugate acids.
• The equilibrium constant for this process is termed the base ionization constant (\( K_b \) ).

For ammonia, when dissolved in water, the reaction is:\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \] This equilibrium is crucial for determining its behavior in buffer solutions. By understanding and using these principal properties, chemists can predict and control the pH levels in various chemical processes.

acid-base equilibria

Acid-base equilibria describe the balance between acids and bases in a solution, which is crucial for understanding the chemical nature of reactions. In the context of chemistry, it is all about how acid and base strengths influence reactions.

In any acidic or basic solution, there is a dynamic equilibrium between the proton donor (acid) and proton acceptor (base).

• Equilibrium implies that the forward and reverse reactions occur at the same rate.
• The concentrations of acids, bases, and their ions determine the exact position of the equilibrium.

Understanding this concept is essential when working with buffer solutions which involve the interplay of weak acids or bases and their conjugate salts to resist pH changes.
In the buffer solution of the exercise, the equilibrium involves:\[ NH_3 + H^+ \rightleftharpoons NH_4^+ \]It's important because it illustrates how adding acids or bases affects the equilibrium, and subsequently the pH. Having insight into acid-base equilibria allows chemists to control pH in industrial processes and biological systems.