Problem 21 - Chemistry and Chemical Reactivity

Question

A buffer solution is prepared by adding 0.125 mol of ammonium chloride to \(5.00 \times 10^{2} \mathrm{mL}\) of \(0.500 \mathrm{M}\) solution of ammonia. (a) What is the pH of the buffer? (b) If 0.0100 mol of \(\mathrm{HCl}\) gas is bubbled into \(5.00 \times 10^{2} \mathrm{mL}\) of the buffer, what is the new \(\mathrm{pH}\) of the solution?

Step-by-Step Solution

Answer

The initial pH of the buffer is 9.55, and after adding HCl, the pH is 9.50.

Step 1: Convert Volume to Liters

First, convert the given volume from milliliters to liters for easier calculations. Since the buffer solution is prepared in \(5.00 \times 10^{2} \text{ mL}\), convert this to liters by dividing by 1000. \[ \text{Volume} = \frac{500}{1000} = 0.500 \text{ L} \]

Step 2: Calculate Concentrations

Calculate the molarity of the ammonium chloride and ammonia. Ammonium chloride: 0.125 moles in 0.500 L gives \(\frac{0.125}{0.500} = 0.250 \text{ M}\). Ammonia is given as 0.500 M since it's already in molarity.

Step 3: Determine Buffer Components

Ammonium chloride acts as the conjugate acid (\(\text{NH}_4^+\)) and ammonia (\(\text{NH}_3\)) acts as the base. We need these to apply the Henderson-Hasselbalch equation.

Step 4: Calculate Initial pH

Use the Henderson-Hasselbalch equation for initial pH: \( \text{pH} = \text{pK}_a + \log{\frac{[\text{Base}]}{[\text{Acid}]}} \). Given \(\text{K}_b\) for \(\text{NH}_3\) is \(1.8 \times 10^{-5}\), calculate \(\text{pK}_a\) using \(\text{pK}_w = 14\): \[ \text{pK}_a = 14 - \log(1.8 \times 10^{-5}) = 9.25 \] Then calculate the pH: \[ \text{pH} = 9.25 + \log\left(\frac{0.500}{0.250}\right) = 9.25 + 0.301 = 9.55 \]

Step 5: Adjust for Addition of HCl

Calculate moles of HCl added: 0.0100 mol. Since \(\text{H}^+\) reacts with \(\text{NH}_3\), the moles of \(\text{NH}_3\) decrease by 0.0100 mol, and \(\text{NH}_4^+\) increases by 0.0100 mol.

Step 6: Calculate New Concentrations

After the reaction, \(\text{NH}_3\) moles = 0.500 - 0.0100 = 0.490 M and \(\text{NH}_4^+\) moles = 0.250 + 0.0100 = 0.260 M. Recalculate using the Henderson-Hasselbalch equation: \[ \text{pH}_{new} = 9.25 + \log\left(\frac{0.490}{0.260}\right) = 9.25 + 0.251 = 9.50 \]

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a crucial tool in chemistry, particularly when dealing with buffer solutions. It helps estimate the pH of a solution containing a weak acid and its conjugate base. This equation is derived from the idea that the pH of a buffer solution depends on the ratio of the concentration of the base to the concentration of the acid. It is expressed as follows:\[\text{pH} = \text{pK}_a + \log{\left(\frac{[\text{Base}]}{[\text{Acid}]\right)}\]In this formula:

• \( \text{pH} \) is the measure of the acidity or basicity of the solution.
• \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant \( \text{K}_a \).
• \([\text{Base}]\) and \([\text{Acid}]\) refer to the molar concentrations of the conjugate base and the acid, respectively.

Using the Henderson-Hasselbalch equation, we can predict how the pH will change when we tweak the concentrations of the acid or base component in the buffer system. It's particularly helpful in understanding how buffers resist changes in pH when small amounts of acids or bases are added.

Equilibrium concentrations

Equilibrium concentrations are pivotal when discussing buffer solutions. When a weak acid or base interacts with its conjugate pair, a dynamic equilibrium is established in the solution. This equilibrium is crucial for maintaining the pH stability that buffers are known for. In an equilibrium state, the rate of the forward reaction (ionization of the weak acid or base) equals the rate of the backward reaction (reformation of the weak acid or base from ions). This balance ensures constant concentrations of all species involved, until an external factor like the addition of more reactant or a change in temperature shifts the equilibrium.To solve buffer problems, we often consider the equilibrium concentrations of reactants and products. For example, when calculating the pH of a buffer with added HCl, the equilibrium concentrations change due to the reaction between HCl and \( \text{NH}_3 \). This interaction results in redux of \( \text{NH}_3 \) and formation of \( \text{NH}_4^+ \), and consequently, these changes alter the equilibrium concentrations accordingly, impacting the overall pH of the buffer.

Acid-base reactions

Acid-base reactions are at the heart of what makes buffers function. A buffer is designed to minimize pH changes in the face of adding small amounts of acids or bases. This ability hinges on acid-base neutralization reactions. In these reactions, acids donate protons (H⁺ ions) to bases, balancing the pH in the process.For example, when HCl is introduced to a buffer containing ammonium ions and ammonia, the HCl dissociates to provide additional \( \text{H}^+ \) ions. These \( \text{H}^+ \) ions react with \( \text{NH}_3 \) to form \( \text{NH}_4^+ \), which means less of the added acid remains free in the solution to affect the pH significantly.Buffers work efficiently in a narrow pH range that is defined by the pK_a of the weak acid involved. They are critical in biological systems, where enzymes and biochemical processes are extremely sensitive to changes in pH. Through these acid-base reactions, buffers can effectively dampen drastic changes, contributing to maintaining homeostasis in biological and chemical environments.