Question

Describe how you would prepare \(1 \mathrm{~L}\) of the buffer \(0.20 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa} / 0.20 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) by (a) mix- ing a solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) with a solution of \(\mathrm{CH}_{3} \mathrm{COONa},\) (b) reacting a solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) with a solution of \(\mathrm{NaOH}\), and (c) reacting a solution of \(\mathrm{CH}_{3} \mathrm{COONa}\) with a solution of \(\mathrm{HCl}\).

Short answer

Buffer can be prepared by directly mixing \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\), or by reacting \(\mathrm{CH}_{3} \mathrm{COOH}\) with \(\mathrm{NaOH}\) or \(\mathrm{CH}_{3} \mathrm{COONa}\) with \(\mathrm{HCl}\). In all methods, 0.20 mol of each component should be used which gives a 0.20 M solution.

Step-by-step solution

Preparation by mixing

To prepare the buffer by mixing, calculate the molar quantities for each component that needs to be included in the buffer. Both \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\) need to be 0.20 M. Each component would therefore require 0.20 mol (since 0.20 mol/L * 1L = 0.20 mol). Prepare the 1L of buffer by mixing these molar quantities together.

Preparation by reacting with NaOH

Reactive preparation can be done by having \(\mathrm{CH}_{3}\mathrm{COOH}\) react with \(\mathrm{NaOH}\) to make \(\mathrm{CH}_{3}\mathrm{COONa}\) and water. Here, 0.20 mol of \(\mathrm{CH}_{3}\mathrm{COOH}\) and 0.20 mol of \(\mathrm{NaOH}\) are required to create the 0.20 M \(\mathrm{CH}_{3}\mathrm{COONa}\). The help of a pH meter might be needed in this case to ensure the right amount of NaOH has been added to the solution.

Preparation by reacting with HCl

For preparation by having \(\mathrm{CH}_{3}\mathrm{COONa}\) react with \(\mathrm{HCl}\), you make \(\mathrm{CH}_{3}\mathrm{COOH\) and \(\mathrm{NaCl}\). This is similar to previous step method. The quantities would be the same as above with 0.20 mol of \(\mathrm{CH}_{3}\mathrm{COONa}\) and 0.20 mol of \(\mathrm{HCl}\) giving you a 0.20 M solution. Like before, a pH meter might be needed to make sure the right amount of HCl is added.

Key concepts

Buffer Solution

A buffer solution is a special type of solution that has the unique ability to resist changes in pH when small amounts of acid or base are added. This is invaluable in many biological and chemical applications where maintaining a stable pH is crucial. For example, in our exercise, a buffer is made using acetic acid ((CH_3COOH)) and its salt, sodium acetate ((CH_3COONa)), both at the same concentration of 0.20 M. This combination works effectively because the acid can neutralize added bases, while the salt can neutralize added acids. The ideal preparation of a buffer involves careful calculation and mixing of precisely measured components, ensuring that the molarity is correct for both the acid and its conjugate base.

Buffers operate on the principle of the acid-base equilibrium between the weak acid (or base) and its conjugate base (or acid). When acidic or basic substances are added to the buffer solution, the equilibrium shifts in a way that minimizes the change in pH. This is known as Le Chatelier's principle. Understanding how to prepare and use buffers is fundamental, particularly in laboratory settings and industries where pH balance is a condition of the process or product.

Molarity

Molarity, symbolized by M, is a measure of concentration in chemistry, representing the number of moles of a solute per liter of solution. This crucial concept allows chemists to quantify and create solutions with precise characteristics. For instance, when preparing a buffer solution as described in our exercise, molarity informs us of the exact amount of each substance required to achieve the desired buffering capacity.

The calculation of molarity is straightforward: take the number of moles of solute and divide by the volume of solution in liters. It's expressed in the equation, \( M = \frac{moles}{liters} \). In practice, if we need 1 L of a 0.20 M (CH_3COOH)/ (CH_3COONa) buffer, we require 0.20 mol of each component for the buffer to function effectively, as seen in the exercise. Accurate measurement and mixing of these molar quantities are crucial for buffer preparation. Being well-versed in molarity calculations is a key skill for any student or professional working with solutions.

Acid-Base Reaction

An acid-base reaction is a chemical reaction that occurs between an acid and a base. In a typical reaction, a hydrogen ion (H+) from the acid combines with a hydroxide ion (OH-) from the base to form water, alongside the formation of a salt. This type of reaction is central to the concept of buffer preparation, which allows for the adjustment of pH levels in a solution.

In the exercise, two methods of buffer preparation involve acid-base reactions. When acetic acid (CH_3COOH) reacts with sodium hydroxide (NaOH), they produce sodium acetate (CH_3COONa) and water (H_2O), creating one component of the buffer. Similarly, when sodium acetate reacts with hydrochloric acid (HCl), it produces CH_3COOH and sodium chloride (NaCl), ensuring the other component's presence in the buffer.

Understanding acid-base reactions is invaluable because it enables one to control the outcomes of such processes, especially in creating a buffer solution with a specific pH range. Ensuring that stoichiometric amounts of acid and base react will determine the successful preparation of a buffer, as both the exercises and real-life applications demand precision and a solid grasp of chemical reaction principles.