Chapter 2: Problem 5
Given \(0.1 M\) solutions of acetic acid and sodium acetate, describe the preparation of \(1 \mathrm{L}\) of \(0.1 \mathrm{M}\) acetate buffer at a pH of 5.4.
Short Answer
Expert verified
To prepare 1L of 0.1M acetate buffer at a pH of 5.4 from 0.1M solutions of acetic acid and sodium acetate, you need to mix 0.81L of the sodium acetate solution with 0.18L of the acetic acid solution.
Step by step solution
01
Calculate the pKa
We start by calculating the pKa of acetic acid. It is commonly known that the pKa of acetic acid is 4.76.
02
Substitute in the Henderson-Hasselbalch equation
Now we can substitute pH, pKa, [A-] and [HA] into the Henderson-Hasselbalch equation. We know pH (5.4) and pKa (4.76). What we are trying to find is the ratio of [A-] to [HA]. Thus, the equation will look like this: 5.4 = 4.76 + log([A-]/[HA]).
03
Solve the equation
After having simplified the equation, we get: log([A-]/[HA]) = 5.4 - 4.76 = 0.64. To get [A-]/[HA], we then take the antilog: [A-]/[HA] = antilog(0.64) = 4.37.
04
Calculate volumes
Given the ratio [A-]:[HA] of 4.37, we can calculate the volume of sodium acetate and acetic acid needed to prepare 1L of 0.1M buffer solution. As the total molarity is 0.1M and the buffer solution is 1L, the total moles required is 0.1 moles. Thus, the moles of sodium acetate (base) and acetic acid (acid) are as follows, sodium acetate: 0.1 moles * 4.37/(4.37+1) = 0.081 moles, acetic acid: 0.1 moles * 1/(4.37+1) = 0.018 moles. As the starting solutions are 0.1M, the volumes required are sodium acetate: 0.081 moles / 0.1M = 0.81L, acetic acid: 0.018 moles / 0.1M = 0.18L.
05
Prepare the buffer
To prepare the buffer, you need to mix 0.81L of the sodium acetate solution with 0.18L of the acetic acid solution. The volumes are less than 1L but when mixed, they make a 1L buffer solution due to the volume addition of two liquids.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a fundamental tool when preparing buffer solutions. This formula relates the pH of a buffer to the pKa (acid dissociation constant) of the acid and the ratio of the concentrations of its conjugate base and the acid itself. It can be expressed as:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
When preparing an acetate buffer, the equation helps us determine the necessary ratio of acetate ions (A-) to acetic acid (HA) to achieve the desired pH. This balance is crucial, as it maintains the pH within a narrow range, buffering against changes when acids or bases are added. The calculated ratio then guides the volumes of acetic acid and sodium acetate needed to create the buffer.
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
When preparing an acetate buffer, the equation helps us determine the necessary ratio of acetate ions (A-) to acetic acid (HA) to achieve the desired pH. This balance is crucial, as it maintains the pH within a narrow range, buffering against changes when acids or bases are added. The calculated ratio then guides the volumes of acetic acid and sodium acetate needed to create the buffer.
pKa Calculation
The pKa value is critical for understanding acid-base chemistry and is essential in buffer preparation. It's defined as the negative logarithm of the acid dissociation constant (Ka). For acetic acid, the pKa value is commonly known to be 4.76. This value signifies the strength of the acid: the lower the pKa, the stronger the acid. In the context of the Henderson-Hasselbalch equation, the pKa allows us to calculate the pH or vice versa if the ratio of the conjugate base to acid concentrations is known. This understanding is fundamental when mixing buffer components to ensure they will maintain the pH under various conditions.
Buffer Solution Ratio
The ratio of buffer components in a solution is directly obtained from the Henderson-Hasselbalch equation. In a buffer system, this ratio dictates how the system can resist pH changes. For an acetate buffer with a desired pH of 5.4, the calculated ratio from the equation is \( \frac{[\text{A}^-]}{[\text{HA}]} = 4.37 \). This tells us that for every mole of acetic acid present, there needs to be 4.37 moles of acetate ions to maintain the desired pH. Maintaining the correct ratio is vital for the buffer's efficacy. If too much of one component is present without the correct proportion of the other, the buffer capacity to resist pH changes will be compromised.
Molarity and Volume Calculations
Molarity (M) is the number of moles of a solute per liter of solution, and it is an essential concept for preparing solutions in chemistry. Understanding molarity allows us to calculate the volumes of different solutions needed to prepare a buffer with specific concentration and volume. For our 0.1 M acetate buffer, we use the molarity and the moles calculated for each component (acetic acid and sodium acetate) to find the volumes required. With the molarity being constant at 0.1 M for both the acetic acid and sodium acetate solutions, the volumes are directly proportional to the number of moles needed for each substance, allowing us to mix these volumes accurately to result in a final 1L solution of the desired concentration and pH.