Question

A biochemist needs 750 \(\mathrm{mL}\) of an acetic acid-sodium acetate buffer with \(\mathrm{pH} 4.50 .\) Solid sodium acetate \((\mathrm{CH}_{3}$$ \mathrm{COONa}\) and glacial acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) are \right. available. Glacial acetic acid is 99\(\% \mathrm{CH}_{3} \mathrm{COOH}\) by mass and has a density of 1.05 \(\mathrm{g} / \mathrm{mL}\) . If the buffer is to be 0.15 \(\mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COOH}\) , how many grams of \(\mathrm{CH}_{3} \mathrm{COONa}\) and how many milliliters of glacial acetic acid must be used?

Short answer

To prepare a 750 mL, 0.15 M acetic acid-sodium acetate buffer with pH 4.50, approximately 7.97 g of sodium acetate (\(CH_3COONa\)) and 11.0 mL of glacial acetic acid (\(CH_3COOH\)) are required.

Step-by-step solution

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] Where: - pH is the desired pH of the buffer solution, - pKa is the acid dissociation constant of the weak acid, - [A^-] is the concentration of the conjugate base (sodium acetate), - [HA] is the concentration of the weak acid (acetic acid).

Find the pKa value for acetic acid

First, we have to find the pKa value for acetic acid. It is given: \(pKa_{CH_3COOH} = 4.74\)

Plug in the values and solve for the concentration of the sodium acetate

We are given the desired pH (4.50) and the concentration of acetic acid (0.15 M). Using the Henderson-Hasselbalch equation, solve for the concentration of sodium acetate: \[4.50 = 4.74 + \log \frac{[CH_3COO^-]}{0.15}\] Subtract 4.74 from both sides of the equation: \[-0.24 = \log \frac{[CH_3COO^-]}{0.15}\] Now, use the inverse log function (antilog) to find the concentration of sodium acetate: \[[CH_3COO^-] = 0.15 \times 10^{-0.24}\] \[[CH_3COO^-] ≈ 0.13 \, M\]

Calculate the amounts of sodium acetate and acetic acid needed

Now that we have found the concentrations of sodium acetate and acetic acid required for the buffer solution, we can calculate the quantities needed. First, we'll calculate the amount of sodium acetate. Recall that the final volume of the buffer solution is 750 mL. Using the formula for moles, mass and molar mass, we can find the amount of sodium acetate required: \[grams \, of \, CH_3COONa = 0.13 \, M \times 750 \, mL \times \frac{82.03 \, g}{1 \, mol}\] \[grams \, of \, CH_3COONa ≈ 7.968 \, g\] Next, calculate the volume of glacial acetic acid. Since glacial acetic acid is 99% \(CH_3COOH\) by mass and has a density of 1.05 g/mL, we can calculate the volume needed for 0.15 M acetic acid in 750 mL buffer solution: \[mL \, of \, glacial \, acetic \, acid = \frac{0.15 \, M \times 750 \, mL \times \frac{60.05 \, g}{1 \, mol}}{0.99 \times 1.05 \, g/mL}\] \[mL \, of \, glacial \, acetic \, acid ≈ 11.013 \, mL\]

Round to an appropriate number of significant figures

Lastly, round the amounts of sodium acetate and glacial acetic acid to an appropriate number of significant figures (2 or 3, in this case): Grams of sodium acetate needed ≈ 7.97 g Milliliters of glacial acetic acid needed ≈ 11.0 mL

Key concepts

Henderson-Hasselbalch Equation

When it comes to preparing a buffer solution, the Henderson-Hasselbalch equation is a vital tool for scientists. It beautifully links the pH of a buffer to the pKa of the weak acid and the ratio of the concentrations of its conjugate base and the acid itself. In simple terms, this equation helps chemists balance the amounts of acid and base to maintain a stable pH environment, which is crucial in many biological and chemical applications.

For those students struggling to understand its application, consider the equation as a recipe that tells you how to mix the right amounts of acid and base. As you saw in the exercise example, knowing the desired pH and the pKa value allows you to work out the ratio of base to acid required. Once you have that ratio, it's all about calculating the actual quantities needed—a vital step for any lab work involving buffers.

It's also important to remember that while the equation is straightforward, accuracy is key. Making precise measurements and calculations will lead to a successful buffer that resists changes in pH, which is especially important in sensitive experiments, like those in biochemistry.

Acid Dissociation Constant (pKa)

The acid dissociation constant (pKa) is a crucial parameter in understanding the strength and behavior of acids in aqueous solutions. Essentially, pKa gives you an idea of how likely an acid is to donate a proton in water—a critical piece of knowledge when crafting a buffer solution.

For students, remembering that a lower pKa value indicates a stronger acid is a helpful tip. Stronger acids dissociate more completely in water, releasing more protons, and thus have lower pKa values. The pKa of acetic acid, used in the exercise, is 4.74, suggesting it's weak since this value is higher than that of strong acids.

When preparing a buffer, knowing the pKa helps predict how the acid will behave in the buffer mixture. As you've seen, the pKa is directly utilized in the Henderson-Hasselbalch equation to determine the right balance between acid and conjugate base, ensuring the buffer maintains the desired pH even when small amounts of other acids or bases are introduced.

Molarity Calculation

The concept of molarity is at the heart of many scientific calculations, especially in chemistry. Molarity describes the number of moles of a solute per liter of solution, and it's represented as moles per liter (M). This measure lets you understand the strength or concentration of a solution.

Whether you're a budding chemist or a student in the middle of lab work, mastering molarity calculations is essential. For example, to prepare our buffer solution, we needed to calculate the molarity of both acetic acid and sodium acetate. Understanding these calculations allows you to measure how much of each compound you need to mix together to form a solution of a desired concentration.

In the exercise, molarity is used to calculate the quantity of acetic acid and sodium acetate in grams and milliliters, respectively. These kinds of practical applications highlight how important it is to be comfortable with molarity calculations—not just for homework problems, but for real-life science as well.