Question

\mathrm{~A}\( biochemist needs \)750 \mathrm{~mL}\( of an acetic acid-sodium acetate buffer with pH 4.50. Solid sodium acetate ( \)\left.\mathrm{CH}_{3} \mathrm{COONa}\right)\( and glacial acetic acid \)\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\( are available. Glacial acetic acid is \)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}_{4} \mathrm{COONa}$ and how many milliliters of glacial acetic acid must be used?

Short answer

To prepare a 750 mL buffer solution with a pH of 4.50 and a 0.15 M concentration of acetic acid, approximately 6.33 grams of sodium acetate (\(\mathrm{CH}_{3}\mathrm{COONa}\)) and 35.80 mL of glacial acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) are needed.

Step-by-step solution

Write down the given information and find molecular weights

We have the following information: - pH = 4.50 - Total volume of the buffer, \(V_{buffer} = 750 \ \mathrm{mL}\) - Concentration of acetic acid, \([\mathrm{CH}_{3}\mathrm{COOH}] = 0.15 \ \mathrm{M}\) - Mass percentage of glacial acetic acid = 99% - Density of glacial acetic acid = \(\rho_{glacial \ acetic \ acid} = 1.05 \ \mathrm{g/mL}\) Now, find the molecular weights of acetic acid and sodium acetate: - \(MW_{\mathrm{CH}_{3}\mathrm{COOH}} \approx 12 + 4 + 16 + 12 + 16 \ \mathrm{g/mol} = 60 \ \mathrm{g/mol}\) - \(MW_{\mathrm{CH}_{3}\mathrm{COONa}} \approx 12 + 4 + 16 + 12 + 16 + 23 \ \mathrm{g/mol} = 82 \ \mathrm{g/mol}\)

Calculate the concentration of sodium acetate using Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the acid and its conjugate base in a buffer solution: \[pH = pKa + \log{\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}}\] For acetic acid, \(pKa = 4.74\). We need to find the concentration of sodium acetate, \([\mathrm{CH}_{3}\mathrm{COO}^-]\). Plug in the known values into the equation: \[4.50 = 4.74 + \log{\frac{[\mathrm{CH}_{3}\mathrm{COO}^-]}{0.15 \ \mathrm{M}}}\] Rearrange the equation to solve for \([\mathrm{CH}_{3}\mathrm{COO}^-]\): \[[\mathrm{CH}_{3}\mathrm{COO}^-] = 0.15 \ \mathrm{M} \times 10^{(4.50 - 4.74)}\] Calculate the concentration of sodium acetate: \[[\mathrm{CH}_{3}\mathrm{COO}^-] \approx 0.15 \ \mathrm{M} \times 10^{-0.24} \approx 0.103 \ \mathrm{M}\]

Determine the needed amount of acetic acid in milliliters

Calculate the moles of acetic acid needed using the given concentration, \([\mathrm{CH}_{3}\mathrm{COOH}]\), and the total volume of the buffer, \(V_{buffer}\): \[moles \ \mathrm{CH}_{3}\mathrm{COOH}= [\mathrm{CH}_{3}\mathrm{COOH}] \times V_{buffer} \times \frac{10^{-3}}{1}\] \[= 0.15 \ \mathrm{M} \times 750 \ \mathrm{mL} \times 10^{-3 \mathrm{L/mL}} \approx 0.1125 \ \mathrm{mol}\] Now we can calculate the volume of glacial acetic acid needed using its given mass percentage and density: \[V_{glacial \ acetic \ acid} = \frac{moles \ \mathrm{CH}_{3}\mathrm{COOH} \times MW_{\mathrm{CH}_{3}\mathrm{COOH}}}{\rho_{glacial \ acetic \ acid} \times mass \ \% \ \mathrm{CH}_{3}\mathrm{COOH}}\] \[= \frac{0.1125 \ \mathrm{mol} \times 60 \ \mathrm{g/mol}}{1.05 \ \mathrm{g/mL} \times 0.99} \] Calculate the volume of glacial acetic acid: \[V_{glacial \ acetic \ acid} \approx \frac{0.1125 \times 60}{1.05 \times 0.99} \approx 35.80 \ \mathrm{mL}\]

Calculate the needed amount of sodium acetate in grams

Calculate the moles of sodium acetate needed using the concentration, \([\mathrm{CH}_{3}\mathrm{COO}^-]\), and the total volume, \(V_{buffer}\): \[moles \ \mathrm{CH}_{3}\mathrm{COO}^{-} \mathrm{(as \ CH}_{3} \mathrm{COONa)} = [\mathrm{CH}_{3}\mathrm{COO}^-] \times V_{buffer} \times \frac{10^{-3}}{1}\] \[= 0.103 \ \mathrm{M} \times 750 \ \mathrm{mL} \times 10^{-3 \mathrm{L/mL}} \approx 0.07725 \ \mathrm{mol}\] Now we can calculate the mass of sodium acetate needed using its molecular weight, \(MW_{\mathrm{CH}_{3}\mathrm{COONa}}\): \[mass \ \mathrm{CH}_{3}\mathrm{COONa} = moles \ \mathrm{CH}_{3}\mathrm{COO}^{-}\mathrm{(as \ CH}_{3} \mathrm{COONa)} \times MW_{\mathrm{CH}_{3}\mathrm{COONa}}\] \[= 0.07725 \ \mathrm{mol} \times 82 \ \mathrm{g/mol} \] Calculate the mass of sodium acetate: \[mass \ \mathrm{CH}_{3}\mathrm{COONa} \approx 0.07725 \times 82 \approx 6.33 \ \mathrm{g}\]

Summarize the results

To prepare a \(750 \ \mathrm{mL}\) buffer solution with a pH of 4.50 and a concentration of acetic acid, \([\mathrm{CH}_{3}\mathrm{COOH}] = 0.15 \ \mathrm{M}\), the following amounts of sodium acetate and glacial acetic acid are needed: - Sodium acetate (\(\mathrm{CH}_{3}\mathrm{COONa}\)): \(\approx 6.33 \ \mathrm{g}\) - Glacial acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)): \(\approx 35.80 \ \mathrm{mL}\)

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a key player when dealing with buffer solutions. It bridges the gap between the pH of a solution and the pKa, which is the acid dissociation constant of the acid in question. Specifically, for a buffer solution consisting of a weak acid and its conjugate base, the equation reads: \[\begin{equation}pH = pKa + \log\left(\frac{{[\mathrm{A}^-]}}{{[\mathrm{HA}]}}\right)\end{equation}\]With\(<[\mathrm{A}^-]>\) representing the molarity of the conjugate base, and\(<[\mathrm{HA}]>\) the molarity of the acid. What this equation essentially does is allow us to calculate the ideal proportions of the two components to achieve a desired pH level, making it critical in the preparation of buffer solutions.

In the context of our example, the equation aids in finding out how much sodium acetate (conjugate base) is needed given the concentration of acetic acid and the desired pH of the buffer.

Acetic Acid-Sodium Acetate Buffer

The acetic acid-sodium acetate buffer system is a classic example of a buffer solution. It works on the principle of an equilibrium between acetic acid (a weak acid) and its conjugate base, the acetate ion, which comes from the sodium acetate salt.

When an acid or base is added to such a buffer, the equilibrium shifts to counteract the pH change. This buffering action makes it a preferred choice in many biochemical applications where pH stability is crucial. Making such a buffer involves combining acetic acid with sodium acetate in precise amounts, in line with the aforementioned Henderson-Hasselbalch equation. It's necessary to consider things like purity and density, especially when using substances like glacial acetic acid, which is undiluted acetic acid, in the preparation process.

pH Calculation

Understanding pH

The pH scale is a measure of the acidity or basicity of an aqueous solution. It's a logarithmic scale based on the concentration of hydrogen ions in the solution, and it runs from 0 to 14, with 7 being neutral. The calculation of pH is fundamental in chemistry, especially when preparing solutions with precise acidic or basic natures.

In the given exercise, calculating the pH is integral to determining how much of each component is required. Once the concentration of the acid and its conjugate base is known, the pH can be calculated using the Henderson-Hasselbalch equation, which is especially useful for buffered solutions like the one in our exercise.

Molarity and Concentration

Key Measurements in Solution Preparation

Molarity is a measure of concentration that indicates the number of moles of a solute (here, acetic acid or sodium acetate) per liter of solution. It allows chemists to quantify the precise amounts of substances needed to create a solution with specific properties. Concentration, more broadly, may also refer to mass per unit volume or percentage composition by mass.

In practical terms, when preparing our buffer solution, understanding molarity and concentration guarantees that the mixture has the exact properties required for its intended use—this includes the buffering capacity and the pH level. Accurate calculations are essential, as even slight deviations can lead to a buffer with ineffective or undesired properties.