Problem 81 - Chemistry: The Central Science

Question

A biochemist needs \(750 \mathrm{~mL}\) of an acetic acid-sodium ?cetate buffer with pH 4.50. Solid sodium acetatè \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) 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 \(\mathrm{CH}_{3} \mathrm{COONa}\) and how many milliliters of glacial acetic acid must be used?

Step-by-Step Solution

Answer

In conclusion, to prepare a $750\,mL$ acetic acid-sodium acetate buffer with pH 4.50 and $0.15\,M$ in $\mathrm{CH}_{3}\mathrm{COOH}$, you should use approximately \(8.29\,g\) of sodium acetate (\(\mathrm{CH}_{3}\mathrm{COONa}\)) and \(6.46\,mL\) of glacial acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH\)).

Step 1: Calculate the pK_a of acetic acid

The acid dissociation constant (K_a) of acetic acid is \(1.8 \times 10^{-5}\). Now, we can calculate pK_a using the formula: \(pK_a = -log(K_a)\) By substituting the value of \(K_a\), \(pK_a = -log(1.8 \times 10^{-5}) \approx 4.74\)

Step 2: Use the Henderson-Hasselbalch equation to find the concentration of sodium acetate

Using the desired pH (4.50) from the exercise and the pK_a value we just found (4.74), we can solve for the concentration of sodium acetate using the Henderson-Hasselbalch equation: \(4.50 = 4.74 + log\frac{[A^-]}{[HA]}\) Since the buffer is supposed to be \(0.15 M\) in acetic acid, we can use this value as the concentration of [HA]. Rearrange the equation to find the concentration of sodium acetate ([A^-]): \(log\frac{[A^-]}{0.15} = -0.24\) Now, we can find the concentration of sodium acetate: \(\frac{[A^-]}{0.15} = 10^{-0.24}\) \([A^-] = 0.15 \times 10^{-0.24} \approx 0.134\,M\)

Step 3: Calculate the mass of sodium acetate needed

Now that we have the concentration of sodium acetate, we can calculate the mass needed for the buffer solution. To do this, we will use the formula: Mass of Sodium Acetate = Molarity × Volume × Molar mass The molar mass of sodium acetate (CH3COONa) is approximately 82.03 g/mol. The volume of the buffer solution is given as 750 mL or 0.750 L (since 1 L = 1000 mL). Plugging in the values: Mass of Sodium Acetate = \(0.134\, M \times 0.750\,L \times 82.03\,g/mol \approx 8.29\,g\) Hence, about 8.29 grams of sodium acetate should be used.

Step 4: Calculate the volume of glacial acetic acid needed

We know that the buffer solution must be \(0.15\,M\) in acetic acid, and we know the glacial acetic acid is \(99\%\) acetic acid by mass with a density of \(1.05\, g/mL\). To find the volume of glacial acetic acid needed, we use the following formula: Volume of Glacial Acetic Acid = \(\frac{Mass\,of\,Acetic\,Acid}{(Density\,of\,Glacial\,Acetic\,Acid) \times (%\,of\,Acetic\,Acid\,by\,Mass)} \) First, we need to calculate the mass of acetic acid itself needed for the buffer solution: Mass of Acetic Acid = Molarity × Volume × Molar mass The molar mass of acetic acid (CH3COOH) is approximately 60.05 g/mol. Using the given buffer volume (0.750 L) and concentration (0.15 M), we get: Mass of Acetic Acid = \(0.15\,M \times 0.750\,L \times 60.05\,g/mol \approx 6.76\,g\) Now, we can find the volume of glacial acetic acid needed: Volume of Glacial Acetic Acid = \(\frac{6.76\,g}{(1.05\,g/mL) \times (0.99)} \approx 6.46\,mL\) Hence, approximately 6.46 milliliters of glacial acetic acid should be used. In conclusion, to prepare the desired buffer solution, you should use 8.29 grams of sodium acetate and 6.46 milliliters of glacial acetic acid.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental formula used to calculate the pH of buffer solutions. A buffer solution is a mixture of a weak acid and its conjugate base, which helps maintain a relatively stable pH. The equation is given by:\[ pH = pK_a + \log \frac{[A^-]}{[HA]} \]Where:

• pH is the hydrogen ion concentration.
• pK_a is the negative logarithm of the acid dissociation constant (K_a).
• [A^-] is the concentration of the conjugate base (in this case, sodium acetate).
• [HA] is the concentration of the weak acid (acetic acid).The Henderson-Hasselbalch equation allows us to determine how much of each component is needed to achieve a desired pH.

This equation is derived from the equilibrium expression for weak acids, making it particularly useful in designing buffer solutions. When applied correctly, it ensures that the desired pH is reached by balancing the ratio of the acid and its conjugate base.
This makes the Henderson-Hasselbalch equation an indispensable tool in many chemical and biochemical applications.

Acetic Acid Dissociation Constant

The dissociation constant (K_a) of a weak acid such as acetic acid provides insight into the acid's strength.It quantifies the acid's tendency to donate protons in solution. For acetic acid (CH₃COOH), the dissociation constant is approximately \(1.8 \times 10^{-5}\).To use the constant in calculations, we often convert it to pK_a. The formula for this conversion is:\[ pK_a = -\log(K_a) \]Substituting the value for acetic acid, we find:\[ pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \]This value is crucial when using the Henderson-Hasselbalch equation to determine the concentrations of components needed.

• Importance of pK_a: A lower pK_a signifies a stronger acid due to its higher tendency to dissociate and release protons.
• Role in Buffers: Knowing the pK_a helps in predicting and controlling the pH of buffer solutions by adjusting the concentrations of the acid and its conjugate base.

Understanding pK_a is essential in chemistry, especially when preparing solutions that require precise acidity, such as in biochemical laboratories or industrial processes.

Sodium Acetate Concentration

In a buffer solution, the concentration of sodium acetate (CH₃COONa) plays a key role in determining the solution's pH.Sodium acetate is the conjugate base of acetic acid (CH₃COOH), and its concentration can be directly calculated using the Henderson-Hasselbalch equation:\[ [A^-] = [HA] \times 10^{(pH - pK_a)} \]For a given pH of 4.50, pK_a of 4.74, and acetic acid concentration of 0.15 M, we find:\[ [A^-] = 0.15 \times 10^{(4.50 - 4.74)} \approx 0.134 M \]

• Adjusting [A^-]: Altering the concentration of sodium acetate can shift the pH, offering control over the buffer system's acidity.
• Importance in Buffers: The balance between [A^-] and [HA] maintains the pH, showing the significance of precise concentration calculation.

Knowing the concentration of sodium acetate is crucial in preparing buffers, as it determines the capacity of the solution to resist pH changes.

Glacial Acetic Acid

Glacial acetic acid is a concentrated form of acetic acid with minimal water content, often used in laboratories.It is 99% pure by mass and has a density of 1.05 g/mL, making it a potent compound for buffer solutions. When calculating the volume of glacial acetic acid needed for a buffer, you first determine the mass of the acetic acid required:\[ \text{Mass of Acetic Acid} = ext{Molarity} \times ext{Volume} \times ext{Molar Mass} \]In this buffer solution, with a 0.15 M concentration over 750 mL using the molar mass of acetic acid (60.05 g/mol), the mass needed is approximately 6.76 g.To determine the required volume of glacial acetic acid:\[ ext{Volume} = \frac{ ext{Mass of Acetic Acid}}{ ext{Density} \times ext{% Acetic Acid by Mass}} \]Using our values, this results in approximately 6.46 mL of glacial acetic acid necessary for the solution.

• Purity and Use: The high purity ensures effective buffering in scientific procedures.
• Role in Calculation: It helps establish the amount of substance needed for accurate buffer preparation.

Understanding glacial acetic acid's properties and applications is critical for effective laboratory experimentation and precise formulation of buffer solutions.