Question

A biochemist needs $750\mathrm{mL}$ of an acetic acid-sodium acetate buffer with $\mathrm{pH}$ 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 $99\mathrm{\%}{\mathrm{CH}}_3\mathrm{COOH}$ 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 750 mL of a pH 4.50 buffer with a 0.15 M acetic acid, use 8.24 grams of sodium acetate and 6.49 mL of glacial acetic acid.

Step-by-step solution

Using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given as: $$\mathrm{pH}=\mathrm{p}{K}_a+\log \frac{[{\mathrm{A}}^{-}]}{[\mathrm{HA}]}$$ where $[{\mathrm{A}}^{-}]$ represents the concentration of the conjugate base (sodium acetate), $[\mathrm{HA}]$ represents the concentration of the weak acid (acetic acid) and 𝑝𝐾𝑎 is the acid dissociation constant of the weak acid. For acetic acid, the 𝑝𝐾𝑎 value is 4.74. Given the pH of the buffer (which is 4.50) and the concentration of acetic acid (0.15 M), we can calculate the concentration of sodium acetate needed as follows: $$4.50=4.74+\log \frac{[{\mathrm{CH}}_3\mathrm{COONa}]}{0.15}$$

Solve for the concentration of sodium acetate

Rearrange the equation to solve for the concentration of sodium acetate: $$\log \frac{[{\mathrm{CH}}_3\mathrm{COONa}]}{0.15}=4.50-4.74$$ $$\log [{\mathrm{CH}}_3\mathrm{COONa}]-\log (0.15)=-0.24$$ $$[{\mathrm{CH}}_3\mathrm{COONa}]={10}^{-0.24}\times 0.15$$ Calculating the concentration of sodium acetate, we get: $$[{\mathrm{CH}}_3\mathrm{COONa}]=0.134\mathrm{M}$$

Find the mass of sodium acetate needed

We know the concentration and the volume of the buffer solution, so we can find the moles and, subsequently, the mass of sodium acetate needed: $$\mathrm{moles}\mathrm{of}\mathrm{sodium}\mathrm{acetate}=[{\mathrm{CH}}_3\mathrm{COONa}]\times \mathrm{volume}\mathrm{of}\mathrm{buffer}$$ $$\mathrm{moles}\mathrm{of}\mathrm{sodium}\mathrm{acetate}=0.134\mathrm{M}\times 0.750\mathrm{L}=0.1005\mathrm{mol}$$ Next, we will find the mass using the molar mass of sodium acetate (82.03 g/mol): $$\mathrm{mass}\mathrm{of}\mathrm{sodium}\mathrm{acetate}=\mathrm{moles}\mathrm{of}\mathrm{sodium}\mathrm{acetate}\times \mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{sodium}\mathrm{acetate}$$ $$\mathrm{mass}\mathrm{of}\mathrm{sodium}\mathrm{acetate}=0.1005\mathrm{mol}\times 82.03\frac{\mathrm{g}}{\mathrm{mol}}=8.24\mathrm{g}$$

Calculate the volume of glacial acetic acid needed

We know the buffer is to be 0.15 M in acetic acid. Convert concentration to moles using the volume of the buffer: $$\mathrm{moles}\mathrm{of}\mathrm{acetic}\mathrm{acid}=[{\mathrm{CH}}_3\mathrm{COOH}]\times \mathrm{volume}\mathrm{of}\mathrm{buffer}$$ $$\mathrm{moles}\mathrm{of}\mathrm{acetic}\mathrm{acid}=0.15\mathrm{M}\times 0.750\mathrm{L}=0.1125\mathrm{mol}$$ Now, we will find the mass of acetic acid using the molar mass of acetic acid (60.05 g/mol): $$\mathrm{mass}\mathrm{of}\mathrm{acetic}\mathrm{acid}=\mathrm{moles}\mathrm{of}\mathrm{acetic}\mathrm{acid}\times \mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{acetic}\mathrm{acid}$$ $$\mathrm{mass}\mathrm{of}\mathrm{acetic}\mathrm{acid}=0.1125\mathrm{mol}\times 60.05\frac{\mathrm{g}}{\mathrm{mol}}=6.756\mathrm{g}$$ Now, let's calculate the volume considering that glacial acetic acid has only 99% acetic acid by mass and density $1.05\frac{\mathrm{g}}{\mathrm{mL}}$: $$\mathrm{mass}\mathrm{of}\mathrm{glacial}\mathrm{acetic}\mathrm{acid}=\frac{\mathrm{mass}\mathrm{of}\mathrm{acetic}\mathrm{acid}}{0.99}=\frac{6.756\mathrm{g}}{0.99}=6.82\mathrm{g}$$ $$\mathrm{volume}\mathrm{of}\mathrm{glacial}\mathrm{acetic}\mathrm{acid}=\frac{\mathrm{mass}\mathrm{of}\mathrm{glacial}\mathrm{acetic}\mathrm{acid}}{\mathrm{density}}=\frac{6.82\mathrm{g}}{1.05\frac{\mathrm{g}}{\mathrm{mL}}}=6.49\mathrm{mL}$$ In conclusion, to prepare 750 mL of a pH 4.50 buffer with a 0.15 M acetic acid, we must use 8.24 grams of sodium acetate and 6.49 mL of glacial acetic acid.

Key concepts

Buffer Solution

A buffer solution is a special type of solution that maintains a relatively constant pH level, even when small amounts of acids or bases are added. This characteristic makes buffer solutions very valuable in scientific and industrial applications. For instance, in biochemical experiments, maintaining a stable pH is crucial as fluctuations could significantly affect the outcome.Buffer solutions are typically made by combining a weak acid with its conjugate base, or a weak base with its conjugate acid. When small amounts of an acid or a base are introduced, the buffer neutralizes them, keeping the pH stable. In this context, an acetic acid and sodium acetate mixture is a classic example of a buffer system, often used in laboratory settings to maintain precise pH levels.

• It contains a weak acid, acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$), and its conjugate base, sodium acetate (${\mathrm{CH}}_3\mathrm{COONa}$).
• The buffer has a target pH of 4.50, making it slightly acidic.

This ability to resist pH change is central to many biological and chemical processes where specific pH levels are needed to maintain the integrity and function of the system.

Acetic Acid

Acetic acid, also known by its chemical formula ${\mathrm{CH}}_3\mathrm{COOH}$, is a weak organic acid that is widely used in buffer solutions. It serves as the acid component in the acetic acid-sodium acetate buffer system. Being a weak acid means it only partially dissociates in water, providing an equilibrium between the acetic acid molecules and their ions.One of the key properties of acetic acid is its $extp{K}_a$, which is a measure of its acid strength. For acetic acid, the $extp{K}_a$ value is 4.74. This value plays a vital role in the Henderson-Hasselbalch equation, a formula used to determine the pH of buffer solutions.

• In the buffer, acetic acid donates protons in addition reactions, helping to neutralize any added bases.
• It is available as glacial acetic acid, known for being mainly pure acetic acid with a density of 1.05 $extg/mL$.

Since the glacial variant is highly concentrated, precise measurements are necessary for creating accurate buffer solutions so the target molarity and pH are reached.

Sodium Acetate

Sodium acetate (${\mathrm{CH}}_3\mathrm{COONa}$) acts as the conjugate base in the acetic acid-sodium acetate buffer system. It supplies the acetate ions needed to counterbalance the hydrogen ions introduced by acetic acid, thereby stabilizing the pH. As a salt of a weak acid and a strong base, it readily dissociates in water.When preparing a buffer solution, the amount of sodium acetate used relative to acetic acid can be calculated using the Henderson-Hasselbalch equation. This formula helps attain the desired buffer pH by balancing the concentrations of the acidic and basic components.

• It is crucial for maintaining the acidic pH in this buffer system.
• The solution requires a careful calculation of sodium acetate to ensure the buffer remains effective.
• In this textbook problem, 8.24 grams of sodium acetate are necessary to achieve the target buffer conditions.

Overall, sodium acetate works in tandem with acetic acid, ensuring the solution behaves consistently as a buffer, providing effective resistance to pH changes.