Question

You are working in a high-powered clinical biochemistry lab. The chief scientist rushes in and announces, "I need \(500 \mathrm{ml}\) of \(0.2 \mathrm{M}\) acetate, \(\mathrm{pH}\) 5.0. STAT! Who is the best and brightest in this room?" All eyes turn toward you. You have solid anhydrous sodium acetate \(\left(\mathrm{MW}=82 \mathrm{g} \mathrm{mol}^{-1}\right)\) and a solution of \(1 \mathrm{M}\) acetic acid. Describe how you would make the buffer. \(\quad 6\)

Short answer

Mix 5.25 g sodium acetate with 36 ml of 1M acetic acid and water to 500 ml.

Step-by-step solution

Understanding the Buffer Components

We need a buffer of 0.2 M acetate at pH 5. Sodium acetate and acetic acid must combine to achieve this. Sodium acetate provides the acetate ion, and acetic acid provides the acetic ion for the buffer.

Calculating Moles Needed

For a 0.2 M solution in 500 ml, we require \(0.2 \times 0.5 = 0.1\) moles of acetate ions in total. This will include sodium acetate and acetic acid.

Using the Henderson-Hasselbalch Equation

Apply the Henderson-Hasselbalch equation: \[\mathrm{pH} = \mathrm{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\] For acetic acid, \(\mathrm{pK}_a = 4.76\). Solve for \(\frac{[\text{A}^-]}{[\text{HA}]}\) at pH 5.

Ratio Calculation

Rearrange the equation to: \[5 = 4.76 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\] \[\log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 0.24\] Thus, \(\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.74\).

Determine Moles of Sodium Acetate and Acetic Acid

Let \([\text{A}^-] = x\), the moles of sodium acetate, then \([\text{HA}] = 0.1 - x\), the moles of acetic acid. From the ratio \(\frac{x}{0.1-x} = 1.74\), solve for \(x\): \(x = 0.064\) moles. Hence, sodium acetate = 0.064 moles, and acetic acid = 0.1 - 0.064 = 0.036 moles.

Convert Moles to Grams for Sodium Acetate

Calculate the mass required for 0.064 moles of sodium acetate: \[\text{mass} = 0.064 \times 82 = 5.25 \text{ grams}.\]

Prepare the Buffer Solution

Dissolve 5.25 grams of solid sodium acetate in water. Add 0.036 liters (or 36 ml) of 1 M acetic acid to the solution. Adjust the final volume to 500 ml with water. Verify the pH and adjust if necessary.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental equation used in buffer solutions to calculate the pH balance between a weak acid and its conjugate base. It is expressed as: \[\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]where \( \text{A}^- \) represents the concentration of the base form of the acid and \( \text{HA} \) symbolizes the concentration of the acid. The value of \( \text{pKa} \) is a measure of the acidity of the acid, directly related to the acid dissociation constant (\( \text{Ka} \)).In this exercise, we have a buffer solution that combines acetic acid (a weak acid) and sodium acetate (its conjugate base). Using the Henderson-Hasselbalch equation, we aim to achieve a pH of 5.0. Given the \( \text{pKa} \) of acetic acid is 4.76, the equation reveals the needed ratio of conjugate base to weak acid to maintain the desired pH level. This ratio is critical in controlling the pH when acids or bases are added to the solution, showcasing the buffer's stability.

Acetate Buffer

An acetate buffer is a solution that maintains a stable pH by combining acetic acid and its conjugate base, sodium acetate. Acetate buffers are commonly used in biochemical experiments where it's critical to sustain an environment's pH.In buffer preparation, we calculate the necessary amounts of acetic acid and sodium acetate to achieve the desired concentration and \([\text{pH}\)The buffer needs to resist changes in pH when small amounts of other acids or bases are introduced. - **Acetic Acid (\( \text{CH}_3\text{COOH} \))** - Acts as the weak acid that donates protons. - Works with its conjugate base to form a buffer solution. - **Sodium Acetate (\( \text{CH}_3\text{COONa} \))** - Provides acetate ions (\( \text{CH}_3\text{COO}^- \)). - Combines with protons to create acetic acid, maintaining a steady concentration of ions.To prepare this acetate buffer, we calculate precise amounts of each component. For example, using the steps outlined in the solution, we determine how much sodium acetate and acetic acid to mix to maintain a pH of 5 and use the Henderson-Hasselbalch equation to find the exact balance.

pH Calculation

Calculating the pH of a buffer solution involves determining the dissociation balance between a weak acid and its conjugate base. While this might sound daunting, the process itself is methodical and mathematical.In our exercise, the aim was to prepare an acetate buffer with a pH level set at 5.0. The initial calculation seeks to find out how many moles of acetate ions are needed in total. This is done by determining the molarity and volume required: \[moles = \text{molarity} \times \text{volume}\]Next, with the Henderson-Hasselbalch equation, you calculate their ratio at the required pH. Rearranging the equation and substituting known values allows us to solve for the number of moles of acetic acid and acetate.Finally, once you have calculated the moles of each component, you can use this information to: - Determine the mass required for sodium acetate - Measure the volume of acetic acid needed, Combine them to adjust the buffer solution to 500 ml. This calculation method is vital for ensuring accuracy in laboratory settings, where maintaining specific pH levels is essential for reliable results. Every calculated increment supports precision in how a buffer resists changes in its hydrogen ion concentration.