Question

You make \(1.00 \mathrm{~L}\) of a buffered solution \((\mathrm{pH}=4.00)\) by mixing acetic acid and sodium acetate. You have \(1.00 M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

Short answer

To create a 1.00 L buffered solution with pH 4.00, mix approximately 0.846 L of 1.00 M acetic acid solution and 0.154 L of 1.00 M sodium acetate solution.

Step-by-step solution

Identify the components of the buffered solution

The buffered solution contains acetic acid (CH3COOH) and sodium acetate (NaCH3COO). The acetic acid will provide the H+ ions while the sodium acetate will provide the CH3COO- ions in the solution.

Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a solution to the pKa of the system and the ratio of the concentrations of the conjugate base and the acid: pH = pKa + log ([conjugate base]/[acid]) We know the pH (4.00) and the pKa of acetic acid, which is 4.74. Thus: \(4.00 = 4.74 + \log\left(\frac{[\text{CH}_3\text{COO}^{-}]}{[\text{CH}_3\text{COOH}]}\right)\) Now we can calculate the ratio.

Calculate the ratio of concentrations

Rearrange the equation from Step 2: \(\frac{[\text{CH}_3\text{COO}^{-}]}{[\text{CH}_3\text{COOH}]} = 10^{4.00 - 4.74}\) After evaluating the power: \(\frac{[\text{CH}_3\text{COO}^{-}]}{[\text{CH}_3\text{COOH}]} \approx 0.183\) Now, we have the ratio of the concentrations.

Analyze the mass balance condition

Let x be the volume (in L) of 1.00 M acetic acid solution to be mixed and y be the volume (in L) of 1.00 M sodium acetate solution to be mixed. Then, we have: Total volume: \(x + y = 1\) (since we're making 1 L of buffered solution) Total acetic acid moles: \(x \times 1\) Total sodium acetate moles: \(y \times 1\) Now apply the ratio calculated in Step 3: \(\frac{y}{x} \approx 0.183\)

Calculate the required volumes

Solve the two equations with two unknowns: \[x + y = 1\] \[\frac{y}{x} \approx 0.183\] Using the second equation, we get \(y \approx 0.183x\) Now, substitute this equation into the first equation: \(x + (0.183x) = 1\) Solving for x and y, we get: \(x \approx 0.846\text{ L}\) (volume of 1.00 M acetic acid solution) \(y \approx 1.00 - 0.846 \approx 0.154 \text{ L}\) (volume of 1.00 M sodium acetate solution) So, to create a 1.00 L buffered solution with pH 4.00, we need to mix approximately 0.846 L of 1.00 M acetic acid solution and 0.154 L of 1.00 M sodium acetate solution.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is crucial when preparing a buffered solution. This equation makes it a breeze to find the right balance between an acid and its conjugate base needed to achieve a desired pH level. In essence, it describes the relationship between the pH of a solution and the pKa value, which is the acid dissociation constant of the weak acid, as well as the ratio of the concentrations of the conjugate base to the acid.
The equation itself is pretty straightforward:
\[\begin{equation}pH = pKa + \text{log}\left(\frac{[conjugate\ base]}{[acid]}\right)\end{equation}\]
This formula is especially handy because it allows you to use the logarithm of the ratio of the two components instead of their actual concentrations. This means, with the pH and pKa known, you can easily calculate the amounts needed for each, making it the go-to equation for scientists and students alike when mixing buffer solutions.

Acetic Acid

Acetic acid ((CH_3COOH)), often recognized for giving vinegar its tangy taste, plays an essential role in buffer solutions. As a weak acid, it doesn't completely dissociate in water, leaving some of the molecules intact while others release hydrogen ions (H^+).
This balance between undissociated acetic acid and the produced hydrogen ions is perfect for creating a stable environment where the pH doesn't fluctuate much when small amounts of acid or base are added. For a student aiming to concoct the perfect buffered solution, getting a hang of how acetic acid behaves in solutions with different concentrations of its conjugate base, acetate, is a valuable skill.

Sodium Acetate

Similarly to acetic acid, sodium acetate (NaCH_3COO) is a significant player in creating a buffer. It's the salt formed from the neutralization of acetic acid with sodium hydroxide (NaOH), and in a buffer solution, it provides the conjugate base, acetate ions (CH_3COO^-).
Sodium acetate can readily dissolve in water and dissociate completely to give acetate ions, which then pair with the hydrogen ions in case any additional acid is put into the solution. This partnership between sodium acetate and acetic acid is pivotal to maintaining a consistent pH as it equips the solution with the ability to neutralize small amounts of other acids or bases added to the mix.

pH Calculation

The very essence of a buffered solution is its pH stability, and calculating the pH is a fundamental skill. To find the pH, as seen in the given exercise, we don't measure the actual hydrogen ion concentration directly. Instead, we rely on the Henderson-Hasselbalch equation.
By taking the known pKa value of the weak acid and the ratio of the concentrations of the conjugate base and the acid, we can punch those numbers into the equation to deduce the pH. It's a classic example of chemistry's puzzle-solving nature — with just a few pieces of information, you can unveil the numeric character (pH) of a solution that defines its acidic or basic nature.

Conjugate Base and Acid Ratio

The ratio of a conjugate base to its corresponding acid is what gives buffered solutions their superpower — the ability to resist changes in pH. In the classroom scenario, students must learn to work out this ratio to create a buffer with the desired pH. As exhibited in the exercise, once you've grasped the target pH and pKa of the acid in question, it's only a matter of some arithmetic to find the ratio using the Henderson-Hasselbalch equation.
A proper understanding of this ratio allows you to predict how the buffer will behave under various conditions, equipping you to create a buffer tailored to your needs. Whether it's for an experiment or an industrial process, nailing this ratio is a crucial step in buffer preparation.