Question

You have to prepare a \(\mathrm{pH} 3.50\) buffer, and you have the following \(0.10 \mathrm{M}\) solutions available: \(\mathrm{HCOOH}, \mathrm{CH}_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}\) \(\mathrm{HCOONa}, \mathrm{CH}_{3} \mathrm{COONa}\), and \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\). Which solutions would you use? How many milliliters of each solution would you use to make approximately 1 L of the buffer?

Short answer

To prepare a pH 3.50 buffer using the available solutions, you would use formic acid (\(\mathrm{HCOOH}\)) and sodium formate (\(\mathrm{HCOONa}\)), since their \(\mathrm{p}K_{a}\) value is closest to the desired pH. To make approximately 1 L of the buffer, you would need to use about 641 mL of the 0.10 M \(\mathrm{HCOOH}\) solution and 359 mL of the 0.10 M \(\mathrm{HCOONa}\) solution.

Step-by-step solution

Identify the Candidate Solutions

The first thing we need to do is to identify which solutions would be appropriate for creating a pH 3.50 buffer. To do this, let's find the \(\mathrm{p}K_{a}\) of the weak acids available in the given solutions, using a \(\mathrm{p}K_{a}\) table: 1. Formic Acid (\(\mathrm{HCOOH}\)): \(\mathrm{pKa} = 3.75\) 2. Acetic Acid (\(\mathrm{CH_{3}COOH}\)): \(\mathrm{pKa} = 4.76\) 3. Phosphoric Acid (\(\mathrm{H_{3}PO_{4}}\)): \(\mathrm{pKa_1} = 2.15\), \(\mathrm{pKa_2} = 7.20\), \(\mathrm{pKa_3} = 12.35\) For a buffer solution, the ideal \(\mathrm{pH}\) should be close to the \(\mathrm{p}K_{a}\) of the weak acid present in the solution. Since the desired \(\mathrm{pH}\) is 3.50, formic acid (\(\mathrm{HCOOH}\)) and its corresponding salt, sodium formate (\(\mathrm{HCOONa}\)), will be the best candidates.

Calculate Buffer Ratio Using the Henderson-Hasselbalch Equation

Now, we will use the Henderson-Hasselbalch equation to determine the concentration ratio of the weak acid and its salt in the buffer solution: \[\mathrm{pH} = \mathrm{p}K_{a} + \log_{10} \left(\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}\right)\] We will plug in the values for the desired pH and the pKa of formic acid: \(3.50 = 3.75 + \log_{10} \left(\frac{[\mathrm{HCOO^{-}}]}{[\mathrm{HCOOH}]}\right)\)

Solve for Buffer Ratio

We will now solve for the buffer ratio \(\frac{[\mathrm{HCOO^{-}}]}{[\mathrm{HCOOH}]}\): \(-0.25 = \log_{10} \left(\frac{[\mathrm{HCOO^{-}}]}{[\mathrm{HCOOH}]}\right)\) \[\frac{[\mathrm{HCOO^{-}}]}{[\mathrm{HCOOH}]} = 10^{-0.25} \approx 0.56\] This means that for every 1 L of solution, the concentration of \(\mathrm{HCOO^{-}}\) (in \(\mathrm{HCOONa}\)) should be approximately 0.56 times the concentration of \(\mathrm{HCOOH}\).

Calculate the Amounts of Each Solution

The given solutions have a concentration of \(0.10 \ \mathrm{M}\), and we need to make a buffer solution with a total volume of approximately 1 L. Using the concentration ratio calculated above, let's determine the volumes of each solution needed: \[\mathrm{Volume \ of \ HCOOH} = V_1\] \[\mathrm{Volume \ of \ HCOONa} = V_2\] \[\frac{0.10 \ \mathrm{M} \cdot V_2}{0.10 \ \mathrm{M} \cdot V_1} \approx 0.56\] \[V_2 \approx 0.56 \cdot V_1\] Since we need a total volume of 1 L, we can write: \[V_1 + V_2 = 1000 \ \mathrm{mL}\] Substitute the expression for \(V_2\): \[V_1 + 0.56 \cdot V_1 = 1000\] \[1.56 \cdot V_1 = 1000\] \[V_1 \approx 641 \ \mathrm{mL}\] \[V_2 \approx 0.56 \cdot 641 \approx 359 \ \mathrm{mL}\] So, you would use approximately 641 mL of the 0.10 M \(\mathrm{HCOOH}\) solution and 359 mL of the 0.10 M \(\mathrm{HCOONa}\) solution to make 1 L of a pH 3.50 buffer.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is essential for preparing pH buffer solutions, as it relates the pH of a solution to the \(pK_a\) of the acid and the ratio of the concentrations of the acid and its conjugate base. The equation is given by:
\[ pH = pK_a + \log_{10} \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
In the context of the exercise, the equation was used to calculate the ratio of sodium formate (conjugate base, \(\text{A}^-\)) to formic acid (weak acid, \(\text{HA}\)) required to achieve a buffer with a pH of 3.50. This understanding is imperative because it allows us to manipulate the quantities of acid and base to create a buffer solution with a target pH, which is a crucial skill in various chemistry and biology applications.

Buffer Capacity

Buffer capacity refers to the ability of a buffer solution to resist changes in pH when acids or bases are added. It is highly dependent on the concentration of the weak acid and its conjugate base within the solution. As a rule of thumb, the closer the molar amounts of the weak acid and its conjugate base, the higher the buffer capacity around the \(pK_a\) value of the acid. In our exercise, we used a 0.10 M concentration for both the weak acid and its conjugate base, providing sufficient buffering around the pH of 3.50. This concept is particularly important in biological systems, where pH must be tightly regulated.

Acid-Base Equilibrium

Acid-base equilibrium is a fundamental concept in chemistry that involves the balance between acids and bases in a solution. This equilibrium is crucial for buffer solutions, as they consist of a weak acid and its conjugate base in equilibrium with each other. Adding an acid or base to the buffer system causes the equilibrium to shift according to Le Châtelier's principle, but the pH remains relatively stable. In the exercised scenario, we have formulated a buffer that can maintain equilibrium, and thus a stable pH, despite small amounts of additional acids or bases. Such equilibria are pervasive in chemistry, from titrations to physiological processes.

Weak Acid and Conjugate Base Pair

The selection of a weak acid and its conjugate base pair is fundamental in buffer preparation, as seen in our exercise with formic acid and sodium formate. A weak acid partially dissociates in solution to yield its conjugate base and a proton (\(H^+\)). The weak acid and its conjugate base are in equilibrium, which allows the buffer to neutralize added acid or base. The pair's ability to moderate pH changes is based on their relative concentrations in solution. Properly choosing the pair with a \(pK_a\) close to the desired buffer pH ensures the most effective buffering action, exemplified by our choice of formic acid with a \(pK_a\) of 3.75 to create a buffer with a pH of 3.50.