Question

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

Short answer

To make a pH 3.50 buffer with the available 0.10 M solutions, use Formic Acid (HCOOH) and Sodium formate (HCOONa) as they have the closest pK_a value to the desired pH. Using the Henderson-Hasselbalch equation, the ratio of salt to acid is found to be 0.5623. To make approximately 1 liter of the buffer, use 640 mL of 0.10 M HCOOH and 360 mL of 0.10 M HCOONa.

Step-by-step solution

Identify the acid and its corresponding salt

Looking at the available solutions, the focus is to identify a weak acid and its corresponding salt that can be used to make the buffer. The pK_a values of the weak acids available are: Formic Acid (HCOOH) pK_a = 3.75, Acetic Acid (CH3COOH) pK_a = 4.76, and Phosphoric Acid (H3PO4) pK_a = 2.15,7.20,12.35 (for each proton). The acid that has a pK_a value closest to the desired pH 3.50 is Formic Acid (HCOOH) with a pK_a = 3.75. Its corresponding salt is Sodium formate (HCOONa).

Use the Henderson-Hasselbalch equation

To determine the ratio of acid to salt needed in the buffer solution, the Henderson-Hasselbalch equation is used. This equation relates the pH, pK_a, and the ratio of the concentrations of the weak acid to its conjugate base (salt) in a solution. To maintain a pH of 3.50, we will have: \(3.50 = 3.75 + \log \left(\frac{[\text{HCOONa}]}{[\text{HCOOH}]}\right)\). By rearranging this equation, we can find the ratio of the salt to the acid: \([\text{HCOONa}]/[\text{HCOOH}] = 10^{(3.50-3.75)}\) = 0.5623.

Determine the volumes of each component needed

The task now is to determine how much of each component, the weak acid and the salt, is needed to make approximately 1 liter of the buffer solution. Given that the available solutions are 0.10 M, you would need: \(\text{Volume of HCOOH} = (1L/0.5623) * (0.10 mol/L) = 1.779 L\) \(\text{Volume of HCOONa} = (1L) * (0.10 mol/L) = 1.0 L\). However, this would give a total solution volume of 2.779 L which is above the 1L required. To adjust this to the 1L requirement, use the ratio of the volumes. This gives \(\text{Volume of HCOOH required for 1L} = 1.779/2.779 = 0.640 L or 640 mL\) and \(\text{Volume of HCOONa required for 1L} = 1/2.779 = 0.360 L or 360 mL\).

Key concepts

pH

The term pH stands for the 'potential of Hydrogen'. It is a scale used to specify how acidic or basic a water-based solution is.
The scale ranges from 0 to 14, with 7 being neutral. A pH less than 7 indicates acidity, while a pH greater than 7 indicates a base or alkaline nature.

Understanding pH is crucial in acid-base chemistry as it helps in determining the acidity or alkalinity of a solution. It is calculated by the expression: \[\text{pH} = -\log_{10}[\text{H}^+]\]where \([\text{H}^+]\) is the concentration of hydrogen ions in the solution. The pH scale is logarithmic, meaning each whole number on the scale represents a tenfold change in acidity or alkalinity.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental formula in acid-base chemistry. It is used to calculate the pH of a buffer solution.
This equation relates the pH to the pK_a (acid dissociation constant) and the concentrations of the proton donor (weak acid) and its conjugate base (salt).

The equation is written as:\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]where \([\text{A}^-]\) represents the concentration of the conjugate base and \([\text{HA}]\) represents the concentration of the acid.
Using this equation, one can determine the ratio of salt to acid needed to achieve the desired pH of the buffer.

Weak acids

Weak acids are a type of acid that do not completely dissociate into their ions in solution.
This incomplete dissociation is a key characteristic that differentiates them from strong acids.

When placed in water, weak acids only partially ionize, releasing some hydrogen ions, which contribute to the acidity of the solution.
This means their ionization is an equilibrium process, where both the undissociated acid and its ionized forms coexist in the solution.

Examples from the exercise include:

• Formic Acid (HCOOH)
• Acetic Acid (CH₃COOH)
• Phosphoric Acid (H₃PO₄)

Due to their ability to maintain a balance between ionized and non-ionized forms, weak acids play a crucial role in buffer solutions.
They help resist changes in pH when small amounts of acids or bases are added.

Acid-base chemistry

Acid-base chemistry is a branch of chemistry that deals with the properties and reactions of acids and bases.
It involves understanding how acids and bases behave in solutions, their strengths, and the equilibrium established in reactions.

Acids are substances that can donate a proton (H⁺), whereas bases are substances that can accept a proton.
This donation and acceptance of protons lead to the formation of conjugate acid-base pairs, which are essential in understanding acid-base reactions.

Key aspects of acid-base chemistry include:

• Acid and base strength
• Equilibrium and dissociation constants (Kₐ for acids)
• Reactions that produce water and salt

Understanding acid-base chemistry is important not only in the preparation of buffer solutions but also in analyzing chemical reactions in biological, environmental, and industrial contexts. Buffer solutions prepared through acid-base interactions resist changes in pH, making them vital in many practical applications.