Question

A particular sample of vinegar has a \(\mathrm{pH}\) of \(2.90\). If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.

Short answer

The concentration of acetic acid in the vinegar is approximately \(6.9 \times 10^{-2}\) M.

Step-by-step solution

Write down the given information

From the exercise, we have: pH = 2.90 \(K_a = 1.8 \times 10^{-5}\) (Acid dissociation constant for acetic acid) Our goal is to find the concentration of acetic acid in the vinegar.

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pK_a + \log_{10} \frac{[A^-]}{[HA]}\] Where: pH is the pH of the solution, \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\), \([A^-]\) is the concentration of the conjugate base of the weak acid, and \([HA]\) is the concentration of the weak acid.

Calculate pKa

In order to use the Henderson-Hasselbalch equation, we need to find \(pK_a\) first. We can calculate pKa using the equation: \[pK_a = -\log_{10}K_{a}\] Substitute the given \(K_a\) value: \[pK_a = -\log_{10}(1.8 \times 10^{-5})\] Calculate the value of pKa: \[pK_a \approx 4.74\]

Use the Henderson-Hasselbalch equation to find acetic acid concentration

Now that we have pH and pKa, we can use the Henderson-Hasselbalch equation to find the concentration of acetic acid in the vinegar. Since acetic acid is the only acid present, the term \(\frac{[A^-]}{[HA]}\) will represent the ratio of conjugate base concentration to acetic acid concentration. Rearrange the Henderson-Hasselbalch equation to find the ratio \(\frac{[A^-]}{[HA]}\): \[\frac{[A^-]}{[HA]} = 10^{(pH - pK_a)}\] Substitute the values of pH and pKa: \[\frac{[A^-]}{[HA]} = 10^{(2.90 - 4.74)}\] \[ \frac{[A^-]}{[HA]} \approx 6.9 \times 10^{-2}\] For dilute solutions as assumed, the concentration of the conjugate base \([A^-]\) will be approximately equal to that of the acetic acid \([HA]\). Let \(x\) be the concentration of acetic acid, \[ \frac{x}{x} \approx 6.9 \times 10^{-2}\] \[ x \approx 6.9 \times 10^{-2} \; \text{M}\]

State the final answer

The concentration of acetic acid in the vinegar is approximately \(6.9 \times 10^{-2}\) M.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a critical tool in acid-base chemistry. This equation is derived from the acid dissociation equilibrium and is used to relate the pH of a solution to the pKa (the negative logarithm of the acid dissociation constant) and the ratio of the concentrations of the conjugate base ([A^-]) to the weak acid ([HA]). The formula is:

• \[ pH = pK_a + \log_{10} \left(\frac{[A^-]}{[HA]}\right) \]

This equation helps predict how the pH of a solution will change with different amounts of weak acid or its conjugate base added. In the exercise, the Henderson-Hasselbalch equation is used to find the concentration of acetic acid in the vinegar by considering the given pH and calculating the pKa from the known association constant (K_a) of acetic acid. Understanding this equation allows students to grasp the delicate equilibrium between acids and bases in solutions, especially in buffer systems.

Acetic Acid Concentration

Acetic acid (CH₃COOH) is a common weak acid found in vinegar. In the exercise, we aim to determine its concentration in a vinegar sample with a known pH. The acetic acid dissociates slightly in water, forming hydrogen ions and acetate ions (CH₃COO^-).To find the concentration of acetic acid, we use the calculated \(\frac{[A^-]}{[HA]}\) ratio from the Henderson-Hasselbalch equation:

• Calculated as \(6.9 \times 10^{-2}\) from the given pH and the pKa of acetic acid.

Assuming that this ratio approximately represents the changes in molar concentrations, we find that the concentration of acetic acid can be closely estimated to be the same as this ratio when dealing with dilute solutions. Hence, the acetic acid concentration in the vinegar is about \(6.9 \times 10^{-2}\) M.

pH Calculation

Calculating pH is an essential skill in understanding acid-base equilibria. pH is defined as the negative log of the hydrogen ion concentration ([H^+]) in the solution. Given the initial information of pH = 2.90 for the vinegar, we back-calculate to derive other characteristics of the solution.The steps involve:

• Recognizing that \(pH = -\log_{10} [H^+]\), which leads to \([H^+] = 10^{-pH}\).
• Using pH to calculate the \(pK_a\), by relating it to the K_a of acetic acid (\[pK_a = -\log_{10} K_a\]), providing us a bridge to further apply the Henderson-Hasselbalch equation.

Understanding how to manipulate these equations and the significance of pH in assessing acidity or alkalinity allows students to predict how changes in concentration impact the system's balance. It's a practical demonstration of the interaction between chemical species in a solution and the importance of equilibrium constants in chemistry.