Question

A particular sample of vinegar has a 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 0.0882 M.

Step-by-step solution

Calculate the concentration of H+ ions

First, we must find the concentration of H+ ions in the solution. This can be done by using the pH formula: \[pH = -\log[H+]\] Where pH is the pH of the solution and \([H+]\) is the concentration of H+ ions in the solution. Since we are given the pH value as 2.90, we can rewrite the formula as follows: \[2.90 = -\log[H+]\]

Solve for the concentration of H+ ions

Now, we'll solve for \([H+]\) to find the concentration of H+ ions in the solution. To do this, simply isolate \([H+]\) by taking the inverse logarithm of both sides of the equation: \[[H+] = 10^{-pH}\] \[[H+] = 10^{-2.90}\] Calculate the concentration of H+ ions: \[[H+] \approx 1.26 \times 10^{-3} M\]

Write the Ka expression for acetic acid

Now that we have the concentration of H+ ions in the solution, we will utilize the \(K_a\) expression relating the concentrations of acetic acid, acetate ion, and H+ ion. The reaction of acetic acid (\(CH_3COOH\)) with water can be written as: \[CH_3COOH \rightleftharpoons H^+ + CH_3COO^-\] And the corresponding \(K_a\) expression for this reaction would be: \[K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\] Given \(K_a = 1.8 \times 10^{-5}\) for acetic acid, we can write the Ka expression as: \[1.8 \times 10^{-5} = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\]

Assume that [H+] is equal to [CH3COO-]

Since acetic acid (\(CH_3COOH\)) is the only acid present in the vinegar, we can assume that the concentration of H+ ions is equal to the concentration of acetate ions (\(CH_3COO^-\)) in the reaction: \[[H^+] = [CH_3COO^-]\]

Substitute the values into the Ka expression

Plug in the values for \([H+] \text{ , } [CH_3COO^-] \text{ and } K_a\) in the Ka expression: \[1.8 \times 10^{-5} = \frac{(1.26 \times 10^{-3})(1.26 \times 10^{-3})}{[CH_3COOH]}\]

Solve for the concentration of acetic acid

Now, we'll solve for the concentration of acetic acid, \([CH_3COOH]\): \[[CH_3COOH] = \frac{(1.26 \times 10^{-3})(1.26 \times 10^{-3})}{1.8 \times 10^{-5}}\] Calculate the concentration of acetic acid: \[[CH_3COOH] \approx 8.82 \times 10^{-2} M\] Hence, the concentration of acetic acid in the vinegar is approximately 0.0882 M.