A particular sample of vinegar has a pH of 2.20 . 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.

The pH equation is: \(pH = -\log_{10} [\mathrm{H}^{+}]\) The dissociation equation for acetic acid is: \(\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^{-}} + \mathrm{H^+}\)

We are given the pH of the solution, which is 2.20. We can use the pH equation to find the concentration of H+ ions: \(2.20 = -\log_{10} [\mathrm{H}^{+}]\) Rearrange the equation to solve for the concentration of H+ ions: \([\mathrm{H}^{+}] = 10^{-2.20}\)

Now, we need to use the dissociation equation for acetic acid and the given Ka value to calculate the concentration of acetic acid. Since the acetic acid is the only acid present in the vinegar, we can assume that the concentration of H+ ions and CH3COO- ions are equal. Let's write the Ka expression for acetic acid dissociation: \(K_a = \dfrac{[\mathrm{CH_3COO^{-}}][\mathrm{H}^{+}]}{[\mathrm{CH_3COOH}]}\) Given that \([\mathrm{CH_3COO^{-}}] = [\mathrm{H}^{+}]\), the Ka expression becomes: \(K_a = \dfrac{[\mathrm{H}^{+}]^2}{[\mathrm{CH_3COOH}]}\) We already calculated the concentration of H+ ions in step 2. Now we can solve for the concentration of acetic acid. Rearrange the equation: \([\mathrm{CH_3COOH}] = \dfrac{[\mathrm{H}^{+}]^2}{K_a}\) Substitute the known values: \([\mathrm{CH_3COOH}] = \dfrac{(10^{-2.20})^2}{1.8 \times 10^{-5}}\)

Now we can compute the concentration of acetic acid using the obtained values: \([\mathrm{CH_3COOH}] = \dfrac{(10^{-2.20})^2}{1.8 \times 10^{-5}} \approx 6.10 \times 10^{-3} \mathrm{M}\) So, the concentration of acetic acid in the vinegar is approximately \(6.10 \times 10^{-3} \mathrm{M}\).