A typical sample of vinegar has a pH of \(3.0\). Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in vinegar.

The pH is related to the concentration of hydrogen ions \([H^{+}]\) in the solution, and can be expressed as: \[pH = -\log_{10} [H^{+}]\] The Ka (acid dissociation constant) is a measure of the strength of an acid in a solution and can be expressed as: \[K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]}\] Where, - \([H^{+}]\) is the concentration of hydrogen ions - \([A^{-}]\) is the concentration of the conjugate base (acetate ions in this case) - \([HA]\) is the concentration of the acid (acetic acid in this case) Since vinegar is assumed to be an aqueous solution of acetic acid, we can presume that the \([A^{-}]\) present in the solution comes entirely from the dissociation of acetic acid and therefore, the concentration of acetate ions equals the concentration of hydrogen ions. Now we can write the Ka expression for this problem as: \[K_{a} = \frac{[H^{+}]^2}{[HA] - [H^{+}]}\]

Given the pH value of the vinegar is 3.0, we can calculate the hydrogen ion concentration using the pH formula: \[pH = -\log_{10} [H^{+}]\] \[3.0 = -\log_{10} [H^{+}]\] To isolate \([H^{+}]\), we take the inverse logarithm of both sides: \[[H^{+}] = 10^{-3.0}\] Now, calculate the numerical value of \([H^{+}]\): \[[H^{+}] = 0.001\,\text{M}\]

Use the Ka expression we derived earlier to solve for the concentration of acetic acid \([HA]\): \[K_{a} = \frac{[H^{+}]^2}{[HA] - [H^{+}]}\] Plug in the values for Ka and the hydrogen ion concentration: \[1.8 \times 10^{-5} = \frac{(0.001)^2}{[HA] - 0.001}\] Rearrange the equation to solve for \([HA]\): \[[HA] - 0.001 = \frac{0.001^2}{1.8 \times 10^{-5}}\] \[[HA] = 0.001 + \frac{0.001^2}{1.8 \times 10^{-5}}\] Now, calculate the numerical value of \([HA]\): \[[HA] \approx 0.056\,\text{M}\] This means that the concentration of acetic acid in the vinegar is approximately 0.056 M.