Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H)in water.Given that the pH for acetic acid is 2.41,the K_a = 1.8 × 10^-5and assuming the density of vinegar to be 1.00 g/cm³,what is the percent dissociation of acetic acid in vinegar? A)0.47% B)1.5% C)4.0% D)5.0%

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Quizplus · Accepted Answer

The percent dissociation of acetic acid in vinegar can be calculated using the given information and the formula for percent dissociation. The dissociation constant (Ka) for acetic acid is given as 1.8 × 10^-5. The pH of the solution is 2.41, which means the concentration of H+ ions ([H+]) is 10^-2.41. Since the density of vinegar is 1.00 g/cm^3 and it is a 5.0% solution by weight, the concentration of acetic acid in the solution is 0.05 g/mL or 50 g/L. Given the molar mass of acetic acid (CH3COOH) is approximately 60 g/mol, the molar concentration of acetic acid is 50/60 = 0.833 mol/L. The percent dissociation is calculated as ([H+]/initial concentration of acetic acid) × 100 = (10^-2.41/0.833) × 100 ≈ 0.47%.

Vinegar Is a 5