Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H)in water.Given that K_a = 1.8 × 10^-5 for acetic acid and assuming the density of vinegar to be 1.00 g/cm³,what is the pH of this vinegar solution? A)2.00 B)2.41 C)2.87 D)4.74

Question

Quizplus · Accepted Answer

First, we need to calculate the molarity of acetic acid in the solution:

Molarity = (5.0 g acetic acid / 60.05 g/mol) / (100 mL vinegar / 1000 mL) = 0.0832 M

Next, we can use the equilibrium expression for acetic acid in water:

CH3COOH + H2O ⇌ H3O+ + CH3COO-

Ka = [H3O+][CH3COO-] / [CH3COOH]

Since we have a weak acid, we can assume that [H3O+] = [CH3COO-]

Therefore, Ka = [H3O+]^2 / [CH3COOH]

Solving for [H3O+]:

[H3O+] = sqrt(Ka*[CH3COOH]) = sqrt(1.8E-5*0.0832) = 1.28E-3 M

Finally, we can calculate the pH:

pH = -log[H3O+] = 2.41

Vinegar Is a 5