Answer:
The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.
Explanation:
4% acetic acid by mass means that 4 gram of acetic acid in 100 g solution.
Given that density of the vinegar is same is that of water = 1 g/mL
Mass of the vinegar solution = 100 g
Volume of the vinegar solution = V
V = 100 mL = 0.1 L
Moles of acetic acid =
The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.
Answer:
[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka
Explanation:
The reaction of dissociation of the benzoic acid in water is given by the following equation:
C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)
The dissociation constant of an acid is the measure of the strength of an acid:
HA ⇄ A⁻ + H⁺ (2)
(3)
<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>
So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:
I hope it helps you!