Answer:
Explanation:Pentanoic acid (C4H8O2) is a weak acid that dissociates in water according to the following equilibrium reaction:
C4H8O2 (aq) + H2O (l) ⇌ C4H7O2- (aq) + H3O+ (aq)
The acid dissociation constant (Ka) is defined by the following equation:
Ka = [C4H7O2-][H3O+] / [C4H8O2]
We are given the pH of the solution, which allows us to calculate the concentration of H3O+:
pH = -log[H3O+]
2.51 = -log[H3O+]
[H3O+] = 10^(-2.51) = 3.548 × 10^(-3) M
We are also given the initial concentration of pentanoic acid, which is 0.67 M. At equilibrium, the concentration of pentanoic acid that has dissociated will be equal to the concentration of C4H7O2-, so we can assume that [C4H7O2-] = x and [C4H8O2] = (0.67 - x).
Substituting these values into the Ka equation, we get:
Ka = (x)(3.548 × 10^(-3)) / (0.67 - x)
At equilibrium, the concentration of pentanoic acid that has dissociated (x) can be calculated using the following equation:
pH = pKa + log([C4H7O2-] / [C4H8O2])
Substituting the given values, we get:
2.51 = pKa + log(x / (0.67 - x))
Solving for x, we get:
x = 0.141 M
Substituting this value back into the Ka equation, we get:
Ka = (0.141)(3.548 × 10^(-3)) / (0.67 - 0.141) = 9.2 × 10^(-5)
Therefore, the acid dissociation constant (Ka) of pentanoic acid is 9.2 × 10^(-5), rounded to 2 significant figures.