What is the equilibrium constant for a weak acid HA that dissociates 1% at an initial concentration of 0.05 M according to the following balanced chemical equation?
HA <----> H+ + A-
To find the equilibrium constant for a weak acid, you first need to know the expression for the equilibrium constant.
The expression for the equilibrium constant (Kc) for the dissociation of a weak acid, HA, can be written as follows:
Kc = [H+][A-]/[HA]
In this case, you are given that the weak acid dissociates 1% at an initial concentration of 0.05 M.
To determine the equilibrium concentrations, you can assume that x (the concentration of H+ and A-) is a small value compared to the initial concentration of HA. Then, use a table to track the concentrations at equilibrium.
Initial concentrations:
[HA] = 0.05 M
[H+] = 0 M (Since no dissociation has occurred initially)
[A-] = 0 M (Since no dissociation has occurred initially)
At equilibrium, let's assume the equilibrium concentration of [H+] and [A-] is x M. The concentration of [HA] at equilibrium would be (0.05 M - x) M because x molecules dissociate.
Substituting these values into the equilibrium expression:
Kc = (x)(x)/(0.05 - x)
Given that the weak acid dissociates 1%, the equilibrium concentration of [HA] is 0.99 times the initial concentration:
(0.05 - x) = 0.99(0.05) = 0.0495 M
Now you can substitute this value into the expression for Kc:
Kc = (x)(x)/(0.0495)
To find x, you can use the quadratic equation since x is in the denominator:
0.0495 * Kc = x^2
x^2 - 0.0495 * Kc = 0
Solving this quadratic equation will yield the value of x. Once you have x, you can substitute it back into the expression for Kc to calculate the equilibrium constant.