What is the [H+] of a 0.075 M solution of the acid HA (Ka = 4.8 × l0–8
)?
The Ka value is the acid dissociation constant, which represents the strength of an acid. To find the [H+] of a 0.075 M solution of the acid HA, we can use the equation for the dissociation of the acid:
HA ⇌ H+ + A-
Since the concentration of HA is given as 0.075 M, we can assume that the initial concentration of H+ is 0 M (as no dissociation has occurred yet). Let's denote the change in concentration of H+ as x.
HA ⇌ H+ + A-
0.075 M 0 M 0 M
At equilibrium, the concentration of HA will decrease by x, and the concentrations of H+ and A- will increase by x. Therefore, the concentration of H+ at equilibrium will be 0 + x M, and the concentration of A- will also be 0 + x M.
Using the Ka expression for the acid dissociation:
Ka = [H+][A-] / [HA]
Substituting the known values:
4.8 × 10^(-8) = (0 + x)(0 + x) / (0.075 - x)
Since x is assumed to be small compared to 0.075, we can neglect it in the denominator:
4.8 × 10^(-8) ≈ x^2 / 0.075
Rearranging the equation:
x^2 ≈ 4.8 × 10^(-8) × 0.075
x^2 ≈ 3.6 × 10^(-9)
Taking the square root of both sides:
x ≈ √(3.6 × 10^(-9))
x ≈ 6 × 10^(-5) M
Therefore, the [H+] of the 0.075 M solution of the acid HA is approximately 6 × 10^(-5) M.