A .35M solution of a newly synthesized acid, HX, has [H+] of 4.1 x 10^-2. What is the value of Ka for this acid?
...............HX ==> H^+ + X^-
initial.....(0.35M)..(0)....(0)
change........ -x.....x......x
final.......(0.35-x)..x......x
Ka = (H^+)(X^-)/(HX)
The problem tells you that x (H^+) = 4.1E-2. Substitute into the Ka expression above and solve for Ka.
To find the value of Ka (acid dissociation constant), we need to use the equation for the ionization of the acid:
HX ⇌ H+ + X-
We know the concentration of the acid solution is 0.35 M, and the concentration of the hydrogen ion ([H+]) is 4.1 x 10^-2 M.
The equation for Ka is as follows:
Ka = ([H+][X-]) / [HX]
We can substitute the known values into this equation and solve for Ka.
Ka = (4.1 x 10^-2)(4.1 x 10^-2) / 0.35
Ka = 1.681 x 10^-3 / 0.35
Ka = 4.803 x 10^-3
Therefore, the value of Ka for the acid HX is 4.803 x 10^-3.
To find the value of Ka for the acid, we need to first understand the relationship between Ka, [H+], and the initial concentration of the acid, [HX].
Ka is the acid dissociation constant, which represents the strength of an acid in its dissociation reaction. It is defined by the equation:
Ka = [H+][X-] / [HX]
In this case, we are given the concentration of the acid, [HX], and the concentration of [H+]. We can assume that the acid completely dissociates and that the concentration of [X-] is equal to the concentration of [H+].
Given:
[H+] = 4.1 x 10^-2 M
[HX] = 0.35 M
We can substitute these values into the equation to find Ka:
Ka = (4.1 x 10^-2)(4.1 x 10^-2) / (0.35)
Simplifying the calculation:
Ka = 1.68 x 10^-3 / 0.35
Ka ≈ 4.8 x 10^-3
Therefore, the value of Ka for the acid HX is approximately 4.8 x 10^-3.