In a 1.070 M aqueous solution of a monoprotic acid, 4.81% of the acid is ionized. What is the value of its Ka?
........HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
This tells you that the 1.070 HA ionized 4.81% which means
(H^) = 1.070 x 0.0481 = ?
(A^-) = same as (H^+).
(HA) = [1.070 - (H^+)]
Substitute these values into the Ka expression and solve for Ka.
To find the value of Ka, we need to use the concentration of the acid, the percent ionization, and some mathematical steps.
First, let's express the percent ionization as a decimal. Since it's given as 4.81%, you divide it by 100: 4.81% / 100 = 0.0481.
Now, let's use the fact that the percent ionization is the ratio of the concentration of the ions produced to the initial concentration of the acid multiplied by 100.
So, we can write the equation: % ionization = (concentration of ions produced / initial concentration of acid) * 100.
Now, the initial concentration of the acid can be represented as M since it's given as 1.070 M. Let's denote the concentration of the ions produced as x M.
Plugging in the values, we get: 0.0481 = (x / 1.070) * 100.
To solve for x, we can rearrange the equation: x = (0.0481 / 100) * 1.070.
Evaluating the expression gives us: x ≈ 0.0005157 M.
Now, let's use the definition of Ka, which is the equilibrium constant for the dissociation of the acid: Ka = [H+][A-] / [HA].
Since the acid is monoprotic, we know that [H+] = [A-] = x, and [HA] = 1.070 - x (initial concentration minus the concentration of ions produced).
Plugging in these values, we get: Ka = (x * x) / (1.070 - x).
Now, let's substitute x = 0.0005157 M into the equation: Ka = (0.0005157 * 0.0005157) / (1.070 - 0.0005157).
Simplifying the expression, we find: Ka ≈ 2.52 * 10^-7.
So, the value of Ka for the monoprotic acid is approximately 2.52 * 10^-7.