An acid HA has Ka = 2.28 x 10 -4. The % ionisation of this acid in a 0.170 M solution of the acid in water is closest to
3.662 %
0.623 %
0.498 %
2.930 %
0.436 %
2.564 %
i have an exam in 1 week and these are the types of questions i don't understand
To find the percent ionization of an acid in a given solution, we need to use the acid dissociation constant (Ka) and the initial concentration of the acid.
The Ka expression for an acid HA is given by:
Ka = [H+][A-] / [HA]
Since we have the Ka value given as 2.28 x 10^-4, we can set up the equation as:
2.28 x 10^-4 = [H+][A-] / [HA]
Now, we need to find the concentration of [H+] (hydrogen ion) at equilibrium when the acid is ionized. Since the value of the initial concentration of HA is given as 0.170 M, we can assume that the concentration of [HA] at equilibrium is (0.170 - x) M, where x is the degree of ionization of the acid.
Now, we can rewrite the equation using the given values:
2.28 x 10^-4 = x^2 / (0.170 - x)
Solving this equation for x will give us the degree of ionization, which can be converted to percent ionization by multiplying by 100.
To solve the quadratic equation, we can assume that x is small relative to 0.170 M, which allows us to simplify the equation:
2.28 x 10^-4 = x^2 / 0.170
Rearranging the equation:
x^2 = 2.28 x 10^-4 * 0.170
x^2 = 3.876 x 10^-5
Taking the square root of both sides:
x = √(3.876 x 10^-5) ≈ 0.0062 M
To find the percent ionization, we multiply the degree of ionization by 100:
% ionization = 0.0062 M * 100 ≈ 0.62 %
Therefore, the closest answer choice is 0.623%.