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?
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
a. 2.930 %
b. 3.662 %
c. 2.564 %
d. 0.498 %
e. 0.436 %
f. 0.623 %
I got b for the answer.. but not sure for it.. please help, thanks !!
I worked this in my head and b looks right to me. Hope I didn't miss anything.
Thanks! Really appreciate it!
To find the percent ionization of the acid HA in a 0.170 M solution, we can use the equation:
% Ionization = (concentration of ionized acid / initial concentration of acid) x 100
In this case, the initial concentration of acid (HA) is 0.170 M.
We need to find the concentration of ionized acid. The ionization reaction for the acid HA can be written as:
HA ⇌ H+ + A-
Since the acid is monoprotic (only donates one proton), the concentration of H+ ions will be equal to the concentration of ionized acid.
To determine the concentration of H+ ions, we can use the equation:
Ka = [H+][A-] / [HA]
We know the value of Ka is 2.28 x 10^-4. Let's assume x is the concentration of H+ ions.
Using the given Ka expression, we can write the equation:
2.28 x 10^-4 = x^2 / (0.170 - x)
Since x is assumed to be much smaller than 0.170, we can simplify the equation to:
2.28 x 10^-4 ≈ x^2 / 0.170
To solve for x, we can rearrange the equation:
x^2 = (2.28 x 10^-4)(0.170)
x^2 ≈ 3.876 x 10^-5
x ≈ √(3.876 x 10^-5)
x ≈ 0.00622
Now, we can calculate the percent ionization using the equation mentioned earlier:
% Ionization = (concentration of ionized acid / initial concentration of acid) x 100
% Ionization ≈ (0.00622 / 0.170) x 100
% Ionization ≈ 3.662 %
Therefore, the closest answer is b. 3.662 %.
To find the percent ionization of an acid, we can use the formula:
% Ionization = (concentration of ionized acid)/(initial concentration of acid) * 100
First, let's determine the concentration of the ionized acid. In a solution of the acid HA, some amount of HA will ionize to give H+ ions. Let's assume that x moles of HA ionize, then the concentration of ionized acid will be x M.
Since the initial concentration of HA is 0.170 M, the concentration of the remaining unionized acid will be (0.170 - x) M.
The equilibrium expression for the ionization of HA is:
Ka = [H+][A-]/[HA]
Substituting the given value of Ka = 2.28 x 10^-4 and the concentrations, we get:
(2.28 x 10^-4) = (x)(x)/(0.170 - x)
We can solve this quadratic equation to find the value of x. However, since Ka is very small, we can assume that x is much smaller than 0.170. This allows us to approximate 0.170 - x to be approximately 0.170.
Now, let's solve the equation:
(2.28 x 10^-4) = (x)(x)/(0.170)
Rearranging the equation, we get:
(2.28 x 10^-4)(0.170) = x^2
x^2 = 3.876 x 10^-5
Taking the square root of both sides, we find:
x ≈ 1.969 x 10^-3
Now, let's calculate the percent ionization:
% Ionization = (concentration of ionized acid)/(initial concentration of acid) * 100
= [(1.969 x 10^-3)/(0.170)] * 100
≈ 1.157 %
Based on the options given, none of them match the calculated percent ionization. Therefore, none of the provided options seem to be the correct answer.
Please double-check your calculations or recheck the options provided.