The Ka of a monoprotic weak acid is 4.47 × 10-3. What is the percent ionization of a 0.184 M solution of this acid?
.............HA ==> H^+ + A^-
I.........0.184......0.....0
C...........-x.......x......x
E.........0.184-x....x......x
Ka = (H^+)(A^-)/(HA)
Substitute the equilibrium amounts into the Ka expression and solve for H^+).
Then % ion = (H^+)*100/0.184
So i got Ka = X^2/(0.184-x)
not sure what to do next.
To find the percent ionization of a weak acid, we need to use the acid dissociation constant (Ka) and the initial concentration of the acid.
The Ka value given here is 4.47 × 10^-3.
First, we need to assume that the acid is not significantly ionized initially, so the initial concentration of the acid (HA) will remain 0.184 M.
Let's denote x as the concentration of the acid that dissociates and becomes H+ ions. This means that [H+] = x.
Using the Ka expression for the dissociation of the weak acid (HA):
Ka = [H+][A-] / [HA]
Since the weak acid is monoprotic, the concentration of the conjugate base (A-) will also be equal to x.
Substituting the values into the Ka expression, we get:
4.47 × 10^-3 = (x)(x) / (0.184 - x)
Simplifying the equation:
4.47 × 10^-3 = x^2 / (0.184 - x)
Now, make an approximation. Since x is expected to be much smaller than 0.184, we can ignore the "-x" term in the denominator.
4.47 × 10^-3 ≈ x^2 / 0.184
To solve for x, we can cross-multiply:
x^2 ≈ 4.47 × 10^-3 * 0.184
x^2 ≈ 8.2148 × 10^-4
Taking the square root of both sides:
x ≈ √(8.2148 × 10^-4)
x ≈ 0.02866
Therefore, the concentration of [H+] is approximately 0.02866 M.
To find the percent ionization, we divide the concentration of the dissociated acid ([H+]) by the initial concentration of the acid ([HA]) and multiply by 100:
Percent ionization = ([H+] / [HA]) x 100
Percent ionization = (0.02866 / 0.184) x 100
Percent ionization = 15.6%
Therefore, the percent ionization of a 0.184 M solution of this acid is approximately 15.6%.