The Ka of a monoprotic weak acid is 3.29 × 10-3. What is the percent ionization of a 0.176 M solution of this acid?
............HA --> H^+ + A^-
I........0.176.....0......0
C.........-x.......x.......x
E.......0.176-x....x.......x
Ka = (H^+)(A^-)/(HA)
Substitute from the ICE chart into the Ka expression and solve for H^+.
%ionization = [(H^+)/0.176]*100 = ?
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 solution.
The equation for the ionization of a weak acid, denoted as HA, can be written as follows:
HA ⇌ H+ + A-
The Ka expression for this reaction is given as:
Ka = [H+][A-]/[HA]
In this case, the Ka value is given as 3.29 × 10-3, and the initial concentration of the acid solution (HA) is 0.176 M.
Let's denote the degree of ionization as "x". This means that [H+] = x and [A-] = x, and [HA] = (0.176 - x).
Plugging these values into the Ka expression, we get:
3.29 × 10-3 = (x)(x)/(0.176 - x)
Now, we need to solve this equation for x using algebraic methods or by using a quadratic solver.
Once we find the value of x, we can calculate the percent ionization using the following formula:
Percent ionization = (x/[HA]) × 100
Substituting the values we obtained, we can find the percent ionization of the weak acid in the given solution.