The Ka of a monoprotic weak acid is 8.11 x 10 ^ -3. What is the percent ionization of a 0.125M solution of this acid?
........HA ==>H^+ + A^-
I....0.125M...0......0
C......-x.....x......x
E....0.125-x...x......x
Ka = ..... Sutstitute the numbers in to Ka expression and solve for x. (It looks like you MAY--check it out---be required to use the quadratic equation.)
Then %ion = [(H^+)/0.125]*100 = ?
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You just have to use the quadratic formula to find x then divide by the initial concentration
To find the percent ionization of a weak acid, you need to know its Ka value and the initial concentration of the acid.
The equilibrium expression for the ionization of a weak acid, HA, is as follows:
HA ⇌ H+ + A-
The Ka equation is given as:
Ka = [H+][A-] / [HA]
We can assume that the initial concentration of the acid, [HA], is equal to its concentration after ionization because the percent ionization is small. Therefore, we can let the concentration of the ionized species (H+ and A-) be represented as x.
Using the given Ka value, we can set up the equation as:
Ka = x^2 / (0.125 - x)
Since the percent ionization is small, we can approximate that x is much smaller than 0.125, allowing us to simplify the equation:
Ka ≈ x^2 / 0.125
Rearranging the equation, we can solve for x (the concentration of the ionized species):
x^2 = Ka * 0.125
x = √(Ka * 0.125)
Now, we can find the percent ionization by calculating the ratio of the concentration of the ionized species to the initial concentration of the acid:
Percent ionization = (x / initial concentration) * 100
Plugging in the values, we have:
Percent ionization = (√(Ka * 0.125) / 0.125) * 100
By substituting the given Ka value, you can now calculate the percent ionization using the steps above.