The Ka of a monoprotic weak acid is 6.56 × 10-3. What is the percent ionization of a 0.151 M solution of this acid?
The Ka of a monoprotic weak acid is 6.56 × 10-3. What is the percent ionization of a 0.151 M solution of this acid?
I'm not sure how to solve this. It says to use the quadratic equation to solve this but I'm not sure how to do that in this equation.
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The first task is to calculate [H+] using the Ka equation and the quadratic equation.
If the acid is HA
This dissociates:
HA ↔ H+ + A-
Ka = [H+] [A-] / [HA]
Because [H+] = [A-] and if you say that [H+] = X
You can say:
Ka = X² / 0.151-X ( note that the [HA] in solution is reduced from 0.151M by the amount of dissociation.
6.56*10^-3 = X² / 0.151-X
X² = (6.56*10^-3) (0.151-X )
X² = 9.90*10^-4 - (6.56*10^-3)X
X² + (6.56*10^-3)X - 9.90*10^-4 = 0
This is the quadratic equation the question calls for. This is not easy to solve without a quadratic equation solver programme - find one on the web or write your own in Excel using the quadratic formula
The value of X = 0.0283
Therefore [H+] = 0.0283M
%dissociation = 0.0283/0.151*100 = 18.7%
In this case you have had to use the quadratic equation because the % dissociation is > 5%. It is an accepted rule that if dissociation is < 5% it is unnecessary to use the complicated quadratic method , and you can use the original [HA] in the equation. This method will not work here.
use an ICE table for the equilibrium HA <=> H+ + A-
this gives
Ka = x^2/(0.151-x)
solve for x then %I = (x/0.151) x 100%