I don't remember how I worked this before. Hopefully what follows will duplicate what I did before. If you have trouble following, post the way your book does it and I'll try to explain it that way.
We'll call the weak acid HA. Then
.........HA ==> H^+ + A^-
initial..0.10M...0......0
change....-x....+x......+x
final....0.10-x...x......x
Ka = (H^+)(A^-)/(HA)
6.4E-6 = (x)(x)/(0.10-x)
Solve for x and convert to pH.
If a weak acid has a dissociation constant of 6.4 x 10^-6, what is the pH of a solution of this acid in a 0.10 molar concentration?
I still don't understand the equation you gave me. My textbook doesn't break it down like this, so I'm confused as to where to plug the 0.10 M in.
6 answers
Thanks, totally still confused. I don't know how to solve for (x) since I don't know what the acid is?
The equation I'm trying to use looks like this:
1.0 x 10^-14 divided by 6.4 x 10^-6 which gives me a solution of 1.5625 x 10^-9 which can't be right!
1.0 x 10^-14 divided by 6.4 x 10^-6 which gives me a solution of 1.5625 x 10^-9 which can't be right!
We'll call the weak acid HA. Then
.........HA ==> H^+ + A^-
initial..0.10M...0......0
change....-x....+x......+x
final....0.10-x...x......x
Ka = (H^+)(A^-)/(HA)
6.4E-6 = (x)(x)/(0.10-x)
Solve for x and convert to pH.
You don't need to know the name of HA, everything is here.
6.4E-6 = (x)(x)/(0.1-x)
6.4E-6 = x^2/(0.1-x)
Often we can avoid a quadratic equation by making the simplifying assumption that 0.1-x = 0.1. Then
6.4E-6*0.1 = x^2 and
x = 8.0E-4 = (H^+)
pH = 3.096 which rounds to 3.10.
The inclusion of Kw in you response makes me wonder if the problem is the 0.1M SALT of a weak acid since the Kb of the salt IS Kw/Ka = 1x10^-4/6.4E-6. If your original post, however, is as above in the first post, pH = 3.1 is correct for this problem. I've gone back and re-read the post and it says nothing about the salt.
.........HA ==> H^+ + A^-
initial..0.10M...0......0
change....-x....+x......+x
final....0.10-x...x......x
Ka = (H^+)(A^-)/(HA)
6.4E-6 = (x)(x)/(0.10-x)
Solve for x and convert to pH.
You don't need to know the name of HA, everything is here.
6.4E-6 = (x)(x)/(0.1-x)
6.4E-6 = x^2/(0.1-x)
Often we can avoid a quadratic equation by making the simplifying assumption that 0.1-x = 0.1. Then
6.4E-6*0.1 = x^2 and
x = 8.0E-4 = (H^+)
pH = 3.096 which rounds to 3.10.
The inclusion of Kw in you response makes me wonder if the problem is the 0.1M SALT of a weak acid since the Kb of the salt IS Kw/Ka = 1x10^-4/6.4E-6. If your original post, however, is as above in the first post, pH = 3.1 is correct for this problem. I've gone back and re-read the post and it says nothing about the salt.
The problem was 6.4 x 10^-6. Thanks
Would that still equal a pH of 3.1?
3 answers