- #1
IntegrateMe
- 217
- 1
Calculate the pH of a 0.0253 M aq solution of a monoprotic acid. The Ka for the acid is 1.79 x 10-2
So a monoprotic acid will have the form HA --> H + A, correct?
I 0.0253 0 0
C -x x x
E (0.0253-x) x x
EDIT: I just realized the formatting is all messed up on my chart, but it shouldn't be too difficult to discern what I'm doing.
So we have:
1.79 x 10-2 = x2/(0.0253 - x)
0 = -x2 - 1.79x10-2x + 4.53x10-4
After using the quadratic formula, we find the acceptable value of x (the one that is positive) to be:
4.10 x 10-2 M
Now my question is: how do I find the pH? I thought I could use -log10(4.10x10-2), but apparently that's wrong.
Any help? Thank you!
So a monoprotic acid will have the form HA --> H + A, correct?
I 0.0253 0 0
C -x x x
E (0.0253-x) x x
EDIT: I just realized the formatting is all messed up on my chart, but it shouldn't be too difficult to discern what I'm doing.
So we have:
1.79 x 10-2 = x2/(0.0253 - x)
0 = -x2 - 1.79x10-2x + 4.53x10-4
After using the quadratic formula, we find the acceptable value of x (the one that is positive) to be:
4.10 x 10-2 M
Now my question is: how do I find the pH? I thought I could use -log10(4.10x10-2), but apparently that's wrong.
Any help? Thank you!