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!