Weak acid-strong base titration?

Question

What is the pH of a 0.7 M solution of a weak acid HA with Ka = 1.78×10-6?

If you show your steps that would be appreciated, thank you.

Answer 1

Note: I'm assuming that the value you've given is the dissociation constant rather than the association constant. If it's the latter, then invert the equation that I show below.

OK, using the Henderson-Hasselbach (or Hasselbalch) equation, we know that Ka = [H+][A-]/[HA]. Since you have a monoprotic acid, H+ = A-.

Let the amount of H+ be x. Then the equation converts to:

Ka = x^2/(1-x). So you have

1.78 x 10^-6(1-x) = x^2.

You can solve this using the quadratic equation.

Answer 2

The previous commenter is correct. I never realized that equation had a name. With a conc of 0.7 M of the weak acid, the x in the denominator of the equation can be ignored, making life much more tolerable.