When an acid reacts by donating a proton, it forms its conjugate base. If the acid, when dissolved in water, produces H^%2B, then the conjugate base, when dissolved in water, produces OH^-.

Consider the example of sodium acetate. Most salts are strong electrolytes -- completely dissociated when dissolved in water. Sodum acetate is a strong electrolye; in water, it is completely dissociated producing sodium ions and acetate ions.

Since the acetate ion is the conjugate base of acetic acid, it can react with water to produce OH^-.

  Although it causes confusion, this reaction frequently is discussed in terms of a special situation, hydrolysis.

It turns out that the K_b for this reaction is related to K_a:

Multiply the above right side by 1, but write that as [H⁺]/[H⁺]:

In the numerator (top), the product of [OH^-] and [H^%2B] is K_w. The other terms are the reciprocal of K_a. (Remember, H₂O is not written in equilibrium expressions for dilute aqueous solutions; its concentration does not change.)

This leads to a general relationship for all acid-conjugate base pairs in water:

More on K_w = K_a x K_b[local].

For more practice, equilibrium problems involving many steps are part of a series of programmed questions. Check this one for drill and practice on solving for the concentrations of ions, acids, and bases of dilute solutions. There isn't an accompanying tutorial for either of the tutorials just mentioned.