[gazeem]

I am wondering if there are certain approximations being made in the way textbooks instruct on how to calculate the change in [OH] of a solution after a weak acid/base salt disassociates. Their instruction seems to be this, for example:

For the salt CH₃COONa, figure out the concentration of of anion dissociated (let's call that x), because x concentration of anion will then react with water, producing x concentration of hydroxide ions. Thus, at equilibrium the concentration of OH will be y, where y = [OH-] in pure water + x.

Is this an overestimation? Won't some of the hydroxide released due to the dissociation of the salt bond with some of the Hydronium ions as the system tries to compensate for this added base?

To use simple numbers for me to keep up with, let's say pure water at equilibrium has 10 hydrogen ions and 10 hydroxides.  And Kw is hydrogens*hydroxides which is always 100.

If we add 5 more hydroxides, we now have 15 OHs and 10 Hs. But since 15*10 does not equal 100, one of two things might happen:

1) The new equilibrium becomes 15 OH and ~6H. Because 15*6 is about 100. (This seems to be the textbook way)

2) The new equilibrium becomes 13 OH and 8H, because 13*8 is about 100. (In this case, to go back to equilibrium, 2 OHs and 2 Hs bonded to form water)

I would expect 2) to be the case because how is the number of hydrogens going to go down besides reacting with some of the OHs! Otherwise, if these hydrogens, like in scenario 1), reacted with water, that would increase OH- even more and defy the Kw constant.

[Borek]

You don't have to think about each equilibrium separately, just use them all to describe the solution and solve for [H⁺].

General case (taking into account water autodissociation):
https://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution

Possible simplifications and their applicability:
https://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base