This rule applies when calculating the pH of weak acids and bases. By the way, if you do not know how to do this, please go here for weak acids and here for weak bases.

The key point is that K_{a} and K_{b} values are typically known only to an accuracy of ±5%. So it is reasonable to make approximations if they remain within the ±5% range of the true answer. (And remember, the true answer is not within our grasp. Another way to view the 5% is that the K_{a} and K_{b} values have a built-in error and that the true answer remains hidden within that ±5% error.)

So, the 5% Rule follows. I'll state it for acids and comment on the parallel expression for bases further down.

If the expression

(x / [HA]_{o}) times 100

is less than or equal to 5%, we consider the approximation valid.

Now, for those of you that are totally confused, my recommendation is to back and study the K_{a} and K_{b} links provided above.
Here's a brief description of the above expression:

'x' is the value being calculated in the linked K_{a}tutorial. It is the [H^{+}], the value that will give us our pH[HA]

_{o}is the original starting concentration of whatever weak acid was used in the problem. This value is generally provided in the problem.The approximation stems from the term '[HA]

_{o}minus x.' The 'minus x' portion is dropped as part of the problem solution, thus resulting in an approximation.

By the way, what is being calculated just above is called the 'percent dissociation.'

For bases, the x represents [OH¯] and the orginal starting concentration would be represented [B]_{o}. In the K_{b} calculations, the term '[B]_{o} minus x' would have the 'minus x' dropped.

The 5% Rule for bases is:

If the expression

(x / [B]_{o}) times 100

is less than or equal to 5%, we consider the approximation valid.

In calculating the pH of a weak acid or a weak base, use the approximation method first (the one where you drop the 'minus x').

Then apply the 5% rule.

If you exceed 5%, then you would need to carry out a calculation that does not drop the 'minus x.' This would result in quadratic equation, which would be solvable. It would just be tedious. Here is an example of a problem where 5% is exceeded:

The K_{a}of HF is 6.46 x 10¯^{4}. Calculate the pH of 0.0100 M solution of HF.

If you solve this by the approximate method, you will exceed 5% by quite a bit (it's 25.4%). To obtain as accurate an answer as possible, the quadratic method must be used.

What could happen in your classroom is that your teacher may mention using the quadratic and possibly show one example. Then, everything after that would use the approximate method. This is what the ChemTeam did when he was in the classroom.