### Kb : Solving Weak Base Problems: Part Three

Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature.

Here's a third typical weak base problem:

What is the pH of a 0.250 M solution of strychnine? Kb = 1.82 x 10¯6

You may have noticed that the solutions in parts one and two were exactly the same. Both ended up with this:

x = √(Kb times starting base concentration)

When you're doing the 'drop subtract x' thing, this above equation always works for weak bases.

I better add a cautionary note here: the equation works for weak MONOBASIC bases. However, the study of dibasic bases is not touched on in high school chemistry nor really in Advanced Placement, so we're safe for the time being.

To carry the similarity one step farther, you may have noticed the similiar wordings in the Ka tutorials and the Kb tutorials. I wrote the Ka files tutorials and then just edited copies with the appropriate changes in wordings. Just remember, when doing a Kb problem, you wind up with the pOH and you have to do one more step involving pH + pOH = 14.

Back to the problem. The solution to the above problem is:

x = √(1.82 x 10¯6 times 0.250) = 6.74 x 10¯4

From this, the pH = 10.829

Checking the 5% rule, we get 0.27% error.

Here's a fourth example. The 5% rule fails.

What is the pH of a 0.150 M solution of piperidine? Kb = 1.7 x 10¯3
x = √(1.7 x 10¯3 times 0.150) = 1.6 x 10¯2

Checking the 5% rule:

(1.6 x 10¯2 / 0.15) x 100

we get 10.6% error.

In order to get an answer, we must turn to the quadratic method. In other words, we cannot ignore the 'subtract x' portion in the denominator.

The equation to use is as follows (I left off the sub b on the K):

x = [-K + √(K2 + 4KC)] / 2

The C stands for the starting concentration of the base.

The solution is left to the reader.