### Ka : Solving Weak Acid 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 acid problem:

What is the pH of a 0.250 M solution of cacodylic acid? Ka = 6.4 x 10¯7
You may have noticed that the solutions in parts one and two were exactly the same. Both ended up with this:
x = √(Ka times starting acid concentration)

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

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

So the solution to the above problem is:

x = √(6.4 x 10¯7 times 0.250) = 4.0 x 10¯4

From this, the pH = 3.40

Checking the 5% rule, we get 0.16% error.

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

What is the pH of a 0.150 M solution of nitrous acid, HNO2? Ka = 4.6 x 10¯4
x = √(4.6 x 10¯4 times 0.150) = 8.31 x 10¯3

Checking the 5% rule:

(8.31 x 10¯3 / 0.15) x 100

we get 5.54% error.

In order to get a correct 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 a on the K):

x = [-K + √(K2 + 4KC)] / 2

The C stands for the starting concentration of the acid.

The solution is left to the reader.