weakacidproblempart2

Weak acid problem part 2

K_a : Solving Weak Acid Problems: Part Two

Here's a third typical weak acid problem:

What is the pH of a 0.250 M solution of cacodylic acid? K_a = 6.4 x 10¯⁷

You may have noticed that the solutions in parts one ended up with this:

x = square root of (K_a 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 = square root of (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 third example. The 5% rule fails.

What is the pH of a 0.150 M solution of nitrous acid, HNO₂? K_a = 4.6 x 10¯⁴

x = square root of (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 + sq. root (K² + 4KC)] / 2

The C stands for the starting concentration of the acid.

The solution is left to the reader.