### Ka : Solving Weak Acid Problems: Part Two

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 second typical weak acid problem:

What is the pH of a 0.300 M solution of benzoic acid? Ka = 6.46 x 10¯5

These are the important equations:

HBz + H2O ⇌ H3O+ + Bz¯
Ka = ( [H3O+] [Bz¯] ) / [HBz] = 6.46 x 10¯5

Bz¯ refers to the benzoate ion. It is completely unimportant what its formula is.

As before, we want the [H3O+]. So we have:

[H3O+] = x

and

[Bz¯] = x

This is because of the one-to-one molar ratio between [H3O+] and [Bz¯] that is created as HBz molecules dissociate.

Remember, the [HBz] started at 0.300 M and went down as HBz molecules dissociated. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.300 - x.

Next is our 'dropping the subtract x' trick:

0.300 - x approximately equals 0.300

We will check the validity of the trick after completing the calculation. If the approximation exceeds 5%, then we have to use the quadratic.

We now have our equation:

6.46 x 10¯5 = {(x) (x)} / 0.300

which is very easy to solve. We move the 0.300 to the other side to get:

x2 = 1.938 x 10¯5

Taking the square root (of both sides!!), we get:

x = 4.40 x 10¯3 M

Remember:

1) The Ka value is unitless, but x is a molarity.
2) Square root both sides. I have had students square root the x2, but not the other side. Weird, but true.

We finish by taking the pH to get a final answer of 2.356.

Checking for 5% we find we had 1.47%, which is acceptable.