Finding PH of carboxylic acids

Question

HCOOH(l) = HCOO–(aq) + H+(aq) Ka= 10^–4

What is the pH of a 10^–2M HCOOH solution?

Answer 1

Carboxylic acids are weak acids, and as such do not ionize completely. This is shown by their Ka values.

The general reaction is HA <===> H⁺ + A^- where HA is the weak acid; H⁺ is a proton (H₃O⁺) and A^- is the conjugate base. Furthermore, one can equate the ionization to the Ka as follows:

Ka = [H⁺][A^-]/[HA]. In the example given in this question, you have....

HCOOH(l) <===> H⁺(aq) + HCOO^-(aq) Ka = 1x10^-4

To find the pH, we first need to find the [H⁺] since pH = -log [H⁺]. To find [H⁺] use the Ka expression.

Ka = [H⁺][A^-]/[HA] and as seen by the ionization reaction above, [H⁺] = [A^-] = x

Ka = 1x10^-4 = (x)(x)/1x10^-2 -x (if we assume that x is small relative to 10^-2 M, we can neglect it in the denominator and thus avoid using the quadratic equation). You then have...

1x10^-4 = x²/1x10^-2

x² = 1x10^-6

x = 1x10^-3 M = [H⁺] = [A^-] (NOTE: 1x10^-3 isn't really negligible compared to 10^-2 M, so you should go back and re-do the calculations using the quadratic. I'm not sure your teacher wants that, so I'm not going to do it.)

Taking [H⁺] = 1x10^-3 M, the pH will be -log 1x10^-3

pH = 3