Question

Calcuate the composition of 0.050 M HCOOH in solution.

I know this should be easy, but I guess I just need a refresher. I know that there are four species in the solution (HCOOH, HCOO-, H+, OH-) and that we need 4 equations.

I have already found:

Mass action: k_w = [H+][OH-] and k_a = [H+][HCOO-]/[HCOOH]

Mass balance: C_t = [HCOOH] + [HCOO]

Charge balance: [H+] = [HCOO-] + [OH-]

However, I'm not sure what my next step is?

Answer 1

We can add some more information

HCOOH is a strong electrolyte so completely ionizes so whatever concentration taken will be ionized.

HCOOH   →   H⁺ + COO⁻

In solution COO^- conc. Is equal to 0.050M

But H⁺ conc. = H⁺ conc. Obtained from HCOOH + H⁺ conc. Obtained Due to self ionization.

0.050M + 1×10^-7M