How do you find the pH, the pOH, [H3O+], and [OH-] in equations?

1 Answer
May 16, 2016

2H_2O(l) rightleftharpoons H_3O^+ + HO^-

K_a = [H_3O^+][HO^-] =10^(-14)

Explanation:

K_a = [H_3O^+][HO^-] =10^(-14)

This ion product, this equilibrium, has been established by experiment at 298 K. At higher temperatures, given that this is a bond-breaking reaction, how would you expect this equilibrium to evolve?

Now, given the equation, we can manipulate it, as long as we do it to both sides of the equation. So we take log_10 of BOTH sides to give:

log_10K_a =log_10[H_3O^+] +log_10[HO^-]. And on rearrangement:

-log_10[H_3O^+] -log_10[HO^-]=-log_10K_a

But by definitions, -log_10[H_3O^+]=pH, -log_10[HO^-]=pOH, and -log_(10)10^-14=14.

And thus pH + pOH =14.

Now if you are at A level, you only have to know this last result. You do have to be able to use it to solve problems. There should be many examples of problems on these boards so get cracking.