Calculating pOH from HA --> H + A

[Natalie Vu 1I]

I was doing this problem: Muscles produce lactic acid, CH3CH(OH)COOH(aq) , during exercise. Calculate the percent ionization (deprotonation), pH, and pOH of a 0.1023 M solution of lactic acid. The acid‑dissociation (or ionization) constant, Ka , of this acid is 8.40×10−4 .

I was confused because doesn't H stand for H3O, so if I find the equilibrium concentration, can't I find pOH from taking the -log? However, I had to get the pH first and subtract 14. I think I'm confused on how to use the equation.

[Aashna Bhandari 1L]

Hi! So because CH3CH(OH)COOH is an acid, the solution will have H+ ions produced, as you see in the general reaction for acids (HA + H2O -> A- + H3O+), and you're right that the H+ basically means H3O+, so when you take the -log([H3O+]), you are solving for the pH, not the pOH. To solve for the pOH directly after the ice table, you would need to be given a base reaction (A- + H2O -> HA + OH-), but because this one is an acidic reaction, we cannot solve for pOH directly. Instead (like you did) we solve for the pH by taking the -log([H3O+]), and then use pOH=14-pH to solve for pOH.
Hope that clears things up! Basically you use these equations after solving the ICE table and getting the equilibrium H3O+ concentration:
pH = -log([H3O+]) ~or~ -log([H+]), they mean the same thing
pOH=14-pH

[Natalie Vu 1I]

Hi! So because CH3CH(OH)COOH is an acid, the solution will have H+ ions produced, as you see in the general reaction for acids (HA + H2O -> A- + H3O+), and you're right that the H+ basically means H3O+, so when you take the -log([H3O+]), you are solving for the pH, not the pOH. To solve for the pOH directly after the ice table, you would need to be given a base reaction (A- + H2O -> HA + OH-), but because this one is an acidic reaction, we cannot solve for pOH directly. Instead (like you did) we solve for the pH by taking the -log([H3O+]), and then use pOH=14-pH to solve for pOH.
Hope that clears things up! Basically you use these equations after solving the ICE table and getting the equilibrium H3O+ concentration:
pH = -log([H3O+]) ~or~ -log([H+]), they mean the same thing
pOH=14-pH

This clears my confusion a lot thank you!