6D.9 - Calculating pH and Ka for benzoic acid

Moderators: Chem_Mod, Chem_Admin

Hope Hyland 2D
Posts: 50
Joined: Wed Feb 20, 2019 12:16 am

6D.9 - Calculating pH and Ka for benzoic acid

Postby Hope Hyland 2D » Tue Jan 28, 2020 10:17 pm

This problem says: The percentage deprotonation of benzoic acid in a 0.110M solution is 2.4%. What is the pH of the solution and the Ka of benzoic acid?

I understand how to use the percentage deprotonation to find the pH, but I am confused on how to find Ka. In the solutions manual, I have the same equation for Ka, but then when plugging in the equilibrium concentration for benzoic acid, I am not sure how they got (1-0.024) x 0.110. Can someone please explain?

Caitlyn Tran 2E
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am

Re: 6D.9 - Calculating pH and Ka for benzoic acid

Postby Caitlyn Tran 2E » Tue Jan 28, 2020 10:38 pm

I don't think the way I solved this was the same as the book, but this is the way I solved it: in this problem, benzoic acid reacts with water to produce its conjugate base and hydronium ions. You can draw out an ICE table for this. At equilibrium, the concentration of the conjugate base and hydronium ions will both increase by +X. Meanwhile, the concentration of benzoic acid at equilibrium is 0.110-X, and we do not care about the concentration of water at equilibrium when calculating Ka. We already know the hydronium concentration since we solved for it using the percent deprotonation and initial concentration of benzoic acid, which means that we know the value of X (hydronium concentration is equal to X). This gives us everything we need to solve for Ka, which is simply the concentration of the products over the concentration of the reactants. Hope this helps!

Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”

Who is online

Users browsing this forum: No registered users and 2 guests