Homework: 12.63, 6th edition

Moderators: Chem_Mod, Chem_Admin

Bianca Barcelo 4I
Posts: 74
Joined: Fri Sep 28, 2018 12:17 am
Been upvoted: 1 time

Homework: 12.63, 6th edition

Postby Bianca Barcelo 4I » Wed Jan 23, 2019 1:12 pm

I am not really too sure how to solve this problem? Does anyone have any suggestions?
Attachments
Screen Shot 2019-01-23 at 1.04.22 PM.png

Kristen Kim 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:16 am

Re: Homework: 12.63, 6th edition

Postby Kristen Kim 2K » Wed Jan 23, 2019 2:39 pm

We know that the formula for percent deprotonation, or percent ionization, is [H+]/[HA] times 100. in this problem, we are trying to find [H+] in order to plug it into the pH formula. The percent deprotonation and the concentration of [HA] (which is in this case benzoic acid) are given to us.
2.4% and 0.110 M
We plug in these values to the percent deprotonation formula and solve for [H+], which we then plug into the pH formula (-log[H+]).
For the Ka part, we need to set up an ICE table and write the formula for the deprotonation of benzoic acid.
C6H5COOH + H20 <--> H3O+ + C6H5CO2-
After we find out the equilibrium expressions for each of the units in the formula (except for H2O), we set up the Ka equation.
Ka = [H3O+][C6H5CO2-] / [C6H5COOH] = (x)(x)/(0.110 - x)
We already found out what x was when we solved for [H+] in the first part, so we just plug that into the Ka equation and solve.

2c_britneyly
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

Re: Homework: 12.63, 6th edition

Postby 2c_britneyly » Wed Jan 23, 2019 2:46 pm

When filling out the ICE table, your initial amt. of benzoic acid is .110 M, and the equilibrium concentration of the products (more importantly, equilibrium concentration of H+) is (.024 x .110 M). Since you know [H+], you can find pH. Then set up an equilibrium expression to find Ka. It should be Ka=(.024x.110)2/(.110-(.024x.110))


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests