### Homework: 12.63, 6th edition

Posted:

**Wed Jan 23, 2019 1:12 pm**I am not really too sure how to solve this problem? Does anyone have any suggestions?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=40561

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Posted: **Wed Jan 23, 2019 1:12 pm**

I am not really too sure how to solve this problem? Does anyone have any suggestions?

Posted: **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.

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.

Posted: **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 K_{a}. It should be K_{a}=(.024x.110)^{2}/(.110-(.024x.110))