"The Kb for an amine is 4.796 x 10^-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.971? Assume that all OH- came from the reaction of B with H2O."
My solution:
pOH = 4.029
percentage protonated = 51.27 %
I've tried this problem over again and I'm not sure what I'm doing wrong!
Achieve HW question
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 100
- Joined: Fri Sep 24, 2021 6:55 am
Re: Achieve HW question
To calculate the percent protonation, I used the expression:
( [BH+] / [BH+] + [B] ) x 100%
Don’t forget that to find [B], we can assume that B - X is just B because Kb is small.
( [BH+] / [BH+] + [B] ) x 100%
Don’t forget that to find [B], we can assume that B - X is just B because Kb is small.
-
- Posts: 110
- Joined: Fri Sep 24, 2021 5:08 am
Re: Achieve HW question
Firstly you need to calculate 10^-4.029 which is equal to the OH- concentration = 9.35x10^-5 and this is the value of x, you then square this value and divide it by the value of kb and then you should add 9.35x10^-5 (according to kb expression to get the original concentration of the amine since (9.3x10^-5 is the value of x)) you should get 2.757x10^-4 , divide (9.35x10^-5/2.757x10^-4) x 100% = 33.9% protonation.
I hope this helps!
I hope this helps!
-
- Posts: 101
- Joined: Fri Sep 24, 2021 5:43 am
Re: Achieve HW question
First, use the pOH to find the OH- concentration. Using an ICE table you know the BH+ will increase by the same factor as OH- so they are the same value. Using that term squared you can use the value of Kb to solve for the amine concentration. Then, finally, you can take the BH+ concentration and divide it by ((BH+) + (B)). This will yield the percent protonation when multiplied by 100%.
Return to “Student Social/Study Group”
Who is online
Users browsing this forum: No registered users and 6 guests