Sapling Week 2 #5

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Jaclyn Schwartz 1I
Posts: 101
Joined: Wed Sep 30, 2020 9:45 pm

Sapling Week 2 #5

Postby Jaclyn Schwartz 1I » Sun Jan 24, 2021 11:14 pm

Hi! I was wondering if someone could double check my work because I can not seem to get the correct answer!
Here is my problem:

The Kb for an amine is 3.107×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.128 ? Assume that all OH− came from the reaction of B with H2O.

Here is my work:

B + H20 = BH+ + OH-
Kb=[BH+][OH-]/[B]=3.107×10^−5
pH=9.128
pOH=4.872
[OH-]=1.34*10^-5=[BH+]
Kb=[BH+][OH-]/[B]=3.107×10^−5
[B]=5.8*10^-6
([BH+]/[B])*100%=230

I know this is wrong but I cant get a right answer. I do not know where my error is. I would appreciate any help!

Hannah Alltucker 3L
Posts: 106
Joined: Wed Sep 30, 2020 9:44 pm

Re: Sapling Week 2 #5

Postby Hannah Alltucker 3L » Sun Jan 24, 2021 11:30 pm

The way I did it was to first find the pOH (14-pH), then find the concentration of the -OH by using 10^-pOH. This will act as the X in the ICE table
Next, we can solve for our missing/needed value by setting up the equation Kb=[BH][OH]/[B] where be is our original base and BH is its conjugate acid, OH the conjugate base. Since this should look like Kb=X^2/[B] since our X is very small so we can just ignore the [B]-x and pretend it does not change significantly. Just plug and chug to solve for [B].
Finally, take what you have and to find the percent and use the equation x/(x+[B]) and multiply it by 100 for a percent.

It looks like you find the pOH just fine, and I think your other values are right just by how similar they were to mine (I had different numbers), so you might be running into issues on your last step! Try the equation I used above and see if it works then!

Jaclyn Schwartz 1I
Posts: 101
Joined: Wed Sep 30, 2020 9:45 pm

Re: Sapling Week 2 #5

Postby Jaclyn Schwartz 1I » Sun Jan 24, 2021 11:33 pm

Hannah Alltucker 3L wrote:The way I did it was to first find the pOH (14-pH), then find the concentration of the -OH by using 10^-pOH. This will act as the X in the ICE table
Next, we can solve for our missing/needed value by setting up the equation Kb=[BH][OH]/[B] where be is our original base and BH is its conjugate acid, OH the conjugate base. Since this should look like Kb=X^2/[B] since our X is very small so we can just ignore the [B]-x and pretend it does not change significantly. Just plug and chug to solve for [B].
Finally, take what you have and to find the percent and use the equation x/(x+[B]) and multiply it by 100 for a percent.

It looks like you find the pOH just fine, and I think your other values are right just by how similar they were to mine (I had different numbers), so you might be running into issues on your last step! Try the equation I used above and see if it works then!



Thank You! x/(x+[B]) this was where my error was. I didn't realize B formal I had to add [BH+] to [B] and then divide. Thanks again!

Sarah Salam 1J
Posts: 55
Joined: Wed Sep 30, 2020 9:31 pm

Re: Sapling Week 2 #5

Postby Sarah Salam 1J » Sun Jan 24, 2021 11:38 pm

When you set your equation equal to Kb, did you arrange it like x^2/M-x (this is how I thought of it)? And when you solved it, did you use the M value to divide the x value when you were getting the percentage? That's the only thing I could think of when I compared my work to yours.


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