Week 2 Sapling Homework - Question #5

[Eliana Carney 3E]

Hey guys! I'm having a little bit of trouble on question #5 in the Week 2 Sapling homework. The question reads "The Kb for an amine is 5.320x10^-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.701?".

-For starters, I wrote out a formula for the chemical reaction that uses B in place of an amine. The formula goes: B(aq) + H2O(l) → BH+(aq) + OH-(aq).
-Then, I calculated the pOH using the formula pOH = 14 - pH = 14 - 9.701 = 4.299. I then used this to calculate [OH-] by calculating 10^-4.299 = 5.032X10^-5.
-Next, I wrote the formula for Kb which is: Kb = [BH+][OH-] / [B]. I then rearranged this formula to solve for [B]. This rearrangement came out to:
[B] = [BH+][OH-] / Kb.
-After that, I plugged in the calculated concentrations ([BH+] = [OH-] because they are present in a 1:1 ratio) and the given Kb value and solved for [B]. When I did this, I got [B] = 4.743x10^-5.
-I then calculated the percentage protonation using the formula: % protonation = ([BH+] / [B]) x 100%. When I plugged in the numbers I calculated above the the percentage protonation formula, I got 105.9% which doesn't make sense.

I can't figure out where I went wrong either in my approach to the problem or my calculations. If someone could possibly point out my mistake, it would be appreciated. Thanks in advance!

[ChihWei Chen 2C]

Hi!

Basically, you used the equilibrium concentration of amine instead of the initial concentration to calculate percent ionization.

if you did an ICE table, the equilibrium concentration of the amine is [ x minus (the [OH-] concentration you calculated)], which is the one you just calculated. x is the original concentration you are trying to solve for. I believe you just have to add 1.205 x 10^-5 to the amine concentration you calculated to get the original amine concentration and use that to calculate the percent ionization.

Hope this helps!

[Steph Du 1H]

For [B] did you make sure that it is ?

[Daniela Santana 2L]

Hi!

Basically, you used the equilibrium concentration of amine instead of the initial concentration to calculate percent ionization.

if you did an ICE table, the equilibrium concentration of the amine is [ x minus (the [OH-] concentration you calculated)], which is the one you just calculated. x is the original concentration you are trying to solve for. I believe you just have to add 1.205 x 10^-5 to the amine concentration you calculated to get the original amine concentration and use that to calculate the percent ionization.

Hope this helps!

Hi! I am also having trouble with this problem and I was wondering from where you got 1.205 x 10^-5. Is this the value of the OH- value you calculated?

[ChihWei Chen 2C]

Hi!

Basically, you used the equilibrium concentration of amine instead of the initial concentration to calculate percent ionization.

if you did an ICE table, the equilibrium concentration of the amine is [ x minus (the [OH-] concentration you calculated)], which is the one you just calculated. x is the original concentration you are trying to solve for. I believe you just have to add 1.205 x 10^-5 to the amine concentration you calculated to get the original amine concentration and use that to calculate the percent ionization.

Hope this helps!

Hi! I am also having trouble with this problem and I was wondering from where you got 1.205 x 10^-5. Is this the value of the OH- value you calculated?

Hi! Yes! That's the equilibrium concentration of OH- I calculated. Apparently sapling changes the values in the questions for different people so we get different versions. The [OH-] I calculated doesn't match the question asked here. The correct [OH-] for this version of the question should be in the original post.