6D.3 part B

Moderators: Chem_Mod, Chem_Admin

Skyllar Kuppinger 1F
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am

6D.3 part B

Postby Skyllar Kuppinger 1F » Sat Jan 18, 2020 9:01 pm

I have no idea why I’m getting this problem wrong. I’ve done it over and over. The problem word for word is “the pH of a 0.10M propylamine, C3H7NH2, aqueous solution was measured as 11.86. What are the values of Kb and pKb of propylamine?”
My work is attached. Thanks to anyone who can help
Attachments
image.jpg

Jacob Puchalski 1G
Posts: 97
Joined: Fri Aug 09, 2019 12:16 am

Re: 6D.3 part B

Postby Jacob Puchalski 1G » Sun Jan 19, 2020 12:21 am

So ultimately we're trying to get from pH to pKb. Since propylamine is a base we know it's best to deal in terms of OH-, so we should convert the pH to pOH (2.14).
Since pOH is just -log[OH-], we can find [OH-] through 10^-2.14, which comes out to be 7.2 x 10^-3.
We know that Kb is the molarity of the products over the molarity of the reactants, and that one mole of C3H7NH2 along with an excess of H20 produces one mole of C3H7NH3+ and one mole of OH- (therefore, since there is 7.2 x 10^-3 M OH-, there is also 7.2 x 10^-3 M C3H7NH3+). Finally, we're given that there is .1M C3H7NH2.
With all of this we can find Kb, which is (7.2 x 10^-3)^2/(.1 - 7.2 x 10^-3) = 5.6 x 10^-4.
pKb = -log(Kb) = -log(5.6 x 10^-4) = 3.25. Hope that helps!


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest