The pH of a 1.00 x 10^-3 M solution of pyrrolidine is 10.82. Calculate Kb.
I know pyrrolidine is acidic because of the pH, and I know that tells me something about the Ka which in turn tells me something about the Kb. What does it tell me? Where should I begin this problem?
Do you think a solution with a pH of 10.82 is acidic?
1. Write the equation. Pyrrolidine is a moderately strong base.
2. Write the expression for Kb
3. Calculate pOH from pH and from that calculate OH^-
4. Plug into Kb expression and solve for Kb.
Post your work if you get stuck. Let me know what you get.
So do I not need the molarity of pyrrolidine? Is that extraneous information? And I meant basic!! I guess I had a brain fart or something. LOL
No, of course you need the molarity. It certainly is not extraneous information. Just follow the steps I listed in my post. Get as far as you can and post what you have.
1. (CH2)4NH==> NCH2(CH2)3+OH- right?
2. Kb=[NCH2(CH2)3][OH-]/[(CH2)4NH] correct?
3. [H+]=1.51 x 10^-11 [OH-]=6.6 x 10^-4 pH=10.82 and pOH=3.18 Correct?
4. This is where I get stuck. What do I plug in where? All I have are [H+] and [OH-].
OK. You've done great. 1,2, and 3 are exactly right (for the (OH^-) I would not round to 6.6 but to 6.61 since you have three places in your other numbers. So, where do you put the numbers? Go back to step 2 and look at Kb expression. Also, look at step 1.
From step 1, if pOH is 3.18, then OH^- is 6.61 x 10^-4 and the HPy is the same. And you started with what? 0.001 M in a liter? So HPy was 0.001 before any ionization, it now is 0.001 - 6.61 x 10^-4. So you have HPy, OH and Py. Calculate Kb.
Here is a corrected equation. It won't change any of your work but it will show what the equation should be
(CH3)4NH + HOH ==> (CH3)4NH2^+ + OH^-
1 answer
1 answer