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^-
1.29e-03?
To calculate Kb, we can use the equation:
Kb = ([OH-] * [Py]) / [HPy]
where [OH-] is the concentration of hydroxide ions, [Py] is the concentration of the conjugate base pyrrolidine, and [HPy] is the concentration of the weak acid pyrrolidine.
Based on the information given, we know the pH of the solution is 10.82. From this, we can calculate pOH by subtracting the pH from 14:
pOH = 14 - 10.82 = 3.18
Next, we can calculate [OH-] using the formula:
[OH-] = 10^(-pOH)
= 10^(-3.18)
= 6.61 x 10^(-4)
Since pyrrolidine is a moderately strong base, we assume that it is completely ionized. Therefore, the concentration of pyrrolidine [Py] will be the same as the initial concentration, which is given as 1.00 x 10^(-3) M.
The concentration of the weak acid [HPy] can be calculated by subtracting the concentration of hydroxide ions from the initial concentration:
[HPy] = [Pyrrolidine] - [OH-]
= (1.00 x 10^(-3)) - (6.61 x 10^(-4))
= 3.39 x 10^(-4)
Finally, we can plug in these values into the Kb expression to calculate Kb:
Kb = ([OH-] * [Py]) / [HPy]
= ((6.61 x 10^(-4)) * (1.00 x 10^(-3))) / (3.39 x 10^(-4))
= 1.948
Therefore, the Kb value for the pyrrolidine solution is approximately 1.948.
To calculate Kb for the pyrrolidine solution, you can follow these steps:
1. Write the balanced equation for the ionization of pyrrolidine in water:
Pyrrolidine (CH3)4NH + H2O → (CH3)4NH2+ + OH-
2. Write the expression for Kb:
Kb = ([Py][OH-])/([HPy])
3. Calculate pOH:
pOH = 14 - pH
pOH = 14 - 10.82
pOH = 3.18
4. Calculate the concentration of OH-:
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.18)
[OH-] = 6.61 x 10^(-4)
5. Calculate the concentration of HPy:
HPy = initial concentration - [OH-]
HPy = 0.001 - 6.61 x 10^(-4)
HPy = 3.39 x 10^(-4)
6. Plug the values into the Kb expression:
Kb = ([Py][OH-])/([HPy])
Kb = (0.001 x 6.61 x 10^(-4))/(3.39 x 10^(-4))
Kb = 1.942
Therefore, the Kb for the pyrrolidine solution is 1.942.