November 29, 2024, 05:02:49 PM
Forum Rules: Read This Before Posting


Topic: Determing pH of aqueous solution  (Read 6329 times)

0 Members and 1 Guest are viewing this topic.

rcoleman

  • Guest
Determing pH of aqueous solution
« on: March 05, 2006, 02:32:40 PM »
I have 2 questions:

1) Which of the following aqueous solutions gives a pH greater than 7?

The answer choices are 10e-8M HCOOH, 10e-8M NH4Cl, 10e-8M CH3COOH, 10e-8M HCL. I was thinking that none of these solutions gives a pH greater than 7 because I don't think any of them are bases.

2) *This is a mathematical problem which I'm pretty sure I did right but I need some clarification.*

Calculate the pH of the solute in an aqueous solution of 0.135M C5H5N (aq) (pyridine) if the Kb is 1.8e-9.

Here's my work:

1.8e-9 = x^2/0.135-x

2.43e-10 - 1.8e-9x - x^2 = 0

Quadratic Formula:

[-(-1.8e-9) +\- sqrt((-1.8e-9)^2-4(1)(2.43e-10))]/2(-1) =

1.8e-9 +\- 5.2650e-5\-2 = 2.6324e-5

pOH = -log(2.6324e-5) = 4.5796

pH = Kw - pOH = 14 - 4.5796 = 9.42035

I thought that was right but when I entered in the answer it was wrong. I believe I did all of the steps right. Can anyone help me?

Thanks.

Offline Albert

  • Lonely Wanderer
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 1050
  • Mole Snacks: +112/-22
  • Gender: Male
  • Half Decent Pharmaceutical Chemist
Re:Determing pH of aqueous solution
« Reply #1 on: March 05, 2006, 02:46:35 PM »
The former is correct.

For what concerns the latter, Cb/Kb>100.

Hence: [OH-] = (Cb*Kb)1/2

I got pH = 9.19 using this method.

I didn't check how you solved the quadratic equation, but, if you want, I can post it.
« Last Edit: March 05, 2006, 02:55:40 PM by Albert »

Sponsored Links