March 29, 2024, 09:02:01 AM
Forum Rules: Read This Before Posting


Topic: Determining the Initial Equilibrium Concentration  (Read 5782 times)

0 Members and 1 Guest are viewing this topic.

Offline !!!

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
  • Gender: Female
Determining the Initial Equilibrium Concentration
« on: February 13, 2008, 10:37:13 PM »

The correct answer is 0.255 M, but I don't know how get to there.
The ICE stands for Initial concentration, Change in concentration, and Concentration at Equilibrium.
Help, please? And thank you!
« Last Edit: February 14, 2008, 01:03:32 AM by !!! »

Offline Sev

  • Full Member
  • ****
  • Posts: 231
  • Mole Snacks: +43/-6
  • Gender: Male
Re: Determining the Initial Equilibrium Concentration
« Reply #1 on: February 13, 2008, 11:54:55 PM »
Hi,

At the step: 4.56*10-3/[(n-0.228)2] = 6.3, just rearrange to: 4.56*10-3/6.3 = (n-0.228)2].

sqrt both sides: n-0.228 = 0.0269, and solve for n  :)

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Determining the Initial Equilibrium Concentration
« Reply #2 on: February 14, 2008, 12:02:06 AM »
Your set up is correct, you just made a small error in the calculations.

When you square root both sides of the equation then multiply by (n-0.228) the RHS of the resulting equation should be 2.51n - 0.572 since Sqrt[6.30]*0.228 = 0.572.

[edit: Sev got to it first]

Offline !!!

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
  • Gender: Female
Re: Determining the Initial Equilibrium Concentration
« Reply #3 on: February 14, 2008, 01:05:19 AM »
Oh, dear. I can't believe I overlooked that mistake.
Thank you very much, guys.

Sponsored Links