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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: !!! on February 13, 2008, 10:37:13 PM

Title: Determining the Initial Equilibrium Concentration
Post by: !!! on February 13, 2008, 10:37:13 PM

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg81.imageshack.us%2Fimg81%2F5047%2Fequilprobtu0.png&hash=3b971ef66025dc058eaf7ad1f51ac77de1a24baa)
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!

Title: Re: Determining the Initial Equilibrium Concentration
Post by: Sev 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  :)

Title: Re: Determining the Initial Equilibrium Concentration
Post by: Yggdrasil 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]

Title: Re: Determining the Initial Equilibrium Concentration
Post by: !!! on February 14, 2008, 01:05:19 AM

Oh, dear. I can't believe I overlooked that mistake.
Thank you very much, guys.