Homework Help: Chem

Posted by Kyo on Friday, October 22, 2010 at 6:33pm.

Estimate the equilibrium constant for the weak base (CH3)2NH, if a 1.59×10-2 M aqueous solution of (CH3)2NH has a pOH 2.58 (make an exact calculation assuming that initial concentration is not equal to the equilibrium concentration).
(CH3)2NH + H2O = (CH3)2NH2+ + OH-

I know how to calculate it if initial concentraton = equilibrium concentration...but how are you suppose to calculate it if they aren't?

Chem - DrBob222, Friday, October 22, 2010 at 9:03pm
I've not seen a problem like this and I'm not sure exactly what it means; however, I think I would try
Keq = [(CH3)2NH2^+)][OH^-]/[(CH3)2NH)]
and plug in (0.00263)^2 for the numerator. For the denominator I would substitute 0.0159 and NOT subtract the OH^- from it (which it would have at equilibrium). However, I don't see that as an exact calculation and I don't understand talk of an exact calculation when the problem starts by saying estimate.
Chem - Kyo, Friday, October 22, 2010 at 9:23pm
Does the .00263 come from antilog (-2.58)?
If so, I plugged it into my calculator and got .00275.
That was the method I initially used (.00275^2)/.0159 and I got 4.35 x 10^-4
but that answer was wrong.
The answer is suppose to be 5.21 x 10^-4.
Thank you for trying though. =]
Chem - DrBob222, Friday, October 22, 2010 at 10:19pm
OK. Yes, that's how I obtained 0.00263 but I just checked and that's the antilog of -2.58 so I suggest you check your calculator again. If you get a different number, perhaps you should look into a new calculator. ;-). The 5.21 number you get by
(0.00263)^2/(0.0159-0.00263) = ?? which is the exact way of doing things; but that is what it is at equilibrium. I'm still confused by the wording of the problem.
Chem - Kyo, Friday, October 22, 2010 at 11:34pm
Ohh. I just tried a similar problem and used the equilibrium way and it was right...
Thank you so much for your help!

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