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August 31, 2014

August 31, 2014

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

(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 9:03pmI'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 9:23pmDoes 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 10:19pmOK. 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 11:34pmOhh. 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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