## Chemical equilibrium, getting molarity given pH and Kb

What should be the molarity in an aquatic dissolution for Trimethylamine, ##(CH_3)_3N## if ##pH=11,2##?

##(CH_3)_3N+H_2O \rightleftharpoons (CH_3)_3NH^++OH^-##, ##K_b=6.3 \times 10^{-5}##

I know that ##pH+pOH=14 \rightarrow pOH=2.8##, then ##[OH^-]=1.58 \times 10^{-3}##

And ##K_b=\displaystyle\frac{ [ (CH_3)_3NH^+] [OH^-]}{[(CH_3)_3N]}## I used the approximation that water remains constant.
I think I should use some other approximation to get the concentration for ##(CH_3)_3N##, but I'm not sure.
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 Admin Many ways to skin that cat. Do you know ICE tables?
 Not really. We see all this really fast, we had only one class for this. Now that I googled it and I know what I mean I think I saw some of these tables in examples in books. But the exercise don't give any information, I would expect to know the initial concentration or something.

## Chemical equilibrium, getting molarity given pH and Kb

In this case original concentration is your unknown. Pretend you use ICE table to calculate pH and use it to derive formula for the concentration of H+ (or OH- as it should make calculations easier) as a function of the concentration of trimethylamine. Then solve this formula for the concentration of trimethylamine - it will be the only unknown.

It is possible to solve the problem using approximate equations, but then you will have no idea what and why you are doing. ICE tables are much more general and can be applied to most typical equilibrium calculations.
 Alright, I would like to try both ways. Where can I find an ICE table? an what approximation should I use?

 Quote by Telemachus Where can I find an ICE table?
ICE stands for Initial/Change/Equilibrium. Ask uncle google or aunt wikipedia for help.
 Alright. I was thinking. Would it work just to suppose some arbitrary initial concentration for each species? lets say k, l, m. I thought of doing that using stoichiometry.
 Admin Why arbitrary? Assume concentration of the trimethylamine to be C, and other concentrations to be 0.
 Nice!
 Ok, this is what I did. I know the concentration for ##[OH^-]## at equilibrium. So I used that I have a concentration ##C_0## for trimethylamine that I want to find. The changes given by the stoichiometry are -X for trimethylamine, +X for ##(CH)_3NH^+## and +X for ##[OH^-]## Then, if initially there is only trimethylamine, the initial concentration ##[OH^-]_0=[(CH)_3NH^+]=0## and then in equilibrium the thrimethylamine concentration is given by ##[(CH)_3N]=C_0-X## and for the products ##[(CH)_3NH^+]=0+X## and ##0+X=[OH^-]\rightarrow X=1.58 \times 10^{-3}## So, using the equilibrium constant: ##K_b=\displaystyle\frac{[(CH)_3NH^+][OH^-]}{[(CH)_3N]}=\frac{X1.58\times10^{-3}}{C_0-X}=6.3\times10^{-5}## Then ##C_0=25.12M ## and at equilibrium ##[(CH)_3N]\approx C_0## Is this fine?
 Admin Check your math. In general you are on the right track, but for some reason you mixed X and 1.58x10-3 in the final equation (even if you know X=1.58x10-3) and your answer is wrong. Whenever you see calculated concentration being higher than 2 or 3 M it should raise red flag for you.
 I've made the numbers again and it gave: ##C_0=1.04## I think I forgot to square ##1.58\times10^{-3}## in the last equality for Kb. Thanks.
 Admin This is still wrong. As a final check you should try to calculate pH of the solution.
 This equation is fine? ##K_b=\displaystyle\frac{[(CH)_3NH^+][OH^-]}{[(CH)_3N]}=\frac{X1.58\times10^{-3}}{C_0-X}=6.3\times10^{-5}## Alright, Iv'e tried again and get to ##C_0=0,04## I've forgot another product last time. Your purpose was that I should calculate pH from the initial concentration as a check? I'm not sure on how to do that, I should look for it in a book. Thank you Borek. BTW, can you help me with this one? http://www.physicsforums.com/showthread.php?t=660731