Homework Help: Chemistry

Posted by Kiana on Friday, March 2, 2012 at 8:13pm.

What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the pH value shown below?
aniline (C6H5NH2) for PH of 9.05

I converted pH to pOH so pOH=4.98
[OH-] =1.122*20^-5

C6H5NH2 + H2O --------> C6H5NH3+ + OH-
I: a N/A 0 0
C: -x N/A +x +x
E: a-x N/A x x

Kb=[C6H5NH3+][OH-]/[C6H5NH2 ]
4.2*10^-10 = x^2/(a-x)

Substituting in the equilibrium values from the ICE table and rearranging I found "a" which is the concentration of C6H5NH2 to be 0.299M but this is the wrong answer. Does anyone know what I did wrong?

Chemistry - DrBob222, Friday, March 2, 2012 at 9:41pm
The only thing I can find is that if pH = 9.05 then pOH is 4.95 and not 4.98 which makes (aniline) = 0.2997 which I would round to 0.300M. I assume you are using Kb found in your text or notes. My OLD OLD text lists it as 3.98E-10 but you should go with the number in your text or notes. The correct answer is 0.300M. See my solution a couple posts above yours under Ashley. It looks almost the same as yours except for the pOH thing.
Chemistry - DrBob222, Friday, March 2, 2012 at 9:59pm
On second thought I do know what the problem is. You have only two significant figures in Kb; therefore, the answer must have no more than two. So the answer should be rounded to 0.03M
Let me know is this is correct. I think that must be the problem.
Chemistry - Kiana, Saturday, March 3, 2012 at 12:17pm
No, that didn't work. I even tried using your Kb value but that still wasn't right.
Chemistry - DrBob222, Saturday, March 3, 2012 at 4:55pm
I made a typo in my response (transposed the numbers). The concn we calculated is 0.298 or so and to two s.f. that rounds to 0.30 (not 0.03 as I posted in my second post). I think 0.30 will get it for you. I would appreciate a new post to DrBob222 since this one is getting buried.
Surely 0.30 is right.
Chemistry DrBob222 - Kiana, Sunday, March 4, 2012 at 5:38pm
No, 0.30 isn't right either. I'm going to ask my prof about it tomorrow.

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