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?
5 answers
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.
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.
Let me know is this is correct. I think that must be the problem.
No, that didn't work. I even tried using your Kb value but that still wasn't right.
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.
Surely 0.30 is right.
No, 0.30 isn't right either. I'm going to ask my prof about it tomorrow.
2 answers