What is the pH of a 0.062 mol L-1 solution of aniline (C6H5NH2)
solution? Kb for this weak base is 4.2 x 10^-10.

Please show me how you get to the pH :)

Cheers.

Hi!!


For a base A that follows the equation:
 A?(aq) + H2O(l) <----> HA(aq) + OH?(aq) 


Kb = [HA]*[OH-]/[A-]

For aniline the ionization equation is:
C6H5NH2 + H2O <----> C6H5NH3+ + OH-

Note that [OH-] = [C6H5NH3+], then for aniline:

Kb = [OH-]^2/[C6H5NH2]


Our first intermediate goal is to find the [OH-]:

[OH-] = sqrt(Kb*[C6H5NH3+]) = 
      = sqrt(4.2 x 10^-10*0.062) =
      = 5.103*10^-6

Now we can obtein the pOH, and with this we can get pH easily:

pOH = -log[OH-] = 5.2922

pH = 14-pOH = 8.7078


For further references see:
"Ionization Constants for Weak Bases":
http://www2.austin.cc.tx.us/chemlab/weakbase.htm


"Calculating the pH of Weak Acid and Base Solutions":
http://dbhs.wvusd.k12.ca.us/webdocs/AcidBase/Weak-AcidBase-Practice.html


"Acid dissociation constant - Wikipedia, the free encyclopedia":
http://en.wikipedia.org/wiki/Acid_dissociation_constant


"Chemistry: Acids and Bases":
http://www.shodor.org/UNChem/basic/ab/


Search strategy:
aniline "ionization equation"
ph kb base
ph calculation base


I hope this helps you. Feel free to request for a clarifiation if you need it.


Regards,
livioflores-ga

Excellent answer thanks a lot livioflores :)