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Q: pH calculation ( Answered 5 out of 5 stars,   0 Comments )
Subject: pH calculation
Category: Science > Chemistry
Asked by: medicine_man-ga
List Price: $9.00
Posted: 19 Nov 2006 22:33 PST
Expires: 19 Dec 2006 22:33 PST
Question ID: 784171
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 :)

Subject: Re: pH calculation
Answered By: livioflores-ga on 27 Nov 2006 21:29 PST
Rated:5 out of 5 stars

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":

"Calculating the pH of Weak Acid and Base Solutions":

"Acid dissociation constant - Wikipedia, the free encyclopedia":

"Chemistry: Acids and Bases":

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.

medicine_man-ga rated this answer:5 out of 5 stars
Excellent answer thanks a lot livioflores :)

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