electrochemistry for webassign

Chemistry
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An aqueous solution contains dissolved C6H5NH3Cl and C6H5NH2. The concentration of C6H5NH2 is 0.37 M and pH is 4.25

(a) Calculate the concentration of C6H5NH3+ in this buffer solution. 
May 1st, 2015

Solution:

pOH = 14 - 4.22 =9.78 
Kb of aniline = 4.0 x 10^-10 
pKb = 9.40 
9.78 = 9.40 + log [C6H5NH3+] / 0.37 
10^0.38= 2.40 = [C6H5NH3+]/ 0.37 
[C6H5NH3+]= 0.89 M 

moles aniline = 0.37 M x 1 L = 0.37 
moles anilinium = 0.89 x 1 L = 0.89 

moles NaOH = 4.6 g / 40 g/mol=0.12 

C6H5NH3+ + OH- >> C6H5NH2 + H2O 
moles anilinium = 0.89 - 0.12 =0.77 
moles aniline = 0.37 + 0.12 = 0.49 

pOH = 9.40 + log 0.77/ 0.49=9.60 
pH = 4.40

May 1st, 2015

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