An aqueous solution contains dissolved c6h5nh3cl and c6h5nh2. the concentration of c6h5nh2 is 0.53 m and ph is 4.17. (a) calculate the concentration of c6h5nh3+ in this buffer 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

Explanation:

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