A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.57 . A. Determine the concentration of C6H5NH+3 in the solution if the concentration of C6H5NH2 is 0.200 M. The pKb of aniline is 9.13. g

[C₆H₅NH₃⁺] = 0.0399 M

Explanation:

This excersise can be easily solved by the Henderson Hasselbach equation

C₆H₅NH₃Cl → C₆H₅NH₃⁺  + Cl⁻

pOH = pKb + log (salt/base)

As we have value of pH, we need to determine the pOH

14 - pH = pOH

pOH = 8.43  (14 - 5.57)

Now we replace data:

pOH = pKb + log ( C₆H₅NH₃⁺/  C₆H₅NH₂ )

8.43 = 9.13 + log (  C₆H₅NH₃⁺ / 0.2 )

-0.7 = log (  C₆H₅NH₃⁺ / 0.2 )

10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2

0.19952 = C₆H₅NH₃⁺ / 0.2

C₆H₅NH₃⁺ = 0.19952 . 0.2  = 0.0399 M

it's c. elements are placed in rows based on how may valencies electrons have.