Answer : The concentration of [tex]C_6H_5NH_3^+[/tex] is, 0.0830 M
Explanation : Given,
[tex]pK_b=9.13[/tex]
Concentration of [tex]C_6H_5NH_2[/tex] = 0.245 M
pH = 5.34
First we have to calculate the value of pOH.
[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-5.34\\\\pOH=8.66[/tex]
Now we have to calculate the concentration of [tex]C_6H_5NH_3^+[/tex]
Using Henderson Hesselbach equation :
[tex]pOH=pK_b+\log \frac{[Salt]}{[Base]}[/tex]
[tex]pOH=pK_b+\log \frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]8.66=9.13+\log (\frac{[C_6H_5NH_3^+]}{0.245})[/tex]
[tex][C_6H_5NH_3^+]=0.0830M[/tex]
Therefore, the concentration of [tex]C_6H_5NH_3^+[/tex] is, 0.0830 M