Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
[tex]n_{acid}=n_{base}=n_{salt}[/tex]
Whereas the moles of the salt are computed as shown below:
[tex]n_{salt}=0.021L*0.68mol/L=0.01428mol[/tex]
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
[tex][salt]=0.01428mol/0.0276L=0.517M[/tex]
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
[tex]C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+[/tex]
Whose equilibrium expression is:
[tex]Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}[/tex]
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
[tex]2.326x10^{-5}=\frac{x^2}{0.517M}[/tex]
Whereas x is:
[tex]x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3[/tex]
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
[tex]pH=-log(3.47x10^{-3})\\\\pH=2.46[/tex]
Regards!
