Answer:
[tex]pH=2.3[/tex]
Explanation:
Hello!
In this case, since pyridinium chloride has a pKb of 8.77 which is a Kb of 1.70x10⁻⁹ and therefore a Ka of 5.89x10⁻⁵ which means it tends to be acidic, we write its ionization via:
[tex]C_5H_5NHCl(aq)+H_2O(l)\rightleftharpoons C_5H_5NCl^-(aq)+H_3O^+(aq)[/tex]
Because it is a Bronsted base which donates one hydrogen ion to the water to produce hydronium. Thus, we write the equilibrium expression with the aqueous species only:
[tex]Ka=\frac{[C_5H_5NCl^-][H_3O^+]}{[C_5H_5NHCl]}[/tex]
In terms of the reaction extent [tex]x[/tex], we write:
[tex]5.89x10^{-5}=\frac{x*x}{0.39-x}[/tex]
Thus, solving for [tex]x[/tex] we obtain:
[tex]x_1=-0.0048M\\\\x_2=0.0048M[/tex]
Clearly the solution is 0.0048 M because to negative values are not allowed, therefore, since it equals the concentration of hydronium which defines the pH, we write:
[tex]pH=-log([H_3O^+])=-log(0.0048)\\\\pH=2.3[/tex]
Best regards!
