Question

Calculate the pH at 25 degrees celsius of a 0.39 M solution of pyridinium chloride (c5h5nhcl) . Note that pyridine (c5h5n) is a weak base with a pkb of 8.77 .
Round your answer to 1 decimal place.

Answer 1

Answer:

$pH=2.3$

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:

$C_5H_5NHCl(aq)+H_2O(l)\rightleftharpoons C_5H_5NCl^-(aq)+H_3O^+(aq)$

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:

$Ka=\frac{[C_5H_5NCl^-][H_3O^+]}{[C_5H_5NHCl]}$

In terms of the reaction extent $x$, we write:

$5.89x10^{-5}=\frac{x*x}{0.39-x}$

Thus, solving for $x$ we obtain:

$x_1=-0.0048M\\\\x_2=0.0048M$

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:

$pH=-log([H_3O^+])=-log(0.0048)\\\\pH=2.3$

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