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.

Explanation:

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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:

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:

In terms of the reaction extent , we write:

Thus, solving for we obtain:

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:

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thats how its supposed to be