Compute for pH of pyridinium chloride.?

the pyridinium ion in solution reacts with water to give an acidic solution 

C5H5NH++H2O⇋C5H5N+H3O+

Ka=[C5H5N][H3O+]/ [C5H5NH+]

Ka = 5.6 x 10-6  a very weak acid so we can use the abbreviated formula for weak acids

[H3O+] = SQRT(Ka * concentration )

= SQRT(  5.6 x 10-6 * 0.458 ) = 1.60E-03

pH = 2.8 >>>>> there is your answer

Refer to: https://ch301.cm.utexas.edu/data/section2.php?target=ka-kb-constants.php
Kb for C₅H₅N (pyridine) = 1.8 × 10⁻⁹
Hence, Ka for C₅H₅NH⁺ (pyrindinium ion) = Kw/Kb(C₅H₅N)

Consider the ionization of aqueous pyridinium ion:
             C₅H₅NH⁺(aq) + H₂O(ℓ) ⇌ C₅H₅N(aq) + H₃O⁺(aq)    Ka = Kw/Kb(C₅H₅N)
Initial:      0.458 M                             0 M             0 M
Change:      -y M                             +y M           +y M
Eqm:    (0.458 - y) M                         y M             y M
                ≈ 0.458 M            (as Ka is very small)
 
At equilibrium:
Ka = [C₅H₅N] [H₃O⁺] / [C₅H₅NH⁺]
(1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁹) = y² / 0.458
y = √[0.458 × (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁹)]
y = 1.6 × 10⁻³

pH of the solution = -log(1.60 × 10⁻³) = 2.8

====
OR:

pH
≈ -(1/2) {log[C₅H₅NH⁺]ₒ + log(Ka)}
= -(1/2) {log(0.458) + log[(1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁹)]}
= 2.8

-log[0.458]= 0.33 M