0.25 M solution of pyridinium chloride C₅H₆N⁺Cl⁻ was found to have a pH of 2.699. if the value of Kₑ of pyridine C₅H₅N is a × 10⁻¹⁰, then the value of a is (given, 10⁻².⁶⁹⁹ = 2 × 10⁻³)

A(g) ->B(g) + $\frac{1}{2}$ (g)

Initial moles             n                         0             0

Eqb. moles               n(1 – a)   $\frac{n\alpha }{2}$           n_a            

total moles =   $n\left(1+\frac{\alpha }{2}\right)$

Eqb. pressure   $\frac{\left(1-\alpha \right)p}{1+\frac{\alpha }{2}}$ $\frac{\alpha p}{1+\frac{\alpha }{2}}$ $\frac{\left(\frac{\alpha }{2}\right)p}{1+\frac{\alpha }{2}}$

$\frac{K_p=\frac{\alpha p}{\left(1+\frac{\alpha }{2}\right)}\times {\left[\frac{\alpha p}{\left(2+\alpha \right)}\right]}^{\frac{1}{2}}}{\frac{\left(1+\alpha \right)p}{1+\frac{\alpha }{2}}}$

$K_p=\frac{{\alpha }^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{p}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(2+\alpha \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(1-\alpha \right)}$