A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Explanation

Pyridinium hydrochloride is the salt of a weak base( pyridine) and a strong acid (HCl) . For the salt of a weak base and strong acid.
` pH= (1)/(2) pK_w - (1)/(20 pK_b -(1)/(2) log_(10) c `
Substituting the values ,we have
` 3.44 = [-(1)/(2) log_(10) 1.0 xx 10 ^(-14) -(1)/(2) xx (- log_(10) K_b) - (1)/(2)xx log_(10) 0.02]`
` or " " 3.44 = - (1)/(2)xx (-14) +(1)/(2) log_(10) K_b- (1)/(2) xx (-1.699)`
` " " 3.44 = 7+ (1)/(2) log_(10) K_b+0.849`
` log_(10) K_b= (3.44 -7-0.849)xx 2 = - 8.82`
` or " " K_b= " antilog"_(10) (-8.82) = 1.5 xx 10^(-9)`