Please calculate the pH of an aqueous solution of pyridinium choloride at the (a)1.0M (b) 1.mM?

Answer 1

Pyridinium Chloride is C5H5NHCl
Pyridinium is the conjugate acid of pyridine.
In one of your other answers you said that for pyridine pKb=8.77
Thus for the conjugate acid, since Ka=Kw/Kb, we have
pKa= pKw-pKb= 14-8.77 =5.23

.. .. .. .. .. .. C5H5NH(+) <=> C5H5N + H(+)
Initial .. .. .. .. .. C
Dissociate .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. x
At Equil. .. .. .. C-x .. .. .. .. .. .. .. x .. .. .. x

Ka=x^2/(C-x) =>
x^2 +Kax -KaC=0=>
x^2 + (10^-5.23)x -(10^-5.23)C=0

x=0.5*(-10^-5.23 + Squareroot( (10^-5.23)^2 + 4(10^-5.23)C))

For C=1 M
x=2.42 *10^-3 and pH=-logx=2.62

For C=1 mM=10^-3M
x=7.38*10^-5 and pH=4.13