Calculate ionisation constant for pyridinium hydrogen Chloride. (given that `H^+` ion concentration is `3.6xx10^(-4)` M and its concentration is 0.02 M)-

Options

A. `6.48xx10^(-2)`
B. `6xx10^(-6)`
C. `1.5xx10^(-9)`
D. `12xx10^(-8)`

Explanation

Pyridinium hydrochloride is a salt of weak base and strong acid.
`pH=-log[H^+]=-log(3.6xx10^(-4))=3.44`
Now, pH=`1/2[logK_w-logK_a+logc]`
or `3.44=-1/2[-14-logK_a+log(2xx10^(-12))]`
or `6.88=14+logK_a+1.70" or "logK_a=-8.82`
or `K_a="antilog"(-8.82)=1.5xx10^(-9)`