0.15 mol of pyridinium chloride is added to 500 mL 0.2 (M) pyridine solution. Determine the pH of the solution. Given: `K_(b)`(pyridine)`=1.5xx10^(-9) and` assume that volume of the solution does not change due to the addition of pyridinium chloride.

Explanation

Solution of pyridine (weak base) and pyridinium chloride (salt) is a buffer solution.
Given, `K_(b)`(pyridine)`=1.5xx10^(-9)`
`pK_(b)=-log_(10)(1.5xx10^(-9))=8.82`
Now, [pyridine]=0.2 (M) and
[pyridiium chloride]`=(0.15xx1000)/(500)=0.3(M)`
`therefore pOH=pK_(b)+"log"(["pyridium chloride"])/(["pyridine"])`
`=8.82+"log"(0.3)/(0.2)=9.0`
`therefore pH` of the solution`=14-pOH=14-9=5`