Question

Pyridine is a conjugate base which reacts with H to form pyridine hydrochloride. The hydrochloride dissociates to yield H' with a pKof 5.36. Describe the preparation of a pyridine buffer at pH 5.2 starting with 0.1M pyridine and 1.OM HCI. You may start with one liter of the 0.1 M pyridine.

Answer 1

Answer:

To one liter of the 0.1 M pyridine you need to add 41 mL of 1,0M HCl to obtain a buffer at 5,2

Explanation:

The reaction is:

pyridine-H⁺ ⇄ pyridine + H⁺ pka = 5,36; k = $10^{-5,36}$

Using Henderson-Hasselbalch formula:

5,2 = 5,36 + log$\frac{[Py-H^+]}{[Py]}$

0,692 = $\frac{[Py-H^+]}{[Py]}$ (1)

As total intial moles are 0,1:

0,1 = Py-H⁺ moles + Py moles (2)

Replacing (2) in (1) final moles of both Py-H⁺ and Py are:

Py: 0,059 moles

Thus:

Py-H⁺: 0,041 moles

Moles in reaction are:

Py: 0,1-x moles

H⁺: Y-x moles Y are initial moles of H⁺

Py-H⁺: x moles

Knowing x = 0,041 moles, pyridine volume is 1L and HCl molarity is 1 mol/L and [H⁺] = $10^{-5,2}$

$10^{-5,2}$ = $\frac{Y-0,041moles}{Y+1L}$

Y = 0,04100605 moles≡ 41 mL of 1,0M HCl

I hope it helps!