Question

Consider the bases in the table. Which base would be the best choice for preparing a pH = 5.00 buffer?
Pyridine is the best choice, however I need help finding the ratio used to make 1.0 L of the buffer

Explain how to make 1.0 L of this buffer.
ratio = ? M base : 1 M acid

Answer 1

We have Henderson's equation for basic buffer as p OH = p K b + log [ Conjugate acid] / [ Base]

For Buffer with good buffer capacity the ratio [ Conjugate acid] / [ Base] must be 1 or close to 1 means buffer solution must contain equal concentration of weak base and its conjugate acid.

If we put [ Conjugate acid] / [ Base] = 1 in above equation we get p OH = p K b.

To identify best choice to prepare buffer we need to calculate p K b values.

Base	K b	p K b = -log K b
Ammonia	1.8 $\times$ 10 -05	4.74
Methyl amine	4.38 $\times$ 10 -04	3.36
Ethyl amine	5.6 $\times$ 10 -04	3.25
Aniline	3.8 $\times$ 10 -10	9.42
Pyridine	1.7 $\times$ 10 -09	8.77

From above table it is clear that p Kb of pyridine is close to p OH = 9 . Hence, Pyridine is best choice to prepare buffer solution with p H =5.

PART 2

We have Henderson's equation for basic buffer as p OH = p K b + log [ Conjugate acid] / [ Base]

$\therefore$ p OH = p K b + log [ C₅H₅NH ⁺ ] / [ C₅H₅NH]

9.0 = 8.77 + log [ C₅H₅NH ⁺ ] / [ C₅H₅NH]

log [ C₅H₅NH ⁺ ] / [ C₅H₅NH] = 9.0 - 8.77 = 0.23

Taking antilog on both sides,we get

[ C₅H₅NH ⁺ ] / [ C₅H₅NH] = 10 ^0.23 = 1.698

[ C₅H₅NH ⁺ ] = 1.698 $\times$ [ C₅H₅NH] or

$\therefore$ 1.0 M $\times$ 1.0 L = 1.698 $\times$ M C₅H₅NH $\times$ 1.0 L

M C₅H₅NH = 1.0 M $\times$ 1.0 L / 1.698 $\times$   1.0 L = 0.5889 M

ANSWER : Molarity of base = 0.5889 M