Question

H2A is a diprotic acid of which pKa values are 3.0 and 7.0. Answer to the following questions.

(d) Please explain how to prepare 0.1 M buffer solution with the pH of 7.4. Assume that the equilibrium concentrations of an acid and its conjugate base remain close to their formal concentration and that the buffer solution behaves ideally at room temperature. (10 pt)

Answer 1

As H₂A is a diprotic acid.

H₂A <-----> HA^- + H⁺, pK_a1 = 3.0

HA^- <-----> A^2- + H⁺, pK_a2 = 7.0

For preparing buffer, second dissiciation will be considered i.e. buffer will comprise of dibasic salt and monobasic salt.

Concentration of buffer = 0.1 M

Let the concentration of monobasic salt (HA^-) = x

Concentration of dibasic salt (A^2-) = 0.1 - x

According to Henderson Hasselbalch equation,

$pH = pKa + log\frac{[A^{^-2}]}{[HA^{-}]}$

$7.4 = 7.0 + log\frac{0.1-x}{x}$

$0.4 = log\frac{0.1-x}{x}$

$10^{0.4} = \frac{0.1-x}{x}$

$2.512 = \frac{0.1-x}{x}$

2.512 x = 0.1 - x

3.512 x = 0.1

x = 0.1/3.512 = 0.0285 M

0.1 - x = 0.0715 M

So, 0.1 M buffer of pH 7.4 can be prepared by taking 0.0715 M solution of dibasic salt and 0.0285 M solution of monobasic salt.