Question

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.140 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?

mL benzoic acid = ?
mL sodium benzoate = ?

Answer 1

Concepts and reason

Concentration of a weak acid is related to its ${\rm{pKa}}$and pH of a solution as shown in the following Henderson -Hasselbalch equation. The equation is shown below.

${\rm{pH = pKa + log}}\frac{{\left[ {{\rm{salt}}} \right]}}{{\left[ {{\rm{acid}}} \right]}}$

Here, Ka is the acid dissociation constant; $\left[ {{\rm{salt}}} \right]$, and$\left[ {{\rm{acid}}} \right]$are the concentration terms of the salt and the acid respectively. This equation is applicable to a system that contains weak acids or weak bases only.

Benzoic acid is a weak acid, so we can apply the equation to this system. Calculate the volumes of benzoic acid and sodium benzoate by substituting the values of${\rm{pKa}}$ and pH in the equation.

For the benzoic acid and for the buffer solution, the equation becomes as shown below.

${\rm{pH = pKa + log}}\frac{{\left[ {{\rm{Sodium benzoate}}} \right]}}{{\left[ {{\rm{Benzoic acid}}} \right]}}$

Fundamentals

An aqueous solution that contains a mixture of a weak acid and its conjugate base or vice versa is called as buffer solution.

Benzoic acid is a weak acid. Because of its poor dissociation, it forms benzoate ions. The Henderson -Hasselbalch equation is applicable to weak acids and bases only.

The value of ${\rm{pKa}}$for the benzoic acid is 4.20. Calculate the fraction of concentrations of acetate and acetic acid.

$\begin{array}{c}\\{\rm{pH = pKa + log}}\frac{{\left[ {{\rm{Sodium benzoate}}} \right]}}{{\left[ {{\rm{Benzoic acid}}} \right]}}\\\\4.0 = 4.20 + {\rm{log}}\frac{{\left[ {{\rm{Sodium benzoate}}} \right]}}{{\left[ {{\rm{Benzoic acid}}} \right]}}\\\end{array}$

$\begin{array}{c}\\{\rm{log}}\frac{{\left[ {{\rm{Sodium benzoate}}} \right]}}{{\left[ {{\rm{Benzoic acid}}} \right]}} = - 0.20\\\\\frac{{\left[ {{\rm{Sodium benzoate}}} \right]}}{{\left[ {{\rm{Benzoic acid}}} \right]}} = {10^{ - 0.20}} = 0.631\\\end{array}$

Assume the volume of the acid is $x$ and volume of the benzoate solution as$y$.

$x + y = 100{\rm{ mL}}$ … $\left( 1 \right)$

$\frac{{\left[ {{\rm{Sodium benzoate}}} \right]}}{{\left[ {{\rm{Benzoic acid}}} \right]}} = 0.631$

$\frac{{y \times 0.140{\rm{ M}}}}{{x \times 0.1{\rm{ M}}}} = 0.631$

$y = 0.451 \times x$ …. $\left( 2 \right)$

By solving the equations $\left( 1 \right)$ and$\left( 2 \right)$,

$\begin{array}{l}\\x = 68.93{\rm{ mL}}\\\\{\rm{y = 31}}{\rm{.07 mL}}\\\end{array}$

Ans:

Volume of benzoic acid is$68.93{\rm{ mL}}$.

Volume of sodium benzoate is${\rm{31}}{\rm{.07 mL}}$.