Question

Write the Henderson-Hasselbalch equation for a solution of propanoic acid (CH3CH2CO2H, pka = 4.874) using HA, A-, and the given Pka value in the expression. Using this equation, calculate the quotient (A]/[HA] at A) pH 4.23 B) pH 4.874 C) pH 530.

Answer 1

Concepts and reason

The concept used is to write the Henderson-Hasselbalch equation for the solution of propionic acid.

Fundamentals

The Henderson-Hasselbalch equation to calculate the ${\rm{pH}}$ of a buffer system is as follows:

${\rm{pH}} = p{K_a} + \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}$

1.

Given,

$p{K_a} = 4.874$

Let, ${\rm{propionic acid}} = {\rm{HA}}$

${\rm{propionate}} = {{\rm{A}}^ - }$

The Henderson-Hasselbalch equation for the solution of propionic acid is as follows:

${\rm{pH}} = 4.874 + \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}$

Given,

$p{K_a} = 4.874$

${\rm{pH}} = 4.23$

Substitute the values in the Henderson-Hasselbalch equation and calculate the quotient $\left[ {{{\rm{A}}^ - }} \right]/\left[ {{\rm{HA}}} \right]$ as follows:

$\begin{array}{l}\\{\rm{4}}{\rm{.23}} = 4.874 + \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}\\\\4.23 - 4.874 = \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}\\\\\frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}} = {10^{ - 0.644}}\\\\\frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}} = 0.23\\\end{array}$

2.

Given,

$p{K_a} = 4.874$

${\rm{pH}} = 4.874$

Substitute the values in the Henderson-Hasselbalch equation and calculate the quotient $\left[ {{{\rm{A}}^ - }} \right]/\left[ {{\rm{HA}}} \right]$ as follows:

$\begin{array}{l}\\{\rm{4}}{\rm{.874}} = 4.874 + \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}\\\\4.874 - 4.874 = \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}\\\\\frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}} = {10^{ - 0}}\\\\\frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}} = 1\\\end{array}$

3.

Given,

$p{K_a} = 4.874$

${\rm{pH}} = 5.30$

Substitute the values in the Henderson-Hasselbalch equation and calculate the quotient $\left[ {{{\rm{A}}^ - }} \right]/\left[ {{\rm{HA}}} \right]$ as follows:

$\begin{array}{l}\\{\rm{5}}{\rm{.30}} = 4.874 + \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}\\\\5.30 - 4.874 = \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}\\\\\frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}} = {10^{0.426}}\\\\\frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}} = 2.7\\\end{array}$

Ans: Part 1

Therefore, the quotient $\left[ {{{\rm{A}}^ - }} \right]/\left[ {{\rm{HA}}} \right]$ at ${\rm{pH}} = 4.23$ is $0.23$ .

Part 2

Therefore, the quotient $\left[ {{{\rm{A}}^ - }} \right]/\left[ {{\rm{HA}}} \right]$ at ${\rm{pH}} = 4.874$ is $1$ .

Part 3

Therefore, the quotient $\left[ {{{\rm{A}}^ - }} \right]/\left[ {{\rm{HA}}} \right]$ at ${\rm{pH}} = 5.30$ is $2.7$ .