Question

How would you use the Henderson-Hasselbalch equation to calculate the $pH$ of a buffer solution that is $0.27M$ in formic acid and $0.50M$ in sodium formate?

Answer 1

The Henderson-Hasselbalch equation is given as $pH = p{K_a} + {\log _{10}}\left( {\dfrac{{[Base]}}{{[Acid]}}} \right)$ . By substituting the values we can get the $pH$ of buffer solution that is prepared by formic acid and sodium formate.

Complete step by step answer:
As, the Henderson-Hasselbalch equation is given as:
$pH = p{K_a} + {\log _{10}}\left( {\dfrac{{[Base]}}{{[Acid]}}} \right)$
As we know that the buffer solution consists of formic acid having molecular formula $HCOOH$ and sodium formate having molecular formula $HCOONa$ ,then formate anion will be $HCO{O^{ - 1}}$ .
The $pH$ of buffer can be calculated by the above given equation. The Henderson-Hasselbalch equation relates the $pH$ with the $p{K_a}$ value of the weak acid and the ratio that exists between the concentrations of the weak acid and conjugate base.
As the $p{K_a}$ value of formic acid is $3.75$ .
So, by substituting the values in the equation:
$pH = p{K_a} + {\log _{10}}\left( {\dfrac{{[Base]}}{{[Acid]}}} \right)$
We will get $pH$ of buffer solution as:
$\ \Rightarrow pH = 3.75 + \log \left( {\dfrac{{0.50M}}{{0.27M}}} \right) \\\ \Rightarrow pH = 4.02 \\\ \$
Hence, the $pH$ of a buffer solution that is $0.27M$ in formic acid and $0.50M$ in sodium formate is $4.02$ .

Note:
The numerical value of the acid dissociation constant ${K_a}$ of the acid is known or assumed. The $pH$ is calculated for given values of the concentrations of the acid, $H_A$ and of a salt, $M_A$ , of its conjugate base, ${A^ - }$ . A simple buffer solution must be a solution of an acid and a salt which is the conjugate base of the acid. For example, the solution may contain formic acid and sodium acetate. The Henderson–Hasselbalch equation relates the $pH$ of a solution containing a mixture of the two components to the acid dissociation constant, ${K_a}$ and the concentrations of the species in solution.