Question

How would you use the Henderson-Hasselbalch equation to calculate the pH of buffer solution that is $0.27M$ formic acid $\left( {HC{O_2}H} \right)$ and $0.50M$ in sodium formate $\left( {HC{O_2}Na} \right)$?

Answer 1

We have to know that while calculating $pH$, we can use Henderson-Hasselbalch equation for calculating the $pH$ of buffer solution and the numerical acid dissociation constant ${K_a}$ of acid that is known and by using this, we can calculate the pH for that specific solution.

Complete answer:
We can calculate the $pH$ of the solution using Henderson-Hasselbalch equation. We can relate the $pH$, $p{K_a}$ and molarity using Henderson-Hasselbalch equation. We have to know that equilibrium between weak acid and its conjugate base permits the solution to prevent the pH change when a certain quantity of strong acid (or) strong base is added and so, we can estimate the pH of this buffer using Henderson-Hasselbalch equation. We can write Henderson-Hasselbalch equation as,
$pH = p{K_a} + \log \left( {\dfrac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}} \right)$
Here, we can say pH is the acidity of the solution.
The negative log of ${K_a}$ is $p{K_a}$.
The concentration of acid is $\left[ {HA} \right]$.
The concentration of conjugate base is $\left[ {{A^ - }} \right]$.
We are provided with a molarity of sodium formate as $0.50M$ and molarity of formic acid as $0.27M$. We can use the $p{K_a}$ of formic acid as $3.75$.
Let us now substitute these values in the Henderson-Hasselbalch equation. By using these values, we can calculate the pH as follows,
$pH = pK + \log \left( {\dfrac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}} \right)$
Now we can substitute the known values we get,
$pH = 3.75 + \log \left( {\dfrac{{\left[ {0.5} \right]}}{{\left[ {0.27} \right]}}} \right)$
$pH = 3.75 + 0.268$
On simplification we get,
$pH = 4.02$
We have calculated the pH of the buffer solution as $4.02$.

Note:
We have to know that the value of pH of the solution is higher when compared to the value of $p{K_a}$ that represents the presence of a more conjugate base when compared to weak base. The logarithm would have position value, which indicates the value of pH would increase gradually.