Question

What volume of $ 0.1M $ HCOONa should be added to $ 50 $ ml of $ 0.05 $ M HCOOH to produce a buffer of $pH = 4 $ ?
(pKa $ = 3.8 $ )

Answer 1

Hint: The pH of the solution is determined by the relative concentrations of weak acids and weak bases. The relationship between the pH of the solution and the concentration of the weak acid and its conjugate base is easily derived using the Handerson-Hasselbach equation.

Complete answer:
To solve this question we need to understand the Handerson-Hasselbach equation. This equation relates the pH of the solution with its pKa and the concentration of an acid and its conjugate base.
So, let’s have a look at the Handerson-Hasselbach equation for a better understanding of the question:
 $ pH = p{K_a} + \log \dfrac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}} $
Where $ {A^ - } $ is a conjugate base and the HA is the weak acid.
So, let’s put the value given in question in this above Handerson-Hasselbach equation:
In the question it is given that the:
Ph $ = 4 $
(pKa $ = 3.8 $ )
So, let's put the values:
So, $ 4.0 = 3.80 + \log \dfrac{{\left[ {HCOONa} \right]}}{{\left[ {HCOOH} \right]}} $
So, here to know the concentration of the conjugate base and the weak acid we will multiply the volume of that chemical species with its molar concentration.
like the concentration of the HCOONa will be $ = V \times M $
 $ = V \times 0.10 $
And the concentration of the HCOOH will be $ = V \times M $
 $ = 50 \times 0.05 $
Now, we will put these concentration values in the above equation.
Thus we get :
 $ 4.0 = 3.80 + \log \dfrac{{V \times 0.10}}{{50 \times 0.05}} $
Now reshuffling the value from RHS to LHs , we will get:
 $ 4.0 - 3.8 = \log \dfrac{{V \times 0.10}}{{50 \times 0.05}} $
Now we will solve the value of the fraction just by multiplying the denominator and then dividing it with the numerator :
 $ 0.2 = \log \dfrac{{V \times 0.10}}{{2.5}} $
 $ 0.2 = \log 0.04V $
Also, we know that value of antilog $ 0.2 = 1.5848 $
Thus by putting the value of antilog we will get:
 $ 0.04V = 1.5848 $
 $ V = 39.62mL $
 So, by putting the values in the question, we get $ V = 39.62mL $
Hence, based on the above discussion we can say that the correct answer is $ 36.92 $ ml of $ 0.1M $ HCOONa should be added to $ 50 $ ml of $ 0.05 $ M HCOOH to produce a buffer of $pH = 4 $

Note:
When the molar concentrations of the acid and its conjugate base are equal then the pH of the solution is numerically equivalent to the pKa of the acid. And at this stage, when the pH equals the pKa of the solution then, the concentration of the deprotonated form of the group of molecules is equal to the concentration of the protonated form. In other words we can say that the deprotonation process is a hallway to completion.