Question

What volume of 0.10 M sodium formate solution should be added to 50ml of 0.05 M formic acid to produce a buffer solution of pH 4.0? [ for formic acid is 3.80]
A) 3.96 ml
B) 25 ml
C) 39.6 ml
D) 100 ml

Answer 1

Buffer solution is basically an aqueous solution which resists the change in pH when a small amount of strong acid or base is added and formed by the mixture of weak acid and its conjugate base or its salt and vice-versa. The above question can be solved by the Henderson-Hasselbach equation.

Formula Used: For acidic buffer-$pH=p{{K}_{a}}+\log \dfrac{\text{Conjugate Base}}{\text{Weak Acid}}$

Complete answer:
First, let us consider the volume of 0.10 M sodium formate added to be x.
Now, with the help of mole concept, we have
No. of moles in x ml of 0.10 M sodium formate = $\dfrac{0.10}{1000}$ $\times$ x= 0.0001x
No. of moles in 50 ml of 0.05 M formic acid = $\dfrac{0.05}{1000}$ $\times$ 50 = 0.0025
In this M represents the molarity of solution.
Thus, $\dfrac{sodium\;formate}{formic\;acid}$ = $\dfrac{0.0001x}{0.0025}$ = 0.04 x, represents the $\dfrac{salt}{acid}$ ratio.
Now, from the Henderson’s equation,
pH = pK$_a$+ log $\dfrac{salt}{acid}$
Given, pH = 4.0, pK$_a$= 3.80;
Put these values in the Henderson’s equation
4.0 = 3.80 + log 0.04 x
log 0.04 x = 0.2
Therefore, x = 39.6 ml

So, we can conclude that the volume of 0.10 M sodium formate added is 39.6 ml. The correct option is (C).

Note: Don’t get confused while solving the log. Here the logarithmic term is with the base 10. The pK$_a$ and pH terms are related to each other. The pH value at which the chemical species accept, or donate a proton is considered to be pK$_a$ value. The pK$_a$ value shows inverse relation with the acid, i.e. lower the pK$_a$, the stronger will be the acid.