1953) Sodium benzoate and benzoic acid are mixed in equimolar ration to form buffer if pKa is 2 what will be the pH?

A. 0
B. 1
C. 2
D. any one

To determine the pH of the buffer solution formed by mixing equimolar amounts of sodium benzoate and benzoic acid with a pKa value of 2, we need to consider the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])

Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant
[A-] = the concentration of the conjugate base
[HA] = the concentration of the weak acid

In this case, since sodium benzoate (NaC7H5O2) is the conjugate base and benzoic acid (C6H5COOH) is the weak acid, the equation becomes:
pH = 2 + log([NaC7H5O2]/[C6H5COOH])

Given that sodium benzoate and benzoic acid are mixed in equimolar ratio, their concentrations will be the same:
[NaC7H5O2] = [C6H5COOH]

Therefore, the equation simplifies to:
pH = 2 + log(1)

The logarithm of 1 is 0, so the equation further simplifies to:
pH = 2

Hence, the pH of the buffer solution will be equal to 2.

Therefore, the answer is C. 2