Answer : The pH of a solution is, 4.50
Explanation : Given,

Concentration of benzoic acid (Acid) = 0.150 M
Concentration of sodium benzoate (salt) = 0.300 M
First we have to calculate the value of
.
The expression used for the calculation of
is,

Now put the value of
in this expression, we get:



Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:


Thus, the pH of a solution is, 4.50