Hello!
To solve this problem we'll use the Henderson-Hasselbach equation, but first we need the vale for the pKa of Benzoic acid, which is pKa= -log(Ka)=4,19
Now, we apply the equation as follows:
[tex]pH=pKa + log ( \frac{[C_6H_5COONa]}{[C_6H_5COOH]} )=4,19+log( \frac{0,15M}{0,25M} )=3,97 [/tex]
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is 3,97
Have a nice day!