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:
![pH=pKa + log ( \frac{[C_6H_5COONa]}{[C_6H_5COOH]} )=4,19+log( \frac{0,15M}{0,25M} )=3,97](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BC_6H_5COONa%5D%7D%7B%5BC_6H_5COOH%5D%7D%20%29%3D4%2C19%2Blog%28%20%5Cfrac%7B0%2C15M%7D%7B0%2C25M%7D%20%29%3D3%2C97%20)
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is 3,97
Have a nice day!
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:
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is 3,97
Have a nice day!