Answer : The pH of a solution is, 4.5
Explanation : Given,
Moles of benzoic acid = 0.15 mol
Moles of sodium benzoate = 0.30 mol
Volume of solution = 1.00 L
The dissociation constant for benzoic acid = ![K_a=6.5\times 10^{-5}](https://tex.z-dn.net/?f=K_a%3D6.5%5Ctimes%2010%5E%7B-5%7D)
First we have to calculate the value of
.
The expression used for the calculation of
is,
![pK_a=-\log (K_a)](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%28K_a%29)
Now put the value of
in this expression, we get:
![pK_a=-\log (6.5\times 10^{-5})](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%286.5%5Ctimes%2010%5E%7B-5%7D%29)
![pK_a=5-\log (6.5)](https://tex.z-dn.net/?f=pK_a%3D5-%5Clog%20%286.5%29)
![pK_a=4.2](https://tex.z-dn.net/?f=pK_a%3D4.2)
Now we have to calculate the pH of a solution.
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:
![pH=4.2+\log [\frac{(\frac{0.30}{1.00L})}{(\frac{0.15}{1.00L})}]](https://tex.z-dn.net/?f=pH%3D4.2%2B%5Clog%20%5B%5Cfrac%7B%28%5Cfrac%7B0.30%7D%7B1.00L%7D%29%7D%7B%28%5Cfrac%7B0.15%7D%7B1.00L%7D%29%7D%5D)
![pH=4.5](https://tex.z-dn.net/?f=pH%3D4.5)
Therefore, the pH of a solution is, 4.5