Answer : The pH of the solution is, 4.9
Explanation : Given,
Dissociation constant for acetic acid = ![K_a=1.8\times 10^{-5}](https://tex.z-dn.net/?f=K_a%3D1.8%5Ctimes%2010%5E%7B-5%7D)
Concentration of acetic acid = 0.05 M
Concentration of sodium acetate = 0.075 M
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 (1.8\times 10^{-5})](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%281.8%5Ctimes%2010%5E%7B-5%7D%29)
![pK_a=5-\log (1.8)](https://tex.z-dn.net/?f=pK_a%3D5-%5Clog%20%281.8%29)
![pK_a=4.7](https://tex.z-dn.net/?f=pK_a%3D4.7)
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)
![pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D)
Now put all the given values in this expression, we get:
![pH=4.7+\log (\frac{0.075}{0.05})](https://tex.z-dn.net/?f=pH%3D4.7%2B%5Clog%20%28%5Cfrac%7B0.075%7D%7B0.05%7D%29)
![pH=4.9](https://tex.z-dn.net/?f=pH%3D4.9)
Therefore, the pH of the solution is 4.9.