Answer : The pH of the solution is, 3.41
Explanation :
First we have to calculate the moles of  .
.


Now we have to calculate the value of  .
.
The expression used for the calculation of  is,
 is,

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



The reaction will be:
                              
Initial moles     0.375     0.100   0.375
At eqm.   (0.375-0.100)      0     (0.375+0.100)
                      = 0.275                    = 0.475
Now we have to calculate the pH of 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)
![pH=pK_a+\log \frac{[F^-]}{[HF]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
Now put all the given values in this expression, we get:
![pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]](https://tex.z-dn.net/?f=pH%3D3.17%2B%5Clog%20%5B%5Cfrac%7B%28%5Cfrac%7B0.475%7D%7B1.50%7D%29%7D%7B%28%5Cfrac%7B0.275%7D%7B1.50%7D%29%7D%5D)

Thus, the pH of the solution is, 3.41