Answer : The pH of the solution is, 3.41
Explanation :
First we have to calculate the moles of
.
![\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HF%7D%3D%5Ctext%7BConcentration%20of%20HF%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%7D)
![\text{Moles of HF}=0.250M\times 1.50L=0.375mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HF%7D%3D0.250M%5Ctimes%201.50L%3D0.375mol)
Now 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 (6.8\times 10^{-4})](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%286.8%5Ctimes%2010%5E%7B-4%7D%29)
![pK_a=4-\log (6.8)](https://tex.z-dn.net/?f=pK_a%3D4-%5Clog%20%286.8%29)
![pK_a=3.17](https://tex.z-dn.net/?f=pK_a%3D3.17)
The reaction will be:
![HF+OH^-\rightleftharpoons F^-+H_2O](https://tex.z-dn.net/?f=HF%2BOH%5E-%5Crightleftharpoons%20F%5E-%2BH_2O)
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)
![pH=3.41](https://tex.z-dn.net/?f=pH%3D3.41)
Thus, the pH of the solution is, 3.41