Answer : The moles of solid NaF is, 1.09 mole.
Explanation : Given,
pH = 3.46
![K_a=7.2\times 10^{-4}](https://tex.z-dn.net/?f=K_a%3D7.2%5Ctimes%2010%5E%7B-4%7D)
Concentration of HF = 0.115 M
Volume of solution = 1.0 L
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 (7.2\times 10^{-4})](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%287.2%5Ctimes%2010%5E%7B-4%7D%29)
![pK_a=4-\log (7.2)](https://tex.z-dn.net/?f=pK_a%3D4-%5Clog%20%287.2%29)
![pK_a=3.14](https://tex.z-dn.net/?f=pK_a%3D3.14)
Now we have to calculate the moles of HF.
![\text{Moles of }HF=\text{Concentration of }HF\times \text{Volume of solution}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DHF%3D%5Ctext%7BConcentration%20of%20%7DHF%5Ctimes%20%5Ctext%7BVolume%20of%20solution%7D)
![\text{Moles of }HF=0.115M\times 1.0L=0.115mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DHF%3D0.115M%5Ctimes%201.0L%3D0.115mol)
Now we have to calculate the moles of NaF.
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{\text{Moles of NaF}}{\text{Moles of HF}}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5Ctext%7BMoles%20of%20NaF%7D%7D%7B%5Ctext%7BMoles%20of%20HF%7D%7D)
Now put all the given values in this expression, we get:
![3.46=3.14+\log (\frac{\text{Moles of NaF}}{0.115})](https://tex.z-dn.net/?f=3.46%3D3.14%2B%5Clog%20%28%5Cfrac%7B%5Ctext%7BMoles%20of%20NaF%7D%7D%7B0.115%7D%29)
![\text{Moles of NaF}=1.09mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaF%7D%3D1.09mol)
Thus, the moles of solid NaF is, 1.09 mole.