Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.
Explanation:
Given: Concentration of hydrogen fluoride = 0.126 M
Concentration of fluoride ions = 0.1 M
Volume of HCl = 9.0 mL
Concentration of HCl = 0.01 M
Volume of HCl = 25.0 mL
Moles of
ions are calculated as follows.
![Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol](https://tex.z-dn.net/?f=Moles%20of%20F%5E%7B-%7D%20%3D%20molarity%20%5Ctimes%20volume%5C%5C%3D%200.1%20M%20%5Ctimes%200.025%20L%5C%5C%3D%200.0025%20mol)
Moles of HF are as follows.
![Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol](https://tex.z-dn.net/?f=Moles%20of%20HF%20%3D%20Molarity%20%5Ctimes%20Volume%5C%5C%3D%200.126%20M%20%5Ctimes%200.025%20L%5C%5C%3D%200.00315%20mol)
Moles of HCl are as follows.
![Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol](https://tex.z-dn.net/?f=Moles%20of%20HCl%20%3D%20Molarity%20%5Ctimes%20volume%5C%5C%3D%200.01%20M%20%5Ctimes%200.009%20L%5C%5C%3D%200.00009%20mol)
Now, reaction equation with initial and final moles will be as follows.
![H^{+} + F^{-} \rightarrow HF](https://tex.z-dn.net/?f=H%5E%7B%2B%7D%20%20%2B%20F%5E%7B-%7D%20%20%5Crightarrow%20%20HF)
Initial: 0.00009 0.0025 0.00315
Equilibrium: (0.0025 - 0.00009) (0.00315 + 0.00009)
= 0.00241 = 0.00324
Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L
Hence, concentration of fluoride ions is calculated as follows.
![Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M](https://tex.z-dn.net/?f=Concentration%20%3D%20%5Cfrac%7Bmoles%7D%7Bvolume%7D%5C%5C%3D%20%5Cfrac%7B0.00241%20mol%7D%7B0.034%20L%7D%5C%5C%3D%200.0709%20M)
Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.