Respuesta :
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 [tex]F^{-}[/tex] ions are calculated as follows.
[tex]Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol[/tex]
Moles of HF are as follows.
[tex]Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol[/tex]
Moles of HCl are as follows.
[tex]Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol[/tex]
Now, reaction equation with initial and final moles will be as follows.
[tex]H^{+} + F^{-} \rightarrow HF[/tex]
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.
[tex]Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M[/tex]
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.
