Question

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. 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 ________ M.
A) 0.0735
B) 0.0762
C) 0.0980
D) 0.0709
E) 0.00253

Answer 1

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 $F^{-}$ ions are calculated as follows.

$Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol$

Moles of HF are as follows.

$Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol$

Moles of HCl are as follows.

$Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol$

Now, reaction equation with initial and final moles will be as follows.

                        $H^{+} + F^{-} \rightarrow HF$

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.

$Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M$

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.