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.

Step-by-step 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 * volume\\= 0.1 M * 0.025 L\\= 0.0025 mol`

Moles of HF are as follows.

`Moles of HF = Molarity * Volume\\= 0.126 M * 0.025 L\\= 0.00315 mol`

Moles of HCl are as follows.

`Moles of HCl = Molarity * volume\\= 0.01 M * 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 = (moles)/(volume)\\= (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.