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

Question

Quizplus · Accepted Answer

First, we need to determine what type of reaction is occurring. HCl is a strong acid and will dissociate completely in water.

HCl + H2O → H3O+ + Cl-

Hydrogen fluoride (HF) is a weak acid and will also dissociate, but not completely.

HF + H2O ⇆ H3O+ + F-

The addition of HCl will increase the concentration of H3O+ ions in the solution. These new H3O+ ions will react with F- ions to form HF, which will shift the HF/F- equilibrium to the left.

H3O+ + F- ⇆ HF + H2O

Initially, we have 0.100 M F- and 0.126 M HF. The added HCl will react completely with the F- ions, because they are a strong base and will neutralize the H3O+ ions from the HCl. The amount of F- initially present in the solution is:

0.100 M F- × 25.0 mL = 0.00250 moles F-

The amount of HCl added is:

0.0100 M HCl × 9.00 mL = 0.0000900 moles HCl

This amount of HCl will react completely with the 0.00250 moles of F- ions, leaving none remaining. The new concentration of F- ions can be calculated using an ICE (initial-change-equilibrium) table for the HF/F- equilibrium:

HF       +       H2O           ⇆       H3O+    +       F-
I       0.126 M                         0.100 M                0            0.100 M
C       -x                                       -x                    +x           +x
E     0.126 - x                          0.100 - x           x           0.100 + x

The equilibrium constant expression for this reaction is:

K = [HF][H2O]/[F-][H3O+]

At equilibrium, we can assume that [H2O] is essentially constant (it is present in a large excess). Thus, we can simplify the expression:

K' = [HF]/[F-][H3O+]

Substituting the equilibrium concentrations from the ICE table into this expression, and using the known value of K' (3.5 × 10^-4), we can solve for x:

3.5 × 10^-4 = (0.126 - x)/[(0.100 + x)(x)]

x = 7.09 × 10^-5 M

This is the concentration of F- ions at equilibrium. To determine the total fluoride concentration, we add the initial F- concentration to the amount that comes from the dissociation of HF:

0.100 M + 7.09 × 10^-5 M = 0.1000709 M

Rounding to the appropriate number of significant figures gives a final answer of 0.0709 M, or choice (D).

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