[AlphaGamma7]

HF_(aq) reacts with NaOH_(aq) according to the reaction represented below:

HF_(aq) +  OH^-_(aq) → H₂O_(l)  +  F^-_(aq)

A volume of 15 mL of 0.40 M NaOH_(aq) is added to 25 mL of 0.40 M HF_(aq).  Assume the volume are additive.

(d)    Calculate the molar concentration of F^-_(aq) in the solution

The answer is 0.15 M.

The given solution is below.

mol F^- formed = mol NaOH added = 0.0060 mol F^-
0.0060 mol F^-/(0.015 + 0.025) L = 0.15 M F^-

However, I get a value of 0.16M because I am calculating the amount of F^- that comes from the remaining HF in addition with the F^- created by the neutralization. Why does the solution omit this step?

[Borek]

However, I get a value of 0.16M because I am calculating the amount of F^- that comes from the remaining HF in addition with the F^- created by the neutralization. Why does the solution omit this step?

I guess you calculated how far the dissociation of the remaining HF went? Can you show how you did it?

[AlphaGamma7]

Ka = [H₃O⁺][F^-]/[HF]
7.2x10^-4 = (x)(0.15M+x)/(0.1-x)

x = 0.000476202

0.15M + 0.000476202M = 0.150476202M = 0.15M

The answer is the same, as I forgot to account for the F^- originally, which gave me a slightly higher value, but my question is this method valid?

Do they just assume the remaining HF barely adds to the concentration of F^-?

[Borek]

my question is this method valid?

Yes.

Do they just assume the remaining HF barely adds to the concentration of F^-?

Yes. Compare values you are adding - 0.15M + 0.000476202M - second term is so small, it can be safely ignored.